Quantum Mechanics

Quantum Mechanics

What I will assume for this paper is that you already know some basic calculus and linear algebra, if you don't, I suggest the book 𝘘𝘶𝘢𝘯𝘵𝘶𝘮 𝘊𝘰𝘮𝘱𝘶𝘵𝘪𝘯𝘨: 𝘍𝘳𝘰𝘮 𝘤𝘰𝘯𝘤𝘦𝘱𝘵𝘴 𝘵𝘰 𝘤𝘰𝘥𝘦 by Andrew Glassner, and for calculus, I suggest the essence of calculus by 3 blue 1 brown. I would also like to warn you that I am really bad at explaining stuff in depth, I hope I make up for this by explaining literally all I know about quantum mechanics in one paper, however to get the full experience and amount of knowledge you do have to check the EXTRAS section at the bottom, as that covers very important topics that I thought were to distracting to add to the main article. Now lets get started!

when talking about a quantum state I mean that as a vector in a Hilbert space, whats a Hilbert space? well its just a space with infinite basis vectors being an option. This property is useful as shown below, (this is a linear combination when (c_i) is the indexed coefficient and |(A_i)> is the corresponding basis)

(∑_𝑖=0^∞)((c_i))*|(A_i)>≠valid vector (no Hilbert space)

(∑_𝑖=0^∞)((c_i))*|(A_i)>=valid vector (Hilbert space)

thats why we need a Hilbert space, as the quantum state vector can and will have infinite discrete sums, just much later on. But now we know that, how do we measure different aspects of the particle (or quantum state)? Well first let me introduce you (a little bit) to eigenvectors, eigenvectors are basically just the vectors that when applied an operator on, don't change direction, Just change magnitude or length, this is rare because most of the time if you apply a matrix on a vector it will change direction. Therefore for the eigenvector |λ> of an operator K, this would be true:

K|λ>=λ|λ>

here, the double use of λ is confusing, but |λ> is the eigenvector of K and λ is the corresponding number that eigenvector is scaled by when operated on by the operator K also known as the eigenvectors eigenvalue.

Next lets talk about Hermitian operators, all a hermitian operator does is satisfy this proposition below (lets say the K operator was hermitian):

K=K†

where the dagger here represents the adjoint of the operator, which is all the components of the operator flipped over the diagonal and conjugated (negative imaginary component), as shown below ( (K_xx) represent the components of the matrix):

[[(K_00),(K_01)],[(K_10),(K_11)]]†=[[(K_00^∗),(K_10^∗)],[(K_01^∗),(K_11^∗)]]

ok, but why is this useful? well the condition that hermitian operators provide ( K=(K^†) ) mean that they form an orthonormal eigenbasis, or a set or orthonormal basis vectors made of an operators eigenvectors. How can we prove this? well first I want to show you that all hermitian operator's eigenvalues are real, as this will be useful later. First lets get the normal operator eigenvector definition:

K|λ>=λ|λ>

conjugate it:

<λ|(K^)†=<λ|*(λ^∗)

since K†=K we have two equations:

K|λ>=λ|λ>

<λ|*K=<λ|*(λ^∗)

multiply by <λ| on the first one, and <λ| on the second:

<λ|*K|λ>=λ <λ|*|λ>

<λ|*K|λ>=(λ^∗) <λ|*|λ>

The only way this statement is true is if λ = (λ^∗), which is only true if λ doesn't have any imaginary component, which means λ is real. Therefore the eigenvalues for a hermitian operator are all real. Now we can use this to show that all hermitian operators form an orthonormal eigenbasis, lets look at the operator acting on two different eigenvectors |(λ_1)> and |(λ_2)> with two different eigenvalues:

K|(λ_1)>=(λ_1)|(λ_1)>

K|(λ_2)>=(λ_2)|(λ_2)>

we can also conjugate the 2nd equation:

K|(λ_1)>=(λ_1)|(λ_1)>

<(λ_2)|*K=(λ_2)<(λ_2)|

inner product trick again!!!:

<(λ_2)|*K|(λ_1)>=(λ_1) <(λ_2)|*|(λ_1)>

<(λ_2)|*K|(λ_1)>=(λ_2) <(λ_2)|*|(λ_1)>

The only way for this to be a true statement (which it is) is for <(λ_2)|*|(λ_1)> to equal zero, as then both would equal zero. What does this mean? Well I wont go into too much detail now, but a bra times a ket right after another is also known as the inner product, which acts kind of like a projection tool in some cases, so if the inner product of two vectors is zero, which it is here, it means that those to kets are not parallel or moving in the same direction at all and therefore have to be perpendicular, forming a basis with normalized vectors that are perpendicular to each other (orthagonal) and made of an operators eigenvectors, or in other words, an orthonormal eigenbasis!

so now that we know that hermitian operators form an orthonormal eigenbasis, how can we use this? Well if we have an operator such as the energy operator H, its orthonormal eigenbasis represents the states that energy can collapse to, so one basis vector would be energy level zero, one energy level one, and so on, since these are basis vectors you can write any vector a a combination of them. How much the state vector of our particle is made of an energy state vector, like energy level 1 vector, will give us the likely hood that that energy state will be collapsed to when the particles energy is measure.

Also the position operator, 𝚡 has a eigenbasis with infinite basis vectors because the particle can be in any position (which of there are infinite). The quantum state vector written in different operators eigenbasis's is shown below by linear combinations (integrals because there are infinite basis vectors):

|𝜓>=∫(ψ(x)*|x>*d(x)) position eigenbasis - operator 𝚡

|𝜓>=∫(𝜑(p)*|p>*d(p)) momentum eigenbasis - operator 𝚙

|𝜓=(∑_i=0^n)((c_i)|(E_i)>)) energy eigenbasis (discrete) - operator H

the functions ψ(x) and 𝜑(p) are coefficient functions as since there are infinite basis vectors, you need infinite coefficients to go with them, so those to wave functions are the functions that output the corresponding coefficients. Integrals are used with continuous basis's because you need continuous or infinite sums. the discrete basis is often used in fields like quantum computing, where you can use something like the spin basis (which only has two basis vectors, corresponding with 1 and 0 ) to model superposition between the 1 and 0 states, also allowing to to make quantum gates to act on those superpositions, these spin states act slightly differently from other types of measurable observable of a quantum state, such as position, We will not touch on the specifics here, but I have a full explanation of spin states in the EXTRAS section.

Now, quickly before we touch on the inner product, we need to know what the dot product is, and what kets and bras are, kets are just quantum notation for (usually complex) vectors, while bras are transposed (horizontal) and conjugated (negative imaginary component), so a conjugated complex number would turn from something like 3 + 2*𝑖 to 3 - 2*𝑖. bras and kets are shown below:

|a> = [[(a_0)],[(a_1)],[(a_2)],[(a_3)]]

|a>(∗𝚃) (conjugated and transposed) = <a| = [[(a_0^*),(a_1^*),(a_2^*),(a_3^*)]]

This is also written as the adjoint of the ket sometimes, as talked about before with operators, (|a>†), Now what is the dot product? well its the inner product, but for real numbered vectors, its written like below:

a ⋅ b = (a_0)*(b_0) + (a_1)*(b_1) … + (a_n-1)*(b_n-1)

Now I won't go crazy deep into what the inner product is, as I will cover what we can use the inner product for, and anything I say can be done with the inner product can be done with the dot product. But the one thing we do need to know is that when you dot product a vector with itself, it will return the magnitude of the vector squared, because you can rewrite the dot product with any two vectors (here its two of the same) as this: |a|*|a|*cos(γ) with γ being the angle between the two vectors, if that angle is zero (and therefore cos(γ) is 1 ) then the dot product equals |a|*|a| or |a|2 or the magnitude of the vector squared. The reason this is important is because to square a complex number (what the inner product is made for) you have to do (a^*)*a not just a*a, so this means that to get the magnitude of the complex vector squared, you would have to have the dot product with that change, how this is used is shown below:

Dot Product = (∑_i=0^n)((a_i)*(b_i)) ← normal

Inner Product = (∑_i=0^n)((α_i^*)*(β_i)) ←changed to match (a^*)*a

(∑_i=0^n)((α_i^*)*(β_i))

(α_0^*)*(β_0)+(α_1^*)*(β_1)+(α_2^*)*(β_2) ← expand

[[(α_0^*),(α_1^*),(α_2^*)]]*[[(β_0)],[(β_1)],[(β_2)]] ← turn into vectors

<α|*|β> ← definition of bras and kets

Now we know two ways to write the inner product! The use of kets and bras will be very used in the very scientific and totally professional paper, so I suggest getting used to them.

Now you already know the inner products definition, in both bra - ket and summation form, however I think it would be fun and good practice to derive the inner product summation form for both the discrete case (the one you already know) and the continuous case, as continuous basis's show up often in this paper. we will derive the summations from the bra - ket notation, (I will give a basic explanation of the δ 's you will see), 𝜓 and 𝜑 are commonly used to represent the average quantum state:

<𝜓|*|𝜑> <𝜓|*|𝜑>

((∑_i=0^n)<(c_i)*(E_i)|) ((∑_j=0^n)|(g_j)*(B_j)>) (∫<ψ(x)*x|*d(x))*(∫|𝜑(y)*y>*d(y))

(∑_i=0^n)((∑_j=0^n)<(c_i)*(E_i)|*|(g_j)*(B_j)>) ∬<ψ(x)*x|*|𝜑(y)*y>*d(x)*d(y)

(∑_i=0^n)((∑_j=0^n)((c_i^*)*(g_j)<(E_i)|*|(B_j)>)) ∬(ψ(x)∗*𝜑(y)) <x|*|y>*d(x)*d(y)

(∑_i=0^n)((∑_j=0^n)((c_i^*)*(g_j)*δ(i,j))) ∬(ψ(x)∗*𝜑(y)*δ(x-y)) d(x) d(y)

(∑_i=0^n)((c_i^*)*(g_i)) ∫(ψ(x)∗*𝜑(x)) d(x)

discrete inner product! continuous inner product!

You might not know what those deltas are, but their definitions are pretty simple. All the Kronecker Delta ( δ(i,j) ) is, is that it will be 1 if those numbers i and j match, the delta is added because the inner product of 2 basis vectors, if they are the same they are 1, otherwise they are zero. This means it is 1 when the basis vector on each side matches, or in this case i=j. while the Dirac Delta ( δ(x-y) ) is basically just the same thing but for continuous sums, singling out a single value just like the Kronecker Delta does, if you want a better explanation, I suggest quantum sense's video on the dirac delta.

One way you can use the inner product is applying it on a basis vector and a quantum state, generating the magnitude of the projection onto that basis vector, I will just show the discrete case here, but the continuous case is basically the same just with the Dirac delta (δ(x-y)) instead of the Kronecker delta (δ(i,j)), and there are wave functions involved. This example shows projections onto any general basis vector (E_j), obviously this only returns the magnitude of the projection not the projection vector as stated above:

<(E_j)|*|𝜓>

<(E_j)|*((∑_i=0^n)((c_i)|(E_i)>)) expand 𝜓

(∑_i=0^n)((c_i)<(E_j)|*|(E_i)>)

(∑_i=0^n)((c_i)*δ(j,i)) <(E_j)|*|(E_i)>=δ(j,i)

(c_j) <- the coefficient for (E_j)

we can also use the inner product for finding out if a vector has a magnitude of 1, which is required for a valid quantum state, when we derived the inner product this is what we based it off of so i'm glad its able to do this, this is shown below:

<𝜓|*|𝜓>

(∑_i=0^n)((c_i^*^*)*(c_i))

(∑_i=0^n)|(c_i)|2

|𝜓|2=1 (if 𝜓 magnitude is 1)

some operators can also commute, or not, I'll skip some of the visuals, because I physically can't show that in a text document, but just trust be on the fact that if two operators communicate, this means that 𝚡*𝚙 |𝜓>=𝚙*𝚡 |𝜓> (for two random commutable operators 𝚡 and (𝚙), then they share an eigenbasis. This means that if you can commute two operators then you can measure the exact quantity of one and the other, but if they do not commute, then you can never know the exact value of both at the same time. It is like this between position and momentum and energy and position.

That is called the Heisenburg uncertainty principle, which I explained really shitty, so if you want an actually good explanation watch one of the many videos on the topic. Or... Look in the EXTRAS section of this paper, where we talk about how all of this relates to the Fourier Transform! Although EXTRAS only explains the uncertainty between the 𝚙 and 𝚡 observables, so don't expect to much.

Now, before we get into the next very long and very important section, I want to state the 3 postulates of Quantum Mechanics, as I know them now at least. These versions of the postulates I got from MIT's 8.04 Quantum Mechanics course, I might update these if I get to 8.05, se postulates change depending on what you are talking about, but Ill try to keep them decently general.

Postulate 𝙸:

A Quantum particle can be fully described by a wave function ψ(x)

This can be slightly misleading in some cases, as the quantum state itself is actually a state vector |𝜓>. But it is true, as you can describe the quantum state vector |𝜓> by knowing its position wave function ψ(x), and often you never use |𝜓> as ψ(x) is easier to work with, and often comes up more in calculations.

Postulate 𝙸𝙸:

The probability of finding a particle between a point x and x+d(x) is given by the expression |ψ(x)|2*d(x)

This is the definition of the probability of measuring any wave function, or any coefficient of a discrete sum. How you get the probability of a certain outcome is described by this postulate, as the probability of an outcome is always the magnitude squared of the coefficient corresponding to the outcome. Since ψ(x) is a function returning coefficients for every position outcome, the magnitude squared is the probability the particle will be in that position.

Postulate 𝙸𝙸𝙸:

Given two valid states (ψ_1)(x) d (ψ_2)(x), there is another valid state that is a superposition of the two states: ψ(x)=α*(ψ_1)(x)+β*(ψ_2)(x)

This is a cool fact, and might seem odd at first, but it is very true and describes exactly what superposition is. Superposition, as described by the postulate, is a linear combination of two states with complex amplitudes or coefficients. These states are usually definite states, or one finite value, such as there being a superposition between one single position value, and another. But as you can see from this postulate, you can also combine two superpositions of definite states into a new super position of definite states.

Now that we know these, lets move on to the most complicated part!:

Next is the most interesting equation in all of quantum mechanics! thats right! its the Schrödinger equation! Now its hard to explain the Schrödinger equation without visual explanations, so I will have to skip over many things and i will ask you to just accept them, if you can't you can watch more videos on it (I suggest Quantum sense's on youtube). Hang in there!

Lets first imagine what the Schrödinger equation is doing. What it does is describe how time evolution changes a quantum state, it does this actually through a connection to classical physics. As in classical Lagrangian mechanics, a change in time to the Lagrangian is resulted from a change in energy, the Lagrangian (ℒ) is like the quantum state but for classical physics, it holds all information about the object. How this connects and equations shown below (finally I can make equations again):

∂(ℒ)/∂(t)=-d(𝙴)/d(t)

(this shows that to generate time change you need to look how energy changes in time)

𝑖*ℏd|𝜓>/d(t)=H|𝜓>

This is one of the things I will ask you to just accept, but trust me this is how the Schrödinger equation is formatted! All the Schrödinger equation is doing is saying the time evolution of a particle is caused by a change in energy (or the Hamiltonian/Energy operator), this assumption comes from experiments and the Lagrangian mechanics equation above. The 𝑖 and ℏ is there for probability conservation and unit type consistency respectively. if we expand the H (total energy) operator into what its made of then we get T + V or the kinetic energy operator plus the potential energy operator, giving us the full energy operator. This lines up with classical physics perfectly as well.

𝑖*ℏd(|𝜓>)/d(t)=(T+V(𝚡))*|𝜓>

Now lets say we wanted to read this equation in the position basis (which is the most common choice of basis for the Schrödinger equation), we would do this by using the inner product, as demonstrated earlier, to project the state |𝜓> into the position basis, (<x|*|𝜓>), but first we will expand the kinetic energy operator into its proper form, which using classical physics as an example, is 𝚙2 / 2*m:

𝑖*ℏd(|𝜓>)/d(t)=((𝚙2)/(2*m)+V(𝚡))*|𝜓> expand the T operator

𝑖*ℏ*<x|d(|𝜓>)/d(t)=<x|(𝚙2)/(2*m)*|𝜓>+<x|*V(𝚡) |𝜓> project onto the x basis

𝑖*ℏ(∂<x|*|𝜓>)/∂(t)=(𝚙2)/(2*m)*<x|*|𝜓>+V(𝚡)*<x|*|𝜓> move all <x|′s over with |𝜓>

𝑖*ℏ∂(ψ(x,t))/∂(t)=-ℏ/(2*m)(∂^2)(ψ(x,t))/(∂^2)(x)+V(x)*ψ(x,t) (i will explain this step)

I would like to point out that what you might see in articles and textbooks is this definition:

𝑖*ℏ∂(ψ(x,t))/∂(t)=(-ℏ/(2*m)∇+V(x))*ψ(x,t)

here, the ∇ is known as the laplacian, which is just the gradient of a function. so

∇ƒ(x)=(∑_i=1^n)(∂(ƒ)/∂((x_i))) and ∇ƒ(x)=(∑_i=1^n)((∂^2)(ƒ)/(∂^2)((x_i)))

this is basically just the derivative in every axis added to make 1 main derivative, in our case this is just 1 derivative of the second order, since ψ(x,t) is 1 dimensional, with just the x axis, so all we have to put is (∂^2)(ψ(x,t))/(∂^2)(x), but with wavefunctions of multiple dimensions, like ψ(x,y,t) we need to put (∂^2)(ψ(x,y,t))/(∂^2)(x)+(∂^2)(ψ(x,y,t))/(∂^2)(y), so all the ∇ s doing is the exact same thing we are doing, but also accounting for higher dimensions which we will use later.

and there you have it! the position-basis Schrödinger equation! To be clear, the derivative turned into a partial derivative because we are now measuring the difference in time only when we also have position as a variable. Now I know i'm not one to explain stuff very deeply, (if you couldn't tell by how horrible my explanations are) but I want to have more space and time devoted to that fourth and final algebraic transformation, so i can explain it better. First we expand all <x|*|𝜓> to ψ(x,t), we do this because <x|*|𝜓> is the same as projecting the quantum state 𝜓 onto the position basis, therefore because projecting a quantum state returns its coefficients in that basis, it should return all the coefficients for the position basis:

<x|*|𝜓>→ projecting the quantum state 𝜓 to the position basis

|𝜓>=∫(ψ((x^′))*|(x^′)>)*d((x^′))→ ψ(x) is a wave function that returns the coefficients

<x|*|𝜓>=<x|*(∫(ψ((x^′))*|(x^′)>)*d((x^′)))→ expand the quantum state 𝜓 in the position basis

∫(ψ((x^′))*<x|*|(x^′)>)*d((x^′))→ pull the <x| into the integral

∫(ψ((x^′))*δ(x-(x^′)))*d((x^′))→ turn <x|*|(x^′)> into the Dirac delta

ψ(x)→ use the definition of the Dirac delta

therefore, we can see that <x|*|𝜓> should return the wave function ψ(x), but then why does it return ψ(x,t) instead? Well, because ψ(x,t) is (basically) ψ(x) ! the only difference is that the ψ(x,t) wave function takes account of time as well, so you can look at the position basis coefficients in the future or past. so ψ(x,0)=ψ(x), as they are both the coefficients of the position basis at the current moment. You can also see that this uses the inner product for projections just like i did earlier when demonstrating what we could use the inner product for, just with a continuous basis and projecting onto a basis instead of a vector.

Next lets talk about the momentum operator 𝚙 we know that the momentum operators eigenbasis forms the momentum basis. But thats not what we are going to use right now, instead we are going to use more Lagrangian mechanics! As a reminder the ℒ function is like the quantum state |𝜓> but for classical physics, it holds all information about the object. And last time we only checked what caused a change in time of ℒ, which it turned out to be energy, so lets check out the other generators of Lagrangian mechanics:

∂(ℒ)/∂(t)=-d(𝙴)/d(t)← time evolution (what we used for SE) ∂(ℒ)/∂(x)=d(p)/d(t)←position change

∂(ℒ)/∂(p)=d(x)/d(t)← momentum change ∂(ℒ)/∂(θ)=d(𝙻)/d(t)←Angle change

Now what i'm about to say i cant really prove with text alone, so I suggest quantum senses's series chapter 14 (youtube). Now what we can do is transform these into there quantum versions just like we used the time evolution formula for the Lagrangian to make the Schrödinger equation. all transformations according to the ℒ (Lagrangian) are shown below:

𝑖*ℏd|𝜓>/d(t)=H|𝜓>← time evolution (SE) 𝑖*ℏd|x>/d(x)=𝚙|x>← position change

𝑖*ℏd|p>/d(p)=-𝚡|p>← momentum change 𝑖*ℏd|θ>/d(θ)=𝙻|θ>← angle change

now it is cool to see the Schrödinger equation is actually just one of multiple "changes in something" equations, you just see it more often because a change in time is more significant than the others. But the important part of this we can use the the generator of spacial or position change, which is momentum, 𝚙, where we can see that if we act 𝚙 on the position basis |x>, we get the equation on the left, we can use this as shown below to find out how the momentum operator acts on a quantum state! (the general idea is integration by parts, but if you don't know that you can just think of it flipping the derivative sign to move to the ψ(x) function)

𝚙|𝜓>← acting on the quantum state

∫(ψ(x)*𝚙|x>)*d(x)← move 𝚙 to |x>

(∫_-*∞^∞)(ψ(x)*(𝑖*ℏd|x>/d(x)))*d(x)← expand 𝚙|x>

[𝑖*ℏ*ψ(x)*|x>]∞(-*(∞^))-(∫_-*∞^∞)((𝑖*ℏd(ψ(x))/d(x))*|x>)*d(x)← integration by parts

-(∫_-*∞^∞)((𝑖*ℏd(ψ(x))/d(x))*|x>)*d(x)← boundary term vanishes to 0

(∫_-*∞^∞)((-𝑖*ℏd(ψ(x))/d(x))*|x>)*d(x)← move the negative inside

therefore ∴ 𝚙|𝜓>=∫((-𝑖*ℏd(ψ(x))/d(x))*|x>)*d(x)

and projecting that onto |x> results in the classic momentum operator formation:

<x|*𝚙|𝜓>=-𝑖*ℏd(ψ(x))/d(x)

now that we know more about the momentum operator we can finally explain the final form of the Schrödinger equation in the position basis:

𝑖*ℏ∂(ψ(x,t))/∂(t)=-ℏ/(2*m)(∂^2)(ψ(x,t))/(∂^2)(x)+V(x)*ψ(x,t)

we already described why derivatives turned partial, because we are now dealing with a position variable as well, we already described why the <x|*|𝜓> turned into ψ(x,t), now we only have two things to tackle, why the is potential energy operator now acting on a position eigenvalue instead of the position operator (V(𝚡)→V(x)), and what the fuck happened to 𝚙/(2*m) ?? well, V changed because |x> is an eigenstate of 𝚡, meaning that when you multiply V(𝚡) by <x|, the 𝚡 will turn into |x> 's eigenvalue.

Next, why does 𝚙 / 2*m turn into... whatever that is? well we have derived what the momentum operator looks like, so lets write that out and try to understand this:

𝚙=-𝑖*ℏd()/d(x)

therefore ∴𝚙2=-ℏ2(d^2)()/(d^2)(x)

so→(𝚙2)/(2*m)=-(ℏ2)/(2*m)(d^2)()/(d^2)(x)

And there you have it! if you also account for we now having to change the derivative to a partial derivative dues to multiple variables, then you have pretty much completely and utterly derived the position basis Schrödinger equation. Which is shown again below so you can check and revisit any parts you are rusty on:

𝑖*ℏ∂(ψ(x,t))/∂(t)=-ℏ/(2*m)(∂^2)(ψ(x,t))/(∂^2)(x)+V(x)*ψ(x,t)

Below I will try to derive the Schrödinger equation in the momentum basis, both for my own practice and yours, see if you can name everything i'm doing!

𝑖*ℏd|𝜓>/d(t)=H|𝜓>

𝑖*ℏd|𝜓>/d(t)=((𝚙2)/(2*m)+V(𝚡))*|𝜓>

𝑖*ℏ*<p|d|𝜓>/d(t)=<p|(𝚙2)/(2*m)*|𝜓>+<p|*V(𝚡)*|𝜓>

𝑖*ℏ∂(𝜑(p,t))/∂(t)=(p2)/(2*m)*𝜑(p,t)+V(𝑖*ℏ∂()/∂(p))*𝜑(p,t)

there! the V transformation was because 𝚡 projected onto the momentum basis is that, proof below:

-𝚡|p>=𝑖*ℏd|p>/d(p)

therefore ∴𝚡|𝜓>=(∫_-*∞^∞)(𝜑(p)*𝚡|p>)*d(p)

(∫_-*∞^∞)(𝜑(p)*(-𝑖*ℏd|p>/d(p)))*d(p)

[𝑖*ℏ*𝜑(p)*|p>]∞(-*∞)-(∫_-*∞^∞)(-𝑖*ℏd(𝜑(p))/d(p))*|p>*d(p)

-(∫_-*∞^∞)(-𝑖*ℏd(𝜑(p))/d(p))*|p>*d(p)

𝚡|𝜓>=(∫_-*∞^∞)(𝑖*ℏd(𝜑(p))/d(p))*|p>*d(p)

therefore ∴<p|*𝚡|𝜓>=𝑖*ℏd(𝜑(p))/d(p)

You might have noticed that when you project some eigenstate (like |p> ) onto something involving that eigenstate's operator (like 𝚙 / 2*m ), you can just change the operator to the eigenvalue of that eigenstate ( p / 2*m ). But when you project a eigenstate onto a expression with a operator that is not the eigenstates operator, like |p> onto V(𝚡), you have to transform the operator into what the operator is in relation to the eigenstate's, operator (like 𝑖*ℏ*∂() /∂(p) ). I certainly noticed this when making quite a few errors writing the equations down, and now I hope you do to!

Now that we know what quantum operators are, the inner product, what quantum states are, how to find quantum operators, and the Schrödinger equation, I believe its time to solve the it! In concept it sounds easy, if we can take the position basis Schrödinger equation:

𝑖*ℏ∂(ψ(x,t))/∂(t)=-ℏ/(2*m)(∂^2)(ψ(x,t))/(∂^2)(x)+V(x)*ψ(x,t)

and solve for ψ(x,t), we can then use ψ(x,t) to find ψ(t):

|ψ(t)>=∫(ψ(x,t)*|x>)*d(x)

and since ψ(t) is the quantum state at a certain time, we can find the quantum state in the future, or predict what it will do. So ψ(0) is the current state and ψ(2) is the quantum state something like 2 seconds in the future. Easy!

But actually it's not that simple... The main problem is solving for ψ(x,t), as differential equations are by nature hard to solve, made even harder with partial and second order derivatives. So what we do instead is transform that position basis Schrödinger equation into the TISE, or Time Independent Schrödinger Equation. This (as you might have guessed) simplifies the Schrödinger equation to a more manageable form based on the assumption that the potential energy operator V(x) is not dependent on time.

The first thing the TISE calls for is the separations of variables, aka turning ψ(x,t) into two functions, shown below:

ψ(x,t) =γ(x)*τ(t)←this makes no partial derivatives needed

(I would like to point out γ(x) is usually 𝜑(x) or ψ(x) in professional work but I don't want confusion with the momentum wave function or position wave function so I use γ(x) ) since we don't need partial derivatives, lets convert the partial derivatives into terms of γ(x) and τ(t), a function that only considers time, and a function that only considers position, some rules do need to be met to make this allowed but I am not going to explain them because separation of variables is one of the things I don't understand very well. the conversion steps are shown below (try to logic your way through the steps):

∂(ψ(x,t))/∂(t)=∂()/∂(t)*[γ(x)*τ(t)] (∂^2)(ψ(x,t))/(∂^2)(x)=(∂^2)()/(∂^2)(x)*[γ(x)*τ(t)]

γ(x)d(τ(t))/d(t) τ(t)(d^2)(γ(x))/(d^2)(x)

substitute:

𝑖*ℏ*γ(x)d(τ(t))/d(t)=-(ℏ2)/(2*m)*τ(t)(d^2)(γ(x))/(d^2)(x)-V(x)*γ(x)*τ(t)

Next, divide by γ*τ (γ(x)*τ(t)):

𝑖*ℏ1/τ(t)d(τ(t))/d(t)=-(ℏ2)/(2*m)1/γ(x)(d^2)(γ(x))/(d^2)(x)-V(x)

Now, what I am about to say is quite weird, but makes sense if you think about it. Since the left side only depends on time, and the right side only depends on position, this means that both sides are constant. This is because if you change t the right side doesn't change (because the right only depends on x) so the left side (where you changed t) shouldn't either, same with changing x, the left side doesn't change so the right shouldn't either. This means that both halves of the equation are constant, even if you change x and t. Since both halves are constant, this means that both sides equal some constant number 𝘌, because:

𝑖*ℏ1/τ(t)d(τ(t))/d(t)=-(ℏ2)/(2*m)1/γ(x)(d^2)(γ(x))/(d^2)(x)-V(x)=𝘌

(since both sides are constant and equal)

Now that we know this, lets solve the left hand side of the equation for τ(t), these steps are shown and explained below:

𝑖*ℏ1/τ(t)d(τ(t))/d(t)=𝘌 ← already proven above

1/τ(t)d(τ(t))/d(t)=(-𝑖*𝘌)/ℏ←move 𝑖*ℏ

(i is negative because 1/𝑖=𝑖/(𝑖2)=𝑖/(-1)=-𝑖)

d(τ(t))/τ(t)=(-𝑖*𝘌)/ℏ*d(t)←move d(t) over

∫(d(τ(t))/τ(t))=∫((-𝑖*𝘌)/ℏ*d(t))←integrate

∫(1/τ(t)*d(τ))=∫((-𝑖*𝘌)/ℏ*d(t))←expand left to make sense

ln|τ(t)|=(-𝑖*𝘌*t)/ℏ+C←integrate (actually)

τ(t)=ℇ(-𝑖*𝘌*t/ℏ+C)←exponentiate (rid of ln())

τ(t)=ℇ(-𝑖*𝘌*t/ℏ)*ℇC←exponent rule

τ(t)=A*ℇ(-𝑖*𝘌*t/ℏ)←ℇC is some constant A

so there are many things i would love to explain in more detail on why we can move the numbers in that way and how we are able to do it, but there are too many things to talk about and it would go off topic, if you want an explanation on the TISE other than this paper, I would suggest my favorite learning technique, find a good youtube video, like https://www.youtube.com/watch?v=Dt_VKsSggAo, and ask some AI or expert on the topic at any points you are confused on.

Now solving for γ(x) is a lot easier, as all that is needed is a little multiplication to turn it into the TISE, steps are shown below:

-(ℏ2)/(2*m)1/γ(x)(d^2)(γ(x))/(d^2)(x)-V(x)=𝘌

-(ℏ2)/(2*m)(d^2)(γ(x))/(d^2)(x)-V(x)*γ(x)=𝘌*γ(x)←multiply each side by γ(x)

And there you have it! the TISE in full glory, How we can use this TISE is we can solve for γ(x) and then use γ(x) (and our definition of τ(t) ) to get ψ(x,t), which can be used to derive our quantum state ψ(t), as ψ(x,t) is just the quantum state in the position basis. How we use gamma to get ψ(x,t) and then get ψ(t) is shown below:

ψ(x,t)=γ(x)*τ(t)←use definition of ψ(x,t)

ψ(x,t)=γ(x)*ℇ(-𝑖*𝘌*t/ℏ)←use definition of τ(t)

|ψ(t)>=∫(ψ(x,t)*|x>)*d(x)←use definition of |ψ(t)>

Notice how the constant of integration A is gone from the τ(t) definition (A*ℇ(-𝑖*𝘌*t/ℏ)), the A was removed because it was a constant of integration, and the normalization rule of any wave function (like ψ(x,t) ) forces it to be a certain value, aka if A isn't set to a certain value the normalization rule is false, which is not true, so we set A to a fixed value where the normalization rule fits, and then just put A built into the γ(x) definition, so we don't have to write it anymore, but its still there:

γ(x)=A⋅γ(x)

...Now I have lied to you slightly, in many practical uses ψ(x,t)=γ(x)*ℇ(-𝑖*𝘌*t/ℏ) is not true, this is because γ(x) is an eigenstate of the eigenbasis the Hamiltonian, or energy operator creates. So for that definition of ψ(x,t) to be true, the energy state would have to be collapsed onto one γ(x), which will not be the case of many many particles. How do I know γ(x) is an eigenvector of the energy eigenbasis? well look back at the TISE:

-(ℏ2)/(2*m)(d^2)(γ(x))/(d^2)(x)-V(x)*γ(x)=𝘌*γ(x)←the TISE

H*γ(x)=𝘌*γ(x)←left is just H

since 𝘌 is a constant, this means that applying the Hamiltonian operator to γ(x) scales it by a constant 𝘌, you know what that means gamma is? an eigenvector / eigenstate of the Hamiltonian operator! This is because... well thats literally the definition of an eigenvector of an operator, applying that operator to that state only scales that state by a factor. This implies really cool things, because if you measure energy, energy collapses onto one eigenvalue of the energy eigenbasis, but what it also does is collapse position into one γ(x) or position wave function, restricting where the particle can be, but back on topic.

But if ψ(x,t)=γ(x)*ℇ(-𝑖*𝘌*t/ℏ) isn's true — in most cases — what is? Well since γ(x) is not usually just 1 value, as that would imply that the particle has collapsed into 1 eigenstate γ(x), all we have to do is write γ(x) as instead of a single value, as a full eigenbasis, which means we can use our definition from the literal beginning of this paper to expand γ(x) to instead be a summation, using coefficients and distributing the ℇ(-𝑖*𝘌*t/ℏ) to each component, which we can then use to derive the quantum state:

ψ(x,t)=(∑_i=0^∞)(((c_i)*γ(x)))*ℇ(-𝑖*𝘌*t/ℏ)

|ψ(t)>=∫(ψ(x,t)*|x>)*d(x)

I would also like to add that the TISE only happens when the Hamiltonian or energy operator doesn't depend on time, aka the energy of a particle doesn't change over time, now in real life this often isn't true, but most of the time for simulations the TISE is just fine for simulating a particle in a perfectly static box, or potential energy well.

Ok. Now that we finally fully understand the TISE, what its used for, what every component means and why the ψ(x,t)=γ(x)*ℇ(-𝑖*𝘌*t/ℏ) is often inaccurate, replacing it with a new term, I think its darn time we finally figure out how to solve the TISE.

Now what you may have noticed is that the wave function we are using is ψ(x,t), only taking in a 1d space x, and trust me we will learn how to solve the 2d TISE with ψ(x,y,t) later, but first we will learn how to solve the 1d TISE, and use what we learned to solve the 2d version.

the first thing we need to establish is how we set up the 1 d walls, so we can simulate the particle in a 1d box, how we can do this is say that the particle would need ∞ potential energy (V(x)) in those regions, aka the particle can't be there. Next we can say the particle has 0 potential energy in the box, as nothing is acting on the particle, meaning we can remove V(x) when solving in the box. Using this information, we can simplify the TISE as shown below when in the walls:

-(ℏ2)/(2*m)(d^2)(γ(x))/(d^2)(x)-V(x)*γ(x)=𝘌*γ(x)

-(ℏ2)/(2*m)(d^2)(γ(x))/(d^2)(x)=𝘌*γ(x) (since V(x) is 0 )

now, what do we do? well -(ℏ2)/(2*m) is a constant, and so is 𝘌, so lets move -(ℏ2)/(2*m) to the right with 𝘌*γ(x) to single out the second derivative:

(d^2)(γ(x))/(d^2)(x)=-(2*m*𝘌)/(ℏ2)*γ(x)

(d^2)(γ(x))/(d^2)(x)=-k2*γ(x) ← where -k2=-(2*m*𝘌)/(ℏ2)

The simplification of -(2*m*𝘌)/(ℏ2) might seem odd, but it will help us in recognizing a pattern. Now we just have to find what function has a second derivative that results in itself times a negative constant squared. This isn't that hard as there is a function that does exactly that, the sin() function, the sin(x) function returns a second derivative of -sin(x), but if we add a constant like 7, can you guess what we get? that's right, that constant (in this case 7 ) negative squared:

(d^2)(sin(7*x))/(d^2)(x) = -72*sin(7*x), so if we put k there, we get this:

(d^2)(sin(k*x))/(d^2)(x)=-k2*sin(k*x)

as you can see, it perfectly matches our equation with a substituted k :

(d^2)(γ(x))/(d^2)(x)=-k2*γ(x)

so γ(x) must be sin(k*x), there is a second solution, its cos(k*x), however this cant work for a 1d box with boundaries at x=0 and x=a. why? …oh boy this is gonna take a while. So the equation we just derived from the TISE when we are measuring a 1d box is a second order differential equation (based off the fact that its highest derivative is second derivative), and the second order differential equations are linear (in general) this includes the TISE! Why is this important? Well first I would like to say that specifically in this case our linear operator is (d^2)() / (d^2)(x)+k2, which is operating on gamma, where gamma is then distributed to the derivative and k2, this operator is made by taking our equation and moving everything to the left as shown below:

(d^2)(γ(x))/(d^2)(x)=-k2*γ(x)

(d^2)(γ(x))/(d^2)(x)+k2*γ(x)=0 ← move stuff over

((d^2)()/(d^2)(x)+k2)*γ(x)=0 ← group

O[γ(x)]=0 ←O is our operator

Now we know that, we need to know that linear operators follow 2 rules, lets call our operator 𝙻[] for now, the two rules are shown below:

𝙻[d*x]=d*𝙻[x]

𝙻[d+x]=𝙻[]+𝙻[x]

Why is this important? Well we know from our deriving of our operator that any valid function inputed into the operator must equal zero, or the equation (d^2)(γ(x))/(d^2)(x)+k2*γ(x)=0 wouldn't be true! But we do know that sin(k*x) (and the second solution cos(k*x) ) are both solutions to this equation, as shown below:

O[sin(k*x)]=0

O[cos(k*x)]=0

But what if we want 1 solution that covers all basis's? well its your lucky day! as proven below, we can linearly combine these two solutions and still get a correct answer:

O[A*sin(k*x)+B*cos(k*x)]

O[A*sin(k*x)]+O[B*cos(k*x)]

A⋅O[sin(k*x)]+B⋅O[cos(k*x)]

A⋅0+B⋅0=0

tada! so now we know that A*sin(k*x)+B*cos(k*x) is also a solution to this equation! So... what about the whole reason I went into this rabbit hole? Why don't we see or use cos(k*x) ? Well this is a general solution to the 1d TISE, but what if we add those impenetrable ∞ potential energy walls? well thats exactly what A and B are meant for, as they are set by boundary conditions. Now we know the particle is not in the walls, aka the wave function is zero there, but what about at the walls? well there it must be zero to, because if the function you pick does not naturally fall down to zero at the walls then you have an infinite slope at the walls, which would break our derivatives of γ(x).

Using this, we can say what functions we can use at x=0 (the first wall) and x=a (the second wall), first lets see what we can prove if we say the at the first wall the function has to be zero:

γ(0)=0

A*sin(k*0)+B*cos(k*0)=0

A*sin(0)+B*cos(0)=0

A⋅0+B⋅1=0

this means that B must equal zero!

Now what about the other side of the box, the wall at x=a does that force A to become a specific value? well, lets see shall we:

γ(a)=0

A*sin(k*a)=0

∴ sin(k*a)=0 ← since A cant be zero or every output would be zero

since sin(k*a) is only zero if k*a is a multiple of π, such as 2*π, we can say:

k*a=n*π←where n = 1, 2, 3, …

k=(n*π)/a←divide by a on both sides

substitue k in A*sin(k*x) :

γ(x)=A*sin((n*π*x)/a)

Now this realization is really cool, as it reveals the principle of quantization, or the fact that a particle can only have discrete energy values, not a continuous set. This proves this because remember how k is related to energy in -(2*m*𝘌)/(ℏ2) being equal to -k2? well that means that k is related to energy, obviously, as 𝘌 is an energy eigenvalue, but our previous definition also says that k is discrete, meaning that the energy of any particle is in discrete values, with no energy levels of a particle allowed between them. This is probably a loose definition but I don't care enough to give a proper one. Now lets get back on topic!

This new definition of k is cool, but what about A? we didn't specify a specific value for it and it didn't seem to force A to be a certain value. And you would be right, but we can force A to be a value, and it comes from normalization, normalization is just the rule that all coefficients magnitude squared add up must be equal to one, this is because the magnitude squared of a coefficient is the probability of measuring the particle in that eigenvector. Since probabilities must add up to 100, or 1, then below is true:

∫(|ψ(x)|2)*d(x)=1

Now I am going to do the math but not do a lot or any explaining at all, so be wary:

(∫_0^a)(|γ(x)|2)*d(x)=1

(∫_0^a)(A2*sin2((n*π*x)/a))*d(x)=1

A2*(∫_0^a)(sin2((n*π*x)/a))*d(x)=1

use trig definition: sin2(θ)=(1-cos(2*θ))/2

A2*(∫_0^a)((1-cos((2*n*π*x)/a))/2)*d(x)

A21/2*(∫_0^a)(1-cos((2*n*π*x)/a))*d(x)

split the integral:

A21/2*((∫_0^a)(1*d(x))-(∫_0^a)(cos((2*n*π*x)/a)*d(x)))

(∫_0^a)(1*d(x))=a ← solve left side

solve right side:

(∫_0^a)(cos((2*n*π*x)/a))*d(x)

u=(2*n*π*x)/a

d(u)/d(x)=(2*n*π)/a

∴ d(u)=(2*n*π)/a*d(x)

∴ d(x)=a/(2*n*π)*d(u)

(∫_0^2*n*π)(cos(u)a/(2*n*π))*d(u)

a/(2*n*π)*(∫_0^2*n*π)(cos(u))*d(u)

a/(2*n*π)*[-sin(u)](2*n*π)0

a/(2*n*π)*(0)=0

substitute into full equation:

A21/2*(a-0)=A2a/2=1 (remember?)

∴ A2=2/a and A=√(,2/a)

Finally! that was a lot of continuous math I am sorry, and there are many points where you might have been confused, but we found a value for A finally! that means we can finally know what the full γ(x) is for the 1d TISE when you have boundary conditions at x=0 and x=a, I will show how to go from general solution to our solution now that we know what each step is, try to remember each step!:

γ(x)=A*sin(k*x)+B*cos(k*x)

γ(x)=A*sin(k*x) ←B=0

γ(x)=A*sin((n*π*x)/a) ← expand k

γ(x)=√(,2/a)*sin((n*π*x)/a) ← expand A

And there we have it! we have a fully working function for γ(x), however this is the position wave function at a single point in time, what about that Time evolution the Schrödinger equation promised? Well... do you remember τ(t)? we can finally use our definition from ages ago to find the dreaded ψ(x,t)! We do have to make a few edits though:

ψ(x,t)=(∑_i=1^∞)((c_i)*γ(x)*ℇ(-𝑖*(𝘌_i)*t/ℏ))

ψ(x,t)=(∑_i=1^∞)((c_i)√(,2/a)*sin((i*π*x)/a)*ℇ(-𝑖*(𝘌_i)*t/ℏ))

you can see n turned to i (because they do the exact same things) and we changed 𝘌 to (𝘌_i) because there are different energy eigenvalues for each energy eigenvector (γ(x)). I would also like to point out that we are summing to infinity, this is because theres technically nothing stopping you from just adding more and more energy to the particle forever, even though energy is discrete and not continuous (this is why we use summation instead of integral, it sums infinitely but is discrete). But in practice, the coefficients and therefore the probability of a high energy state goes down very fast, so summing to 100 is good enough for pretty much everything:

ψ(x,t)=(∑_i=1^100)((c_i)√(,2/a)*sin((i*π*x)/a)*ℇ(-𝑖*(𝘌_i)*t/ℏ))

Now thats how we find the time evolution of a particle in a 1 d box, but what about a 2d box/square? well thats actually pretty easy, first lets look what the 1d TISE looks like and the 2d TISE looks like:

1d TISE: -(ℏ2)/(2*m)(d^2)(γ(x))/(d^2)(x)=𝘌*γ(x) (assuming no V(x) in the box)

2d TSIE: -(ℏ2)/(2*m)*((∂^2)(γ(x,y))/(∂^2)(x)+(∂^2)(γ(x,y))/(∂^2)(y))=𝘌*γ(x,y)

Now we can turn the 2d version into 2 1d TISE's, how we do this is the same thing we did to get the TISE out of the normal Schrödinger equation. The separation of variables! We can turn γ(x,y) into Y(y) and X(x), so γ(x,y)=X(x)*Y(y), we use this to get two different 1d TISE's below:

-(ℏ2)/(2*m)*((∂^2)(γ(x,y))/(∂^2)(x)+(∂^2)(γ(x,y))/(∂^2)(y))=𝘌*γ(x,y)

-(ℏ2)/(2*m)*(Y(y)(d^2)(X(x))/(d^2)(x)+X(x)(d^2)(Y(y))/(d^2)(y))=𝘌*X(x)*Y(y)

-(ℏ2)/(2*m)*(1/X(x)(d^2)(X(x))/(d^2)(x)+1/Y(y)(d^2)(Y(y))/(d^2)(y))=𝘌←divide by X(x)*Y(y)

-(ℏ2)/(2*m)*(1/X(x)(d^2)(X(x))/(d^2)(x))+-(ℏ2)/(2*m)*(1/Y(y)(d^2)(Y(y))/(d^2)(y))=𝘌←distribute -(ℏ2)/(2*m)

the only way for 2 sides (both depending on different variables) added together can equal a constant 𝘌, is for those two sides to also be constants, (𝘌_x) and (𝘌_y). Using these new constants, we can split this equation into two 1d TISE's and solve them!

-(ℏ2)/(2*m)1/X(x)(d^2)(X(x))/(d^2)(x)=(𝘌_x)

-(ℏ2)/(2*m)(d^2)(X(x))/(d^2)(x)=(𝘌_x)*X(x)←multiply both sides by X(x)

1d TISE!!

-(ℏ2)/(2*m)1/Y(y)(d^2)(Y(y))/(d^2)(y)=(𝘌_y)

-(ℏ2)/(2*m)(d^2)(Y(y))/(d^2)(y)=(𝘌_y)*Y(y)←multiply both sides by Y(y)

1d TISE!!

Now, lets say that the x based TISE's boundary positions are x=0 and x=(L_x), while the y based TISE's boundary conditions are y=0 and y=(L_y). Using this we already know the solution to these 1d TISE's, this is because we already know what the solution to the TISE looks like! so we can just substitute in our boundary conditions and bam! we have a solution:

solution for x based TISE: √(,2/(L_x))*sin((n*π*x)/(L_x))

solution for y based TISE: √(,2/(L_y))*sin((n*π*y)/(L_y))

we can then combine the two square root values, and find γ(x,y), then using that to find ψ(x,y,t):

γ(x,y)=X(x)*Y(y)

γ(x,y)=√(,2/(L_x))*sin((n*π*x)/(L_x))⋅√(,2/(L_y))*sin((n*π*y)/(L_y))

combine square root values:

γ(x,y)=2/√(,(L_x)*(L_y))*sin((n*π*x)/(L_x))*sin((n*π*y)/(L_y))

then use γ(x,y) to find ψ(x,y,t):

ψ(x,y,t)=(∑_i=0^100)((c_i)*γ(x,y)*ℇ(-𝑖*(𝘌_i)*t/ℏ))

ψ(x,y,t)=(∑_(n_x)=0^100)((∑_(n_y)=0^100)((c_(n_x),(n_y))2/√(,(L_x)*(L_y))*sin((i*π*x)/(L_x))*sin((n*π*y)/(L_y))*ℇ(-𝑖*(𝘌_(n_x),(n_y))*t/ℏ)))

We used a double summation because we need to allow the y and x coordinates to have different energy values, creating different eigenstates, like 1, 5, low x energy and high y energy. But there you have it (once you understand what ive just said) a particle moving through time and space in a stationary box! now you can probably guessed based on the pattern how the 3d form looks:

1d :

ψ(x,t)=(∑_n=1^100)((c_i)√(,2/a)*sin((n*π*x)/a)*ℇ(-𝑖*(𝘌_n)*t/ℏ))

2d :

ψ(x,y,t)=(∑_(n_x)=0^100)((∑_(n_y)=0^100)((c_(n_x),(n_y))2/√(,(L_x)*(L_y))*sin(((n_y)*π*x)/(L_x))*sin(((n_x)*π*y)/(L_y))*ℇ(-𝑖*(𝘌_(n_x),(n_y))*t/ℏ)))

3d :

ψ(x,y,z,t)=(∑_(n_z)=0^100)((∑_(n_y)=0^100)((∑_(n_x)=0^100)((c_(n_x),(n_y),(n_z))(2√(,2))/√(,(L_x)*(L_y)*(L_z))*sin(((n_x)*π*x)/(L_x))*sin(((n_y)*π*y)/(L_y))*sin(((n_z)*π*z)/(L_y))*ℇ(-𝑖*(𝘌_(n_x),(n_y),(n_z))*t/ℏ))))

therefore, (for the people that just want an absurd equation) this is how to find the quantum state itself |𝜓(t)> in 3d space, which you don't need to find, as ψ(x,y,z,t) is the quantum state, just in the position basis, aka ψ(x,y,z,t) is just <ψ(x,y,z)|*|𝜓(t)>:

|𝜓(t)>=∭((∑_(n_z)=0^100)((∑_(n_y)=0^100)((∑_(n_x)=0^100)((c_(n_x),(n_y),(n_z))(2√(,2))/√(,(L_x)*(L_y)*(L_z))*sin(((n_x)*π*x)/(L_x))*sin(((n_y)*π*y)/(L_y))*sin(((n_z)*π*z)/(L_y))*ℇ(-𝑖*(𝘌_(n_x),(n_y),(n_z))*t/ℏ))))*|x*y*z>*d(x)*d(y)*d(z))

or simplified:

∭(ψ(x,y,z,t)*|x*y*z>)*d(x)*d(y)*d(z)

tada! all the position and quantum state information and equations you will ever need! If you want to find out how to solve the Schrödinger equation with the hamiltonian depending on time (aka the particle is allowed to change energy levels through time)... I'm not gonna tell you! Go find it yourself. Ive done enough math, and you probably have to, thanks for reading! and I hoped you understood at least a little about what Im saying, you can use the 1 d, 2 d, or 3 d position-time equations to simulate a particle or atom in a perfectly static box, with the x boundaries being 0 to (L_x), y boundaries being 0 to (L_y), and z boundaries being 0 to (L_z), this is what I am planning to do, however I don't know any 3d software thats good for this. Problems, problems...

Oh! if you where wondering, this doesn't just limit to 3d 2d or 1 d, you can go as high as you want, therefore I present the totally stupid big 10 dimensional equation for finding the position wave function evolution over time of a 10 dimensional particle and projecting it to make the quantum state vector:

|𝜓(t)>

=∫(∫(∫(∫(∫(∫(∫(∫(∫(∫(((∑_(n_o)^∞)((∑_(n_k)^∞)((∑_(n_ℏ)^∞)((∑_(n_g)^∞)((∑_(n_i)^∞)((∑_(n_j)^∞)((∑_(n_q)^∞)((∑_(n_z)^∞)((∑_(n_y)^∞)((∑_(n_x)^∞)((c_(n_x),(n_y),(n_z),(n_q),(n_j),(n_i),(n_g),(n_ℏ),(n_k),(n_o))√(,210)/√(,(L_x)*(L_y)*(L_z)*(L_q)*(L_j)*(L_i)*(L_g)*(L_ℏ)*(L_k)*(L_o))*sin(((n_x)*π*x)/(L_x))*sin(((n_y)*π*y)/(L_y))*sin(((n_z)*π*z)/(L_z))*sin(((n_q)*π*q)/(L_q))*sin(((n_j)*π*j)/(L_j))*sin(((n_i)*π*i)/(L_i))*sin(((n_g)*π*g)/(L_g))*sin(((n_h)*π*h)/(L_h))*sin(((n_k)*π*k)/(L_k))*sin(((n_o)*π*o)/(L_o))*ℇ(-𝑖*(𝘌_(n_x),(n_y),(n_z),(n_q),(n_j),(n_i),(n_g),(n_h),(m_k),(n_o))*t/ℏ))))))))))))*|x*y*z*q*j*i*g*h*k*o>*d(x)*d(y)*d(z)*d(q)*d(j)*d(i)*d(g)*d(h)*d(k)*d(o)))))))))))

EXTRAS

here is stuff I learned after the fact and was too lazy to integrate into the main paper, enjoy!

Spin States:

The thing I mentioned at the start of this paper! it is true these spin states behave very differently from other observables, such as energy or position.

first, spin states only have two possible outcomes, +1 or -1, up or down spin if you want to be fancy. Now lets do what we would normally do, take our quantum state vector |𝜓> and write it in the spin operators eigenbasis, however, there are multiple spin operators, (σ_z) (σ_y) and (σ_x), so whichever do we choose??? Well lets just go with (σ_z) for now, since its the most commonly chosen and represents the up and down ket vectors.

now the simple 2 dimensional quantum state looks like this:

|𝜓> = α|𝓊>+β|𝒹>

now we measure and lets say we get |𝓊>

|𝜓>=|𝓊>

now we have prepared |𝜓> in the |𝓊> or up state, nothing special here, now lets measure (σ_x) instead while |𝜓> is still in the |𝓊> state, if spin states act similar to other states such as position, measuring the x shouldn't effect the z output, we can see this in the position operators observables. Yes, plural, the position operator is often thought as just one entity but its actually 3, one for each dimension: 𝚡 𝚢 𝚣

the reason we haven't talked about these much is because the commute, now we already talked on what commuting means, but not very well. To be simple, this means they share an eigenbasis, so when you measure one observable 𝚡 you can also measure the observable 𝚢 without moving the state vector off of the 𝚡 measurement, allowing you to measure both at once. If the operators do not share an eigenbasis, and therefore don't commute, you can not measure both at once with certainty.

now since all position observables commute, you can measure all 3 at once and be certain about all 3, so you can know the x, y, and z positions of a particle with certainty. In other words, you can measure the 𝚡 observable, then measure the 𝚢 observable, and not disturb your previous 𝚡 measurement, so most of the time, we just use all 3 observables as just one observable 𝚡.

However, with spin observables (σ_x) (σ_y) and (σ_z), they do not commute, which means if you measure the (σ_z) observable, just like we did earlier, then measure the (σ_x) observable, you can be certain about what (σ_x) 's spin state is, but now the (σ_z) observable has been thrown back into superposition, and we cant be certain about its measurement. Meaning if you check (σ_z) again, there is a 50 chance for +1 or |𝓊> spin and a 50 chance for -1 or |𝒹> spin.

This makes measurement in the (σ_x) (σ_y) and (σ_z) directions, slightly weird sure, but what makes things really really weird is you need to find a new observable for every other direction you want to measure in. For example, lets say we measure the (σ_z) observable again, and prepare the state vector in the |𝓊> direction, then lets say we move our measuring apparatus 𝒜 in a random direction n and measure, what would be the probability of measuring +1? well if this were position, there would be a eigenvector shared between the position operators, 𝚡 (𝚡,𝚢,𝚣), representing that direction , so all we would do is project the current state vector |𝓊> onto that eigenvector, then square that magnitude to get the probability. However, because unlike the position set of observables 𝚡 𝚢 and 𝚣, the spin observables can't commute, meaning they don't share an eigenbasis or any eigenvectors, meaning there is not a shared vector representing that direction.

so how do we measure the probability of +1 or -1 in an arbitrary direction if we cant just combine all three direction observables? Well we make a new observable for that specific direction, lets assume (σ_) (which is the total vector operator) acts like a regular vector or ket. This means that if we want to find the spin amount in the n direction, we should just be able to dot product the two vectors together, as they are normalized, so that will return cos(θ) where θ is the angle between the two vectors:

(σ_)⋅n=(n_x)*(σ_x)+(n_y)*(σ_y)+(n_z)*(σ_z)=cos(θ)

now what we have to do is find out what the matrixes are for the 3 spin operators, But, how? Well the (σ_z) is easy, logic for it is shown below:

(σ_z)|𝓊>=|𝓊>

(σ_z)|𝒹>=-|𝒹>

<𝓊|*(σ_z)|𝓊>=<𝓊|*|𝓊>=1 <𝒹|*(σ_z)|𝓊>=<𝒹|*|𝓊>=0

<𝓊|*(σ_z)|𝒹>=-<𝓊|*|𝒹>=0 <𝒹|*(σ_z)|𝒹>=-<𝒹|*|𝒹>=-1

∴ (σ_z)=[[1,0],[0,-1]]

next the (σ_x) operator, first, this operator is in a different direction which means its "up" state and "down" state is different from the (σ_z) 's states, so we will call (σ_x) up state right or |𝓇> and the down state left or |𝓁>. we can write these states in terms of the classic |𝓊> and |𝒹> using probabilities, because if you put the state vector as |𝓇> then measure the (σ_z) |𝓊> state, there is a 50 chance you get the |𝓊> or |𝒹> state. meaning, since the probabilities are the magnitude squared of the amplitudes, the amplitudes for the |𝓇> and |𝓁> states should be 1/√(,2). so when we square it we get a probability of 1 half. However the |𝓇> and |𝓁> states should be orthonormal, since they are an eigenbasis, so they cant be the same. easy way to do this is just turn the second amplitude negative for |𝓁>, this keeps the probabilities the same but the (σ_x) states orthonormal:

|𝓇>=1/√(,2)*|𝓊>+1/√(,2)*|𝒹>=[[1/√(,2)],[1/√(,2)]]

|𝓁>=1/√(,2)*|𝓊>-1/√(,2)*|𝒹>=[[1/√(,2)],[(-1)/√(,2)]]

Now, to solve for the matrix elements in the up and down basis, we need to find how (σ_x) acts of |𝓊> and |𝒹>, to do this we need to write |𝓊> and |𝒹> in the right and left basis, we can solve for them as shown below:

|𝓇>=1/√(,2)*|𝓊>+1/√(,2)*|𝒹>

add this:

|𝓁>=1/√(,2)*|𝓊>-1/√(,2)*|𝒹>

result:

|𝓁>+|𝓇>=√(,2)*|𝓊>

(|𝓁>+|𝓇>)/√(,2)=|𝓊>

now that we know up, lets subtract to get down:

|𝓇>=1/√(,2)*|𝓊>+1/√(,2)*|𝒹>

subtract this:

|𝓁>=1/√(,2)*|𝓊>-1/√(,2)*|𝒹>

result:

|𝓁>-|𝓇>=√(,2)*|𝒹>

(|𝓁>-|𝓇>)/√(,2)=|𝒹>

Now we can cleanly solve for the (σ_x) operator in the z basis:

<𝓊|*(σ_x)|𝓊>=(<𝓁|+<𝓇|)/√(,2)*(σ_x)(|𝓁>+|𝓇>)/√(,2)

1/2 <𝓁|+<𝓇|*(σ_x)*|𝓁>+|𝓇>

1/2 <𝓁|+<𝓇|-|𝓁>+|𝓇>

1/2 -<𝓇|*|𝓇>+<𝓁|*|𝓁>

1/2 (-1+1)

1/2 0

0

Now I will leave it to you as an exercise to prove what the other components of the matrix is mathematically, but don't worry if you don't want to, just as I don't want to type it out, because if you do all the math, you just end up with this matrix:

(σ_x)=[[0,1],[1,0]]

Now we know two of the spin matrixes, (σ_z) and (σ_x), but what about (σ_y)? well it to must have a 1/2 probability of measuring up or down when in its eigenstates, so using our logic from last time the two eigenvectors of (σ_y) should be the same as (σ_x), right? Well no, because then 1. if that were true we would just use (σ_x) instead of (σ_y), and 2. (σ_y) 's eigenvectors must have a 1/2 chance to measure (σ_x) 's eigenvectors as well, so they HAVE to be different. But what other combinations could we have? well we still have our good old pals complex numbers with us, as we can, instead of making the second amplitude negative, times it by 𝑖 rotating it in complex plane, making it distinct from all other eigenvectors but keep all the probabilities we need (we are using in and out for (σ_y) eigenvectors):

|𝒾>=1/√(,2)*|𝓊>+𝑖/√(,2)*|𝒹>

|ℴ>=1/√(,2)*|𝓊>-𝑖/√(,2)*|𝒹>

Now what I could do is show you the exact process of getting the (σ_y) operator, but... I have typed plenty, so I will give you the answer and if you would like to do the math yourself feel free! I present the (σ_y) operator:

(σ_y)=[[0,-𝑖],[𝑖,0]]

Now, what the fuck were we doing again? Oh yea! trying to find the spin operator for an arbitrary direction n instead of a basis direction, now that we know what the three operators are we can plug in our operators and solve for the new one:

(σ_)⋅n=(n_x)*(σ_x)+(n_y)*(σ_y)+(n_z)*(σ_z)

(σ_)⋅n=(n_x)*[[0,1],[1,0]]+(n_y)*[[0,-𝑖],[𝑖,0]]+(n_z)*[[1,0],[0,-1]]

(σ_)⋅n=[[(n_z),((n_x)-𝑖*(n_y))],[((n_x)+𝑖*(n_y)),-(n_z)]]

and since (σ_)⋅n s just the amount of spin in the n direction, or the observable corresponding to the direction, I will just rewrite it as (σ_n):

(σ_n)=[[(n_z),((n_x)-𝑖*(n_y))],[((n_x)+𝑖*(n_y)),-(n_z)]]

now that we have how to get the operator for any direction, lets specify the n direction as a direction in the x - z plane, in other words, these are its coordinates where θ is the angle between n and the z axis:

(n_z)=cos(θ)

(n_x)=sin(θ)

(n_y)=0

now solve the matrix (σ_n):

(σ_n)=[[cos(θ),sin(θ)],[sin(θ),-cos(θ)]]

Now apperently theres a way to directly calculate eigenvectors and eigenvalues of operators, but im tired and might tell you about that as a note later, so for simplicity, here are the eigenvectors and eigenvalues of the operator (σ_n) when lying in the x - z plane:

(λ_1)=1

|(λ_2)>=[[cos(θ/2)],[sin(θ/2)]]

and:

(λ_1)=-1

|(λ_2)>=[[-sin(θ/2)],[cos(θ/2)]]

its good to see that the eigenvalues of the eigenvectors are +1 and -1, as those are the eigenvalues needed for spin observables. Now lets recap what we are trying to do here with this information as you might have forgot.

The original question was if we moved our apparatus 𝒜 along the z axis, and measured |𝓊>, that means that the state vector would be |𝓊> now. Now what if we moved 𝒜 in a random direction n and measured? what would we get? As stated before, spin states are weird, and don't commute, which means they don't share an eigenbasis or any eigenvectors, so unlike position's 3 operators, there is not an eigenvector shared between all 3 spin operators that represents this direction. This not commutable behavior also means if you measure one observable, you can't be sure about the measurements of the other 2 . because there is no shared eigenvector between the spin operators that represents this direction, we have to make a new operator with its own up and down eigenvectors, and then project our state |𝓊> onto those.

The operator blueprint we made allows us to make a spin operator for any direction:

(σ_n)=[[(n_z),((n_x)-𝑖*(n_y))],[((n_x)+𝑖*(n_y)),-(n_z)]]

however we decided to go the simpler route and use a direction of n in the x - z plane, making this simple matrix the operator for our direction n:

(σ_n)=[[cos(θ),sin(θ)],[sin(θ),-cos(θ)]]

Where θ is the angle between n and the z axis.

from this we can determine the eigenvectors and values of our observable:

(λ_1)=1

|(λ_2)>=[[cos(θ/2)],[sin(θ/2)]]

and:

(λ_1)=-1

|(λ_2)>=[[-sin(θ/2)],[cos(θ/2)]]

Now that we have the eigenvectors of the observable for the direction n, all we have to do to predict the probability of what we will get on measurement in that direction n, Is project the current state vector, |𝓊>, onto the operator (σ_n) 's eigenvectors, then find the magnitude squared of that just like any other measurement of the quantum system:

P(+1)=|<(λ_1)|*|𝓊>|2 probability of measuring "up" or +1

P(-1)=|<(λ_2)|*|𝓊>|2 probability of measuring "down" or -1

And that is all I know about the quirky - ness of spin states and how to measure the spin of a quantum system in every direction you want. Thank you for looking in the EXTRAS Section! I put this info down here because it didn't really fit anywhere into the article without causing a huge distraction.

Fourier Transform:

oh shit, Quantum mechanics uses this to!?! Yes it does, however, I will not be explaining what the Fourier transform or Fourier series is, as there are much better ways to learn it, such as videos. If you don't know what the Fourier transform is, thats fine, but it does help to know what it is. My recommendation is 3 Blue 1 brown's video, which provides (in my opinion) the best explanation: https://www.youtube.com/watch?v=spUNpyF58BY

Now how does this relate to quantum mechanics? Well first lets say we will constrain a particles position basis wave function (and therefore the particle in general) to a circle, just because. But because the circle and therefore the wave function wrap back on itself, the wave function needs to have the same starting value as ending value. Or in other words, by the time the wave function has a made a full loop around the circle, and therefore traveled the circles circumference, the wave function should have the same value its started as:

ψ(x) = ψ(x+2*π*R) ← must be true

where R is the radius of the circle, so 2*π*R is the circumference.

The only functions that fit this description are periodic functions, like sin or cos functions, that cycle once for every 2*π*R distance, or in other words have a period of 2*π*R. Now here comes Fourier with his big head, as we can write any periodic function as a sum of cos and sin's in a Fourier series, as shown and simplified below:

ψ(x)=(∑_n=0^j)((A_n)*cos((n*x)/R)+(B_n)*sin((n*x)/R))

or in the complex exponential version:

ψ(x)=(∑_n=0^j)((𝜓_n)*ℇ(𝑖*n*x/R))

now the reason we use (n*x)/R instead of just x is because of our periodic needs as shown below where we are solving for k:

ψ(x) = ψ(x+2*π*R)

ℇ(𝑖*k*x)=ℇ(𝑖*k*(x+2*π*R))

ℇ(𝑖*k*x)=ℇ(𝑖*k*x)*ℇ(𝑖*k*2*π*R)

to make these sides equal, ℇ(𝑖*k*2*π*R) has to be 1, which means 𝑖*k*2*π*R has to be a multiple of 2*π, because of the definition of Eulers formula, full solution for k shown below:

ℇ(𝑖*k*2*π*R)=1

k*2*π*R=n*2*π (where n is some integer)

k=(n*2*π)/(2*π*R)

k=n/R

Sweet! Now we know that we can write the position wave function as a Fourier series if its restricted to a circle, but what's even cooler is that ℇ(𝑖*k*x) is actually a eigenfunction of the momentum basis, or in other words, the position of a particle is described by a momentum superposition. This eigenfunction ℇ(𝑖*k*x) corresponds to a wave with a definite momentum value p=ℏ*k. proof of ℇ(𝑖*k*x) being an eigenfunction or wave is shown below:

𝚙*ℇ(𝑖*k*x)=-𝑖*ℏd(ℇ(𝑖*k*x))/d(x)

𝚙*ℇ(𝑖*k*x)=-𝑖*ℏ*𝑖*k*ℇ(𝑖*k*x)

𝚙*ℇ(𝑖*k*x)=ℏ*k*ℇ(𝑖*k*x)

Therefore ℇ(𝑖*k*x) is an eigenfunction of momentum with the eigenvalue ℏ*k. This is a really cool fact and allows us to properly define the uncertainty principle later on, but we need one more equation for the wave function on a circle. How to get the coefficients (𝜓_n), the steps and explanation are below:

ψ(x)=(∑_n=0^j)((𝜓_n)*ℇ(𝑖*n*x/R))

(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)=(∑_n=0^j)((𝜓_n))*(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*ℇ(𝑖*n*x/R)*d(x))

(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)=(∑_n=0^j)((𝜓_n))*(∫_-π*R^π*R)(ℇ(𝑖*(n-𝚖)*x/R)*d(x))

(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)=(∑_n=0^j)((𝜓_n)) ⋅ {[0↔n≠𝚖],[2*π*R↔n=𝚖]) 

(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)=(𝜓_𝚖)*2*π*R

(𝜓_𝚖)=1/(2*π*R)*(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)

Now i'm not the best at explaining this stuff but we just integrated on both sides, combined the exponentials, and if you look at it, thats actually an edge case, so we swapped it with the case brackets, those case brackets broke the sum down to 1 term times 2*π*R and we divided that over, so get a solution to the coefficients. If you want a better explanation look at Physics Elliot's video "To Understand the Fourier Transform, Start From Quantum Mechanics" which this information is based off of.

so now we have two equations defining stuff on a circle:

ψ(x)=(∑_k=n/R^)((𝜓_n)*ℇ(𝑖*n*x/R)) or ψ(x)=(∑_k=n/R^)((𝜓_n)*ℇ(𝑖*k*x))

(𝜓_𝚖)=1/(2*π*R)*(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)

As you might notice in the first one we are summing over every single possible value of k, this will be useful very soon, and also makes sense. Now lets turn that circle math into the infinite x axis, we can do this by making the circumference of the circle we are confined to infinite, how do we do this? well we take the limit as R →∞ turning 2*π*R infinite, and removing our confines of this damn circle! So lets take the limit right now! ... but first to help us later we are going to transform the (𝜓_n) coefficients into 1/(√(,2*π)*R)*ψ(k), I could explain how and why we can do this, but for me its to complicated of a detour (and I didn't bother to understand it fully), so we are just going to accept it as a fact here. The part I understand and you should to is that because we are no longer confined to a circle, the momentum of the particle could be a continuous set of values, thats why the coefficients turned into a function. Completing this swap is shown below:

ψ(x)=(∑_k=n/R^)(1/(√(,2*π)*R)*ψ(k)*ℇ(𝑖*k*x))

1/(√(,2*π)*R)*ψ(k)=1/(2*π*R)*(∫_-π*R^π*R)(ℇ(-𝑖*𝚖*x/R)*d(x))*ψ(x)

Now lets take the limit of both equations as R → ∞ to find what these equations for the position wave functions mean when applied to a full axis of values. Lets first try to solve for the first equation and then tackle the second one:

ψ(x)=(∑_k=n/R^)(1/(√(,2*π)*R)*ψ(k)*ℇ(𝑖*k*x))

now theres actually one thing we need to do before getting into the algebra manipulation. You see the 1/R factor, right? well we can actually rewrite that, because k=n/R and therefore the difference between each k value is 1/R, because n is any integer, so when n cycles through each integer k goes up by 1/R each time, this means that we can actually write this as the differential d(k) because it is the difference between each k value. Or if your fancy with it Δ(k), but it means the same thing. But lets substitute our find into our equation:

ψ(x)=(∑_k=n/R^)(d(k)1/√(,2*π)*ψ(k)*ℇ(𝑖*k*x))

Now, as we take the limit as R→∞, d(k) gets smaller and smaller, becoming an infinitesimal, just like in an integral or derivative! Also, notice what this is now telling us to do, its telling just to sum over all possible values of k, with a function based on (k^′)s value, multiplied by the width between each k value, d(k). Sounds a lot like an integral eh? Thats because it is! we can transform this sum into an integral integrating over all possible values of k, as shown below:

ψ(x)=(∑_k=n/R^)(d(k)1/√(,2*π)*ψ(k)*ℇ(𝑖*k*x))

and because our reasoning from before this equals:

ψ(x)=(∫_-∞^∞)(d(k)1/√(,2*π)*ψ(k)*ℇ(𝑖*k*x))

ψ(x)=1/√(,2*π)*(∫_-∞^∞)(d(k)*ψ(k)*ℇ(𝑖*k*x))

and if you'll notice, this looks like the Fourier Transform formula. Thats because it is! And since ψ(k) is the function to generate the coefficients of the momentum eigenstates, that means its just the momentum wave function! Thats literally the definition of it!, so I will now continue to swap ψ(k) for my regular notation of the momentum wave function 𝜑(k). So with all this information and the fact that we can clearly see the equation is a Fourier transform, as shown below:

our equation:

ψ(x)=1/√(,2*π)*(∫_-∞^∞)(d(k)*𝜑(k)*ℇ(𝑖*k*x))

the Fourier transform of a function g(t):

(∫_-∞^∞)(d(t)*g(t)*e(-𝑖*2*π*ƒ*t))

We can definitively say that the position wave function ψ(x) is equal to the inverse Fourier Transform of the momentum wave function 𝜑(k), or 𝜑(p). What about the second equation? where we solved for the coefficients, which we now know to be the momentum wave function in disguise. Well thats much simpler:

1/(√(,2*π)*R)*𝜑(k)=1/(2*π*R)*(∫_-π*R^π*R)(ℇ(-𝑖*k*x)*d(x))*ψ(x)

𝜑(k)=(√(,2*π)*R)/(2*π*R)*(∫_-π*R^π*R)(ℇ(-𝑖*k*x)*d(x))*ψ(x)

𝜑(k)=1/√(,2*π)*(∫_-π*R^π*R)(ℇ(-𝑖*k*x)*d(x))*ψ(x)

𝜑(k)=1/√(,2*π)*(∫_-∞^∞)(ℇ(-𝑖*k*x)*d(x))*ψ(x)

Basically the only thing that changed was the fact that the boundaries for the integral are infinite. But now you can clearly see that the momentum wave function 𝜑(k) and the position wave function ψ(x) of a particle are Fourier transform pairs, as one is the Fourier transform of the other and one is the inverseFourier transform of the other:

ψ(x)=1/√(,2*π)*(∫_-∞^∞)(d(k)*𝜑(k)*ℇ(𝑖*k*x))

𝜑(k)=1/√(,2*π)*(∫_-∞^∞)(d(x))*ψ(x)*ℇ(-𝑖*k*x)

Whichever one you consider to be the inverse Fourier and the normal Fourier transform doesn't really matter, as long as the imaginary components in their exponential have the opposite sign.

I would also like to point out that in our main article we used p when applying the momentum wave function, as in 𝜑(p), but here we used k, why is that? Well as we have seen before the eigenvalues of the momentum operator are ℏ*k. Therefore when we measure the momentum of a particle, we will get some value ℏ*k, that means that p=ℏ*k, where k is changing.

So why do we use both? well because they are basically the same thing, as k known in the real world as the angular spatial frequency of the particle, or how many times it rotates through phase in space. ℏ is known as the reduced Planck's constant, what multiplying by ℏ does is basically convert the units of the equation is to quantum units used for small scales, just like in quantum mechanics. So in short, what p=ℏ*k saying, is that the momentum of a particle is equal to the angular spatial frequency of the particle converted to quantum scales using ℏ.

But what about the normal time based frequency of a particle? is there anything cool involved with that? Yes! As the energy of a particle 𝘌=ℏ*ω where omega is the angular time based frequency of the particle. If you think about it, these two conjectures make a lot of sense, as the momentum of a particle would be related to some part of the spatial frequency, and the energy makes sense to be connected to the time based frequency, or how many times it oscillates or rotates per second.

anyway, back to the Fourier transforms we found with the position and momentum wave functions. Now that we know what p equals, we can say what the Fourier transform for 𝜑(p) and 𝜑(k) is, along with the position Fourier transform:

ψ(x)=1/√(,2*π)*(∫_-∞^∞)(d(k)*𝜑(k)*ℇ(𝑖*k*x))

𝜑(k)=1/√(,2*π)*(∫_-∞^∞)(d(x))*ψ(x)*ℇ(-𝑖*k*x)

𝜑(p)=1/√(,2*π*ℏ)*(∫_-∞^∞)(d(x))*ψ(x)*ℇ(-𝑖*p*x/ℏ)

A cool fact about the Fourier transform, when you feed it a sound wave with a short duration (also known as concentrated in one area), it won't be able to tell you with as much confidence what frequency that is. Which means the function outputted will be more spread out around the answer.

Fourier transform (somewhat):

and since the momentum wave function is the Fourier transform of the position wave function, if you know the position of a particle, which is the same as its wave function being concentrated to a point, you will not know the momentum of the particle, as its momentum wave function will spread out, and vise versa. This is the uncertainty principle that I poorly explained in the main article!

Hope you enjoyed! thats all my Fourier transform knowledge, for now.

Deriving the Schrödinger equation another way

We derived the time dependent Schrödinger equation below by connecting it to Lagrangian mechanics very loosely:

-𝑖*ℏ d|𝜓(t)>/d(t)=H|𝜓>

But what if we wanted to come to the conclusion more mathamatically? Well look no further! as this is how to derive that equation using only algebra and derivatives!

Lets say theres a operator called the time evolution operator U(t) and if you apply this operator to the initial quantum state, it evolves the quantum state by time t:

U(t)*|𝜓(0)>=|𝜓(t)>

and by quantum principles, if two states 𝜓(0) and 𝜑(0) are distinguishable, they will stay distinguishable, and that means they will be and stay orthonormal quantum states:

<𝜓(0)|*|𝜑(0)>=0

<𝜓(t)|*|𝜑(t)>=0

For this to be true, because of our declaration from earlier, this will also be true:

<𝜓(t)|*|𝜑(t)>=0

<𝜓(0)|U†*(t)*U(t)*|𝜑(0)>=0

For this to also be true, then U†*(t)*U(t) must do nothing and act like the identity operator 𝘐

U†*(t)*U(t)=𝘐

This is a cool property called unitarity, it can be used to derive the Schrödinger equation like we are about to do, if ε (an infinitesimal) is put into the U(t) operator this should still be true:

U†*(ε)*U(ε)=𝘐

Now its pretty evident that when ε=0, U(ε) is equal to the 𝘐 operator, so when ε is slightly above zero (an infinitesimal) it should be the identity operator subtracted by a value proportional to ε:

U(ε)=𝘐-𝑖*ε*H

I will ignore why 𝑖 the hamiltonian H is there, but they are there. Now lets apply this operator to the quantum state to see what happens!:

U(ε)*|𝜓(0)>=|𝜓(ε)>=|𝜓(0)>-𝑖*ε*H|𝜓(0)>

Now we are almost there! Just do some algebraic manipulation and we have derived the time dependent Schrödinger equation!:

|𝜓(ε)>=|𝜓(0)>-𝑖*ε*H|𝜓(0)>

|𝜓(ε)>-|𝜓(0)>=-𝑖*ε*H|𝜓(0)> ←subtract |𝜓(0)>

(|𝜓(ε)>-|𝜓(0)>)/ε=-𝑖*H|𝜓(0)> ←divide by ε

since on the left side its equivalent to the derivative formula, we can just replace it with a time derivative as ε → 0 eating the time dependent Schrödinger equation (for real this time):

d|𝜓>/d(t)=-𝑖*H|𝜓(0)>

move 𝑖:

𝑖d|𝜓>/d(t)=H|𝜓(0)>

add ℏ for unit compliance:

𝑖*ℏd|𝜓>/d(t)=H|𝜓(0)>

tada! hope you enjoyed another section of the EXTRAS chapter!

Expectation Values

Expectation values are written as the average value of an observables measurement, using the statistics formula for expectation values, we get this below:

<H>=(∑_j=1^n)((λ_j)*P((λ_j)))

or for a continuous observable:

<𝚡>=(∫_-∞^∞)(x*P(ψ(x))*d(x))

this is just the formula for the summation of all possible measurement values times their probability of occurring. Since the probability of measuring a certain value of the position wave function is just the position wave function at that point magnitude squared, we can replace P(ψ(x)) with |ψ(x)|2:

<𝚡>=(∫_-∞^∞)(x*|ψ(x)|2*d(x))

Another way to write this for any continuous sum is to write it in the position basis exclusively, instead of in the operators respective basis, for example, this is what we would normally do for the position basis:

<𝚙>=(∫_-∞^∞)(p*|𝜑(p)|2*d(p))

or

<𝚙>=(∫_-∞^∞)(|𝜑(k)|2*ℏ*k*d(k))

because k and p are proportional and p is just k but with quantum scale and units,

and this is what we would do for representing that in the position basis:

<𝚙>=(∫_-∞^∞)(ψ(x)∗*𝚙*ψ(x)*d(x))

This makes sense because ψ(x)∗⋅ψ(x)=|ψ(x)|2, now how can this be useful? well take one of the problems taken off of MIT's 8.04 Quantum Physics 1, which is this:

use these:

ψ(x)=1/√(,2*π)*(∫_-∞^∞)(𝜑(k)*ℇ(𝑖*k*x)*d(k))

𝚙=-𝑖*ℏd()/d(x)

<𝚙>=(∫_-∞^∞)(ψ(x)∗*𝚙*ψ(x)*d(x))

to prove that this equation is true:

<𝚙>=(∫_-∞^∞)(|𝜑(k)|2*ℏ*k*d(k))

now I really encourage you to try to solve this yourself, if you need a hint:

first plug in the momentum operator and the Fourier transform equations into the expectation value function, then try to combine the derivatives and the plane waves.

But if you are too lazy, or just want to know how to solve it, then heres the solution!:

<𝚙>=(∫_-∞^∞)(ψ(x)∗*𝚙*ψ(x)*d(x))

<𝚙>=(∫_-∞^∞)((1/√(,2*π)*(∫_-∞^∞)(𝜑(k)*ℇ(𝑖*k*x)*d(k)))∗*(-𝑖*ℏd()/d(x))*(1/√(,2*π)*(∫_-∞^∞)(𝜑(k)*ℇ(𝑖*k*x)*d(k)))*d(x))

<𝚙>=(∫_-∞^∞)((1/√(,2*π)*(∫_-∞^∞)(𝜑(q)∗*ℇ(-𝑖*q*x)*d(q)))*(1/√(,2*π)*(∫_-∞^∞)(𝜑(k)*(-𝑖*ℏd()/d(x))*ℇ(𝑖*k*x)*d(k)))*d(x))

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*ℇ(-𝑖*q*x)*(-𝑖*ℏd()/d(x))*ℇ(𝑖*k*x)*d(q))*d(k)))*d(x)

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*((∫_-∞^∞)(ℇ(-𝑖*q*x)*(-𝑖*ℏd()/d(x))*ℇ(𝑖*k*x)*d(x)))*d(q))*d(k))

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*((∫_-∞^∞)(ℇ(-𝑖*q*x)*ℏ*k*ℇ(𝑖*k*x)*d(x)))*d(q))*d(k))

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*ℏ*k*((∫_-∞^∞)(ℇ(-𝑖*q*x)*ℇ(𝑖*k*x)*d(x)))*d(q))*d(k))

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*ℏ*k*((∫_-∞^∞)(ℇ(𝑖*(q-k)*x)*d(x)))*d(q))*d(k))

<𝚙>=1/(2*π)*(∫_-∞^∞)((∫_-∞^∞)(𝜑(q)∗*𝜑(k)*ℏ*k*2*π*δ(q-k)*d(q))*d(k))

<𝚙>=(2*π)/(2*π)*(∫_-∞^∞)(𝜑(k)∗*𝜑(k)*ℏ*k*d(k))

<𝚙>=(∫_-∞^∞)|𝜑(k)|2*ℏ*k*d(k)

bam! Integral mania! I won't explain all of this, but I will explain the main steps, hopefully that should be enough for you to understand what this means.

First important thing is we plugged the definition of the momentum operator 𝚙 and the definition of the position wave function ψ(x) into the definition MIT gave us for the momentum operators expectation value. Then applied the conjugation on one of the Fourier transforms, pulled out the constants, 1/√(,2*π) and multiplied them, getting 1/(2*π), Then moved around the integrals.

Second, we integrated over the plane waves times the momentum operator. Which turned into a Dirac Delta function times 2*π times ℏ*k. This is because multiplying the plane wave by the momentum operator gives ℏ*k, the eigenvalue of the plane wave in the momentum basis, and the integral of a plane wave involving subtraction in its phase like it does here (ℇ(𝑖*(q-k)*x)), is equal to the Dirac Delta times 2*π.

Finally, we pull out 2*π, cancelling it with 1/(2*π) and use the Dirac delta to collapse the two variables down to 1. Then we use the definition of a complex number times its conjugate to get the result we wanted!

If you understand what I just said then you can say you have solved an MIT level quantum physics problem! I do also recommend checking out MIT 8.04, as its actually a pretty good free lecture course on quantum mechanics, no need to do the problem sets fully though, as some of the problems rely on previous knowledge form other lecture courses, or are just based on things you don't know, but the lectures themselves is very very good either way.

Thanks for reading another EXTRAS chapter!