Solve the particle in a box problem using matrix mechanics?

[mike1000]

How do we solve the particle in a box (infinite potential well) problem using matrix mechanics rather that using Schrodingers Equation? Schrodingers Equation for this particular problem is a simple partial differential equation and is easy for me to follow. The solution has the following form,\begin{equation}\psi_n(x)= \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x\end{equation}
However, I am interested in solving this problem using the methods of matrix mechanics using bras and kets. As I understand, the dirac notation, $\langle \vec{x}|\psi\rangle$ is equivalent to the above notation for $\psi_n(x)$ or \begin{equation}\langle \vec{x}|\psi\rangle\equiv\psi_n(x)\end{equation}
Both notations express the idea of a function, however, the term on the left in Equation (2) (the Dirac notation) indicates (to me) a dot product of two vectors whereas the term on the right is an expression for a continuous function obtained by solving the Schrodinger Equation.

I would like to attempt to solve this problem using the methods of matrix mechanics and not the Schrodinger Equation.

Here is a description of the problem...

The potential energy is 0 inside the box (V=0 for 0<x<L) and goes to infinity at the walls of the box (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box.

Here is a link to the solution found by solving the Schrodinger Equation.
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/05.5:_Particle_in_Boxes/Particle_in_a_1-dimensional_box

To get the state vector(s) $\psi$ for the particle in the box, using the methods of matrix mechanics, how should I start?

Here is the Hamiltonian
\begin{equation}-\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x)\end{equation}

And here is the momentum operator \begin{equation}\hat{p}=-\imath\hbar \frac{\partial}{\partial x} = -\imath\hbar\nabla\end{equation}

And I think this is the Hamiltonian Operator
\begin{equation}\hat{H}=-\dfrac{\hbar^2}{2m} \nabla^2- E\end{equation}

I think what I have to do at this point is find the eigenvalues and eigenvectors of the Hamiltonian Operator.(Equation (5))

This is where I get lost. How do I set up that matrix for the particle in the box?

 

[vanhees71]

I don't think that you can solve this academic problem in matrix mechanics since it's not clear how to incorporate the boundary conditions, which is easy in the position representation (wave mechanics).

 

[mike1000]

I don't think that you can solve this academic problem in matrix mechanics since it's not clear how to incorporate the boundary conditions, which is easy in the position representation (wave mechanics).

Here are the boundary conditions...

The box is one-dimensional and has length = L. The probability of finding the particle outside the box is zero, which implies that \begin{equation}\psi(x=0)=\psi(x=L) = 0\end{equation} or using brakets \begin{equation}\langle x=0|\psi\rangle = \langle x=L|\psi\rangle = 0 \end{equation}Can you be more specific about what it is, about this problem, which makes it not possible to solve using matrix mechanics? Is it because it is only one-dimensional? How could we change it to make it solvable using matrix mechanics?

How would I implement those boundary conditions in matrix mechanics?

 

[PeterDonis]

the term on the left in Equation (2) (the Dirac notation) indicates (to me) a dot product of two vectors

No, it indicates an infinite family of such dot products, since the bra $\langle \vec{x} \vert$ does not refer to a single (dual) vector, but to the infinite family of such vectors that constitute the position basis.

Here is the Hamiltonian

That is not the Hamiltonian. It's the eigenvalue equation for the free particle Hamiltonian (which is not the Hamiltonian for this problem, since there is a nonzero potential).

I think this is the Hamiltonian Operator

Not quite. The first term on the LHS is correct, but the second is not. The potential energy as a function of $x$ is what should appear there, not the eigenvalue $E$.

At least, that's what the Hamiltonian should look like in the Schrodinger formalism. But in the matrix formalism, you need a matrix, not a differential operator. The key question is this: how many rows/columns should this matrix have? Answering this question will tell you why this particular problem is not easily solved in the matrix mechanics formalism.

 

[vanhees71]

Here are the boundary conditions...

The box is one-dimensional and has length = L. The probability of finding the particle outside the box is zero, which implies that \begin{equation}\psi(x=0)=\psi(x=L) = 0\end{equation} or using brakets \begin{equation}\langle x=0|\psi\rangle = \langle x=L|\psi\rangle = 0 \end{equation}Can you be more specific about what it is, about this problem, which makes it not possible to solve using matrix mechanics? Is it because it is only one-dimensional? How could we change it to make it solvable using matrix mechanics?

How would I implement those boundary conditions in matrix mechanics?

Of course, these are the boundary conditions, and in the position representation it's a pretty easily solved problem (although there's more mathematics in it than usually taught in QM1, e.g., the fact that there exists no momentum observable, i.e., $-\mathrm{i} \partial_x$ is only Hermitean but not self-adjoint).

As I said, I have no idea, how to solve this in matrix mechanics, because you cannot so easily implement these boundary conditions, but why would one like to solve it with matrix mechanics? The choice of the right coordinate system is already key to the solution of classical-mechanics problems. So it is in quantum mechanics: Here the position representation is the natural choice, and thus the problem is solved with it most easily.

 

[mike1000]

No, it indicates an infinite family of such dot products, since the bra $\langle \vec{x} \vert$ does not refer to a single (dual) vector, but to the infinite family of such vectors that constitute the position basis.
That is not the Hamiltonian. It's the eigenvalue equation for the free particle Hamiltonian (which is not the Hamiltonian for this problem, since there is a nonzero potential).
Not quite. The first term on the LHS is correct, but the second is not. The potential energy as a function of $x$ is what should appear there, not the eigenvalue $E$.

At least, that's what the Hamiltonian should look like in the Schrodinger formalism. But in the matrix formalism, you need a matrix, not a differential operator. The key question is this: how many rows/columns should this matrix have? Answering this question will tell you why this particular problem is not easily solved in the matrix mechanics formalism.

Since there are an infinite number of possible energy states I would say that there should be an infinite number of rows and columns in the matrix. Are you trying to tell me that Matrix Mechanics cannot solve a problem when there are an infinite number of eigenvalues? If so, why not?

 

[mike1000]

Of course, these are the boundary conditions, and in the position representation it's a pretty easily solved problem (although there's more mathematics in it than usually taught in QM1, e.g., the fact that there exists no momentum observable, i.e., $-\mathrm{i} \partial_x$ is only Hermitean but not self-adjoint).

As I said, I have no idea, how to solve this in matrix mechanics, because you cannot so easily implement these boundary conditions, but why would one like to solve it with matrix mechanics? The choice of the right coordinate system is already key to the solution of classical-mechanics problems. So it is in quantum mechanics: Here the position representation is the natural choice, and thus the problem is solved with it most easily.

There are two reasons why I would like to solve this with Matrix Mechanics(MM)...

First, I have read, many times, that the Schrodinger method and the Heisenberg Method(Matrix Mechanics) are equivalent methods for solving quantum mechanical problems. If that is true, then I think I should expect to be able to solve the simplest of all problems using both methods. However, I was not able to find any example of solving the particle in a box problem using Matrix Mechanics...

The second reason is, I am still perplexed by the Matrix Mechanics method in general. I still have absolutely no idea where the initial state vector $|\psi\rangle$ comes from. As I said in a previous thread, it looks to me like the MM method depends upon state vector rather than discovers what it is. I know this is wrong, but I do not see, yet, how to start the MM method without already knowing what $\psi$ is. Starting with the particle in the box problem, I thought, would be a good way to expose the MM method from its roots.

What is the simplest problem that can be solved using the MM method? If you point me in that direction I will look that up.

 

[PeterDonis]

Since there are an infinite number of possible energy states I would say that there should be an infinite number of rows and columns in the matrix

Yes, exactly.

Are you trying to tell me that Matrix Mechanics cannot solve a problem when there are an infinite number of eigenvalues?

No, I'm just saying it's not at all easy.

What is the simplest problem that can be solved using the MM method?

Try an example where there are just two eigenvalues, such as the spin of a spin-1/2 particle.

 

[vanhees71]

There are a few eigenvalue problems you can solve with the abstract Dirac formalism only, i.e., by only using the commutator relations of the involved observables. The key to the solution are the underlying symmetry principles (imho 99% of physics is applied Lie-group theory anyway ;-)). The examples that come to my mind are:

(a) position and momentum: You use the fact that the momentum component in a fixed direction generates translations, i.e., assuming that there's one eigen vector for $\hat{x}$ with eigenvector $x$, you get eigenvectors for any eigenvalue $x \in \mathbb{R}$. One should of course be aware, that this is the typical physicist's abuse of mathematics: The position and also the momentum eigenvectors in QT are not real vectors in the Hilbert space, but they are in the dual of the domain of these operators, i.e., they are generalized vectors in the sense of distributions.

(b) energy eigenvalue problem for the harmonic oscillator: Here you find ladder operators, such that you can build the energy eigenvectors from the ground state, and you get also the eigenvalues $E_{n}=(n+1/2)\hbar \omega$, where $n \in \mathbb{N}_0$.

(c) eigenvalue problem of the angular-momentum algebra, where you can show that you can diagonalize simultaneously $\hat{\vec{J}}^2$ and one component, usually $\hat{J}_z$ is used, where also ladder operators come into use (this is not an accident but due to the fact that the two-dimensional symmetric harmonic oscillator has a dynamical SU(2) symmetry, and the Lie-algebra of SU(2) is just the Liealgebra of the rotation group, and that's nothing else than the angular-momentum algebra, because angular momentum is generating rotations).

(d) eigenvalue problem of the Hamiltonian of the hydrogen atom: It's solved due to the fact that it has an SO(4) dynamical symmetry, whose Lie algebra is equivalent to a direct sum of to su(2) algebras, and you can just reuse what you've learned in considering (c). Funnily enough for $E<0$ you get a representation of SO(4), for $E=0$ of the Galilei group, and for $E>0$ the proper orthochronous Lorentz group.

 

[Nugatory]

You'll find a worked example for a spin-1/2 particle (PeterDonis is right about that being one of the simplest examples) here: http://scipp.ucsc.edu/~dine/ph101/bands.pdf

The second reason is, I am still perplexed by the Matrix Mechanics method in general. I still have absolutely no idea where the initial state vector $|\psi\rangle$ comes from.

It's given to you, in exactly the same way that is given to you when you're solve a problem using the Schrodinger formalism. Either way, the problem is "Given a quantum system in this initial state at time zero and subject to this Hamiltonian, what is its future state?". You'll see this in section 2 of the notes I linked above, in the sentence that starts "At $t=0$, ...".
For that matter, it's the same in classical physics: If I ask you to calculate where a fired cannonball is going to land, Newton's laws give you all the physical knowledge and mathematical machinery you need to calculate the trajectory from arbitrary initial conditions, but I still have to give you the initial position and velocity of the cannonball as it leaves the cannon before you can calculate where it will land.

 

[mike1000]

You'll find a worked example for a spin-1/2 particle (PeterDonis is right about that being one of the simplest examples) here: http://scipp.ucsc.edu/~dine/ph101/bands.pdfIt's given to you, in exactly the same way that is given to you when you're solve a problem using the Schrodinger formalism. Either way, the problem is "Given a quantum system in this initial state at time zero and subject to this Hamiltonian, what is its future state?". You'll see this in section 2 of the notes I linked above, in the sentence that starts "At $t=0$, ...".
For that matter, it's the same in classical physics: If I ask you to calculate where a fired cannonball is going to land, Newton's laws give you all the physical knowledge and mathematical machinery you need to calculate the trajectory from arbitrary initial conditions, but I still have to give you the initial position and velocity of the cannonball as it leaves the cannon before you can calculate where it will will land.

Ah, I finally understand! The initial state vector represents the initial condition. That makes sense. The matrix mechanics method has to start somewhere.

And I suppose this is part of the problem with using MM to solve the particle in the box problem? We do not know what is the initial condition, ie what is the initial energy state for the particle in the box? (That is a question not a statement.)

With the Schrodinger Equation we can use boundary conditions instead of initial conditions to solve for the wave equation.

How do we translate the boundary conditions for the particle in a box to an initial state vector?

If we knew the initial state vector could we then solve the particle-in-the-box problem easily using MM?

(I am reading the article you linked to)

 

[Nugatory]

And I suppose this is part of the problem with using MM to solve the particle in the box problem? We do not know what is the initial condition, ie what is the initial energy state for the particle in the box? (That is a question not a statement.)

You don't know it when you use the Schrodinger formalism either. See below.

With the Schrodinger Equation we can use boundary conditions instead of initial conditions to solve for the wave equation.
How do we translate the boundary conditions for the particle in a box to an initial state vector?

You don't, whether you're using the Schrodinger or the Heisenberg formalism. Review the steps you went through to solve this problem using the Schrodinger formalism: First you solved the time-independent Schrodinger equation to find the eigenvalues and eigenfunctions of the Hamiltonian operator, and you used the boundary conditions for that. But that didn't give you the state of the particle, it gave you the base vectors that you can use to write any given state of the particle; in general that state will be a superposition of those vectors.

Next, you took the initial state (which you've been given as part of the complete problem statement) and you wrote it in terms of those eigenvectors. Sometimes this is really easy. For example in the particle-in-a-box problem we're often told that the particle has been prepared with a particular energy (or equivalently, momentum); then the initial state is just the corresponding energy eigenfunction and its time evolution is trivial. However, the initial condition might be something like: The particle is localized at the left side of the box and moving to the right at a given speed (both of which can be only specified to within the limits allowed by the uncertainty principle). In this case, we would treat the initial state as a Gaussian superposition of many different energy eigenfunctions, and its time evolution would show it moving back and forth in the box.

 

[mike1000]

You don't know it when you use the Schrodinger formalism either. See below.

You don't, whether you're using the Schrodinger or the Heisenberg formalism. Review the steps you went through to solve this problem using the Schrodinger formalism: First you solved the time-independent Schrodinger equation to find the eigenvalues and eigenfunctions of the Hamiltonian operator, and you used the boundary conditions for that. But that didn't give you the state of the particle, it gave you the base vectors that you can use to write any given state of the particle; in general that state will be a superposition of those vectors.

Next, you took the initial state (which you've been given as part of the complete problem statement) and you wrote it in terms of those eigenvectors. Sometimes this is really easy. For example in the particle-in-a-box problem we're often told that the particle has been prepared with a particular energy (or equivalently, momentum); then the initial state is just the corresponding energy eigenfunction and its time evolution is trivial. However, the initial condition might be something like: The particle is localized at the left side of the box and moving to the right at a given speed (both of which can be only specified to within the limits allowed by the uncertainty principle). In this case, we would treat the initial state as a Gaussian superposition of many different energy eigenfunctions, and its time evolution would show it moving back and forth in the box.

Yes, I agree solving the Schrodinger Equation did not give us the initial state, it gave us the eigenvalues and eigenvectors with which we can express any given state of the particle. Thats a lot. (You are not suggesting that we have to solve the Schrodinger Equation first to get the eigenvalues and eigenvectors before we can use the methods of Matrix Mechanics, are you?)

But the question I am struggling with, is why can't we get the same eigenvalues and eigenvectors using Matrix Mechanics? Why is it so difficult to solve this simple problem using Matrix Mechanics? You have shown me that we did not need the initial state to be able to solve the Schrodinger Equation and we do need the initial state vector to solve it using MM. Is that what is missing?

In MM we have a matrix of simultaneous equations of some sort, one equation per row. What does each row in the matrix represent? Why is it so difficult to fill in the values for this matrix?

 

[PeterDonis]

Since there are an infinite number of possible energy states I would say that there should be an infinite number of rows and columns in the matrix.

This is almost correct. The only wrong word is the word "energy". In the Schrodinger formalism you were using the position representation. If you use the matrix formalism with the same representation, which is implied by the bra $\langle \vec{x} \vert$ you wrote down in the OP, then that implies that the rows/columns of the matrix should not represent an infinite number of possible energy states, but an infinite number of possible...what kind of states?

What does each row in the matrix represent?

Answering the above question will tell you.

 

[mike1000]

This is almost correct. The only wrong word is the word "energy". In the Schrodinger formalism you were using the position representation. If you use the matrix formalism with the same representation, which is implied by the bra $\langle \vec{x} \vert$ you wrote down in the OP, then that implies that the rows/columns of the matrix should not represent an infinite number of possible energy states, but an infinite number of possible...what kind of states?
Answering the above question will tell you.

Your answer (or rather question) is not what I am looking for. I suppose you are trying to say that the rows represent positions in some way. I am not looking for such a general answer. I am looking for the actual values that we would calculate to start to populate the matrix. We do not have to calculate all the values, just the first few elements of the first few rows. That will tell me everything I need to know. The problem is the particle in the well. We know the Hamiltonian for that problem. And we know the matrix equation $H\psi=E\psi$ How do we form the matrix which will lead us to the eigenvalues for the problem? What are the knowns and what are the unknowns? I think, almost everything, except the eigenvalues has to be known in order to solve this problem using the Matrix Method. How do we incorporate the boundary conditions into the matrix equation?

In order to use the MM method, I am coming to believe that we have to have some actual measurements of the particles position and I guess momentum. Using the Schrodinger Equation, we were able to determine the eigenvalues in an abstract way, knowing only the boundary conditions. The Schrodinger equation allowed us to predict the eigenvalues. The MM method does not seem to me to be a way to predict, but rather a way to confirm. If this sounds vague, you are right. I know something is missing but I do not know what it is yet...

If I assume the MM method requires us to measure the position and momentum of the particle in the box, how many measurements would we have to make and how dense would they have to be? And what would happen if we only "sampled" the positions and momentums, say only 100 measurements. I am guessing that if we took 100 measurements that would result in a 100x100 matrix. But what would that do for us? We know there are infinite number of eigenvalues. So would we learn anything if we only took 100 measurements? Would that matrix eigenvalues relate to the real eigenvalues in any meaningful way?

As I think about the measurement scenario, if we sampled the positions and momentums of the particle in the box they would have to be made when the particle was in different energy states for the measurements to have any value. If all the measurements are made with the particle in the same energy state, the rows in the resulting matrix would be "degenerate" . Each row in the matrix would have to correspond to a measurement made when the particle was in a different energy state. This seems to me to be a very difficult requirement for a single particle in the well.

 

[hilbert2]

The 1D time independent Schrödinger equation can actually be converted to a tridiagonal matrix eigenvalue equation by replacing the wavefunction $\psi (x)$ with a vector of discrete values $\psi_n = \psi(n\Delta x )$ and then changing the second derivative operator $\frac{d^2}{dx^2}$ to its finite difference version $\frac{\psi_{n-1}-2\psi_n + \psi_{n+1}}{[\Delta x]^2}$. As far as I know, this kind of discretization comes with a built-in boundary condition that the wavefunction vanishes at the endpoints of its domain (as it does in the infinite well problem).

http://www.dartmouth.edu/~pawan/final%20project.pdf

 

[vanhees71]

I think we first should get an understanding, what's meant by "matrix mechanics". To understand different representations of QT, it's good to use the representation free Hilbert-space formalism, i.e., for physicists, the Dirac bra-ket formalism.

It starts with an abstract Hilbert space with vectors written as $|\psi \rangle$ and a scalar product written as $\langle \phi|\psi \rangle$. The $|\psi \rangle$ can be used to represent a pure state of a quantum system. Then you have observables, which are represented in the formalism as self-adjoing operators $\hat{A}$ with a domain (which by definition equals their co-domain), which is in general only a dense subspace of Hilbert space. A (perhaps generalized) eigenvector of $\hat{A}$ with eigenvalue $a \in \mathbb{R}$ is denoted as $|a \rangle$. It obeys by definition
$$\hat{A} |a \rangle=a |a \rangle.$$
To define a representation you choose an arbitrary complete set of commuting self-adjoint operators (also called a complete compatible set of observables), $\hat{A}_1,\ldots, \hat{A}_N$. There are common eigenvectors of such sets of operators and complete means that giving the eigenvalues $(a_1,\ldots,a_N)$, the eigenvector is (up to a non-zero factor) uniquely defined (in other words the common eigenspaces to given $(a_1,\ldots,a_N)$ eigenvalues are one-dimensional).

These are also complete sets of orthonormal eigenvectors (with the proper normalization). Matrix mechanics now is defined in the case of a completely discrete set of eigenvalues, and then all the eigenvectors are proper normalizable eigenvectors, i.e., you can choose them such that
$$\langle a_1,\ldots,a_N|a_1',\ldots,a_N' \rangle=\delta_{a_1,a_1'} \cdots \delta_{a_N,a_N'}$$
and
$$\sum_{a_1,\ldots,a_N} |a_1,\ldots a_N \rangle \langle a_1,\ldots a_N|=\hat{1}.$$
This means you can uniquely describe any state by it's components
$$\psi_{a_1,\ldots,a_N} = \langle a_1,\ldots,a_N|\psi \rangle$$
and any operator by its matrix elements
$$A_{a_1,\ldots,a_N;a_1',\ldots,a_N'} = \langle a_1,\ldots,a_N|\hat{A}|a_1',\ldots,a_N' \rangle.$$
It is then easy to show by using the completeness relation several times that the components of $|phi \rangle=\hat{A} |\psi \rangle$ are given by
$$\phi_{a_1,\ldots,a_N}= \sum_{a_1',\ldots,a_N'} A_{a_1,\ldots,a_N;a_1',\ldots, a_N'} \psi_{a_1',\ldots,a_N'}.$$
This is like matrix-vector multiplication, and the composition of two operators maps to matrix products, and so on. That's why it's called "matrix mechanics".

So for your energy-eigenvalue problem for a particle in a rigid box to define it in terms of matrix mechanics you first have to find an appropriate basis, in which you work, and then you should be able to calculate the Hamiltonian's matrix elements with respect to this basis. This is not so easy. In fact the only "natural" basis that comes to my mind are indeed the energy eigenstates themselves, and to find them here the representation in the position representation is good. Here the basis are the generalized position eigenvectors $|x \rangle$ (I consider the 1D case for simplicity). These are living in the dual of the domain of the position operator and $x$ can take all real values (or here in this pretty artificial problem restricted to the interval $[0,L]$). Then the "orthonormalization and completeness relations" have to be generalized to
$$\langle x|x' \rangle=\delta(x-x'), \quad \int_0^L \mathrm{d} x |x \rangle \langle x|=\hat{1},$$
and you have to use that in the position representation (i.e., matrix mechanics) the Hamiltonian is given as the differential operator
$$\hat{H}=-\frac{\hbar}{2m} \partial_x^2.$$
It's living on a proper subset of the square-integrable functions, which represent the states $\psi(x)=\langle x|\psi \rangle$. The Hilbert space is further specified (and that's what you can do only in the position representation) by $\psi(0)=\psi(L)=0$. With this you can solve the eigenvalue problem for $\hat{H}$, and then use the eigenvectors of $\hat{H}$ for other matrix-mechanics calculations.

 

[hilbert2]

Here's an R-Code that forms the Hamiltonian matrix for the discretized square well problem and solves the eigenstate n=3 (the ground state is n=1), plotting the probability density as an output:

L <- 1                                               # Length of the domain
N <- 100                            # Number of discrete points
dx <- L/N
A = matrix(nrow = N, ncol = N)                # Hamiltonian matrix
n_state <- 3                        # Number of the eigenstate to be calculated

for(m in c(1:N))                    # Fill the Hamiltonian matrix with elements appropriate for an infinite square well problem
{
for(n in c(1:N))
{
A[m,n]=0
if(m == n) A[m,n] = 2/dx^2
if(m == n-1 || m == n+1) A[m,n]=-1/dx^2
}
}

v = eigen(A)                       # Solve the eigensystem
vec = v$vectors

soln = c(1:N)
xaxis = c(1:N)*L/N

for(m in c(1:N))
{
soln[m] = vec[m,n_state]               # Fill the vector "soln" with the wavefunction values
}

jpeg(file = "plot.jpg")                   # Plot the probability density
plot(xaxis,abs(soln)^2)
lines(xaxis,abs(soln)^2)
dev.off()

The plot that this code produces looks just like you'd expect from a square of a sine function.

 

[mike1000]

Thank you for the time it took to compose your reply. I have made comments to express (I hope) what I think you mean. I am sure I misinterpreted a lot but I think I also got much of it. Thanks.

I think we first should get an understanding, what's meant by "matrix mechanics". To understand different representations of QT, it's good to use the representation free Hilbert-space formalism, i.e., for physicists, the Dirac bra-ket formalism.

The Dirac bra-ket formalism is a representation free way to talk about quantum mechanics.

It starts with an abstract Hilbert space with vectors written as $|\psi \rangle$ and a scalar product written as $\langle \phi|\psi \rangle$. The $|\psi \rangle$ can be used to represent a pure state of a quantum system.

You are defining, in simple terms the properties of a Hilbert Space.

Then you have observables, which are represented in the formalism as self-adjoing operators $\hat{A}$ with a domain (which by definition equals their co-domain), which is in general only a dense subspace of Hilbert space. A (perhaps generalized) eigenvector of $\hat{A}$ with eigenvalue $a \in \mathbb{R}$ is denoted as $|a \rangle$. It obeys by definition
$$\hat{A} |a \rangle=a |a \rangle.$$

The observables are represented by operators which are square matrices. The eigenvalues of the matrices(observables) are the allowed values for the observable. The associated eigenvector is the quantum state associated with that eigenvalue.

To define a representation you choose an arbitrary complete set of commuting self-adjoint operators (also called a complete compatible set of observables), $\hat{A}_1,\ldots, \hat{A}_N$. There are common eigenvectors of such sets of operators and complete means that giving the eigenvalues $(a_1,\ldots,a_N)$, the eigenvector is (up to a non-zero factor) uniquely defined (in other words the common eigenspaces to given $(a_1,\ldots,a_N)$ eigenvalues are one-dimensional).

This is where you choose the basis isn't it?

These are also complete sets of orthonormal eigenvectors (with the proper normalization). Matrix mechanics now is defined in the case of a completely discrete set of eigenvalues, and then all the eigenvectors are proper normalizable eigenvectors, i.e., you can choose them such that
$$\langle a_1,\ldots,a_N|a_1',\ldots,a_N' \rangle=\delta_{a_1,a_1'} \cdots \delta_{a_N,a_N'}$$

The eigenvectors are orthogonal.

$$\sum_{a_1,\ldots,a_N} |a_1,\ldots a_N \rangle \langle a_1,\ldots a_N|=\hat{1}.$$

and normalized.

This means you can uniquely describe any state by it's components
$$\psi_{a_1,\ldots,a_N} = \langle a_1,\ldots,a_N|\psi \rangle$$

We can express any state as a superposition of the eigenvectors of the observables.

and any operator by its matrix elements
$$A_{a_1,\ldots,a_N;a_1',\ldots,a_N'} = \langle a_1,\ldots,a_N|\hat{A}|a_1',\ldots,a_N' \rangle.$$

Eureka!
This is where the matrix elements are calculated! The matrix elements are obtained from the eigenvectors that were defined by the observable operators.

It is then easy to show by using the completeness relation several times that the components of $|phi \rangle=\hat{A} |\psi \rangle$ are given by
$$\phi_{a_1,\ldots,a_N}= \sum_{a_1',\ldots,a_N'} A_{a_1,\ldots,a_N;a_1',\ldots, a_N'} \psi_{a_1',\ldots,a_N'}.$$
This is like matrix-vector multiplication, and the composition of two operators maps to matrix products, and so on. That's why it's called "matrix mechanics".

So for your energy-eigenvalue problem for a particle in a rigid box to define it in terms of matrix mechanics you first have to find an appropriate basis, in which you work, and then you should be able to calculate the Hamiltonian's matrix elements with respect to this basis. This is not so easy.

I think I know what basis means, but I am not sure I know what appropriate basis means.

In fact the only "natural" basis that comes to my mind are indeed the energy eigenstates themselves, and to find them here the representation in the position representation is good.

I do not understand what you are saying here.

Here the basis are the generalized position eigenvectors $|x \rangle$ (I consider the 1D case for simplicity). These are living in the dual of the domain of the position operator and $x$ can take all real values (or here in this pretty artificial problem restricted to the interval $[0,L]$). Then the "orthonormalization and completeness relations" have to be generalized to
$$\langle x|x' \rangle=\delta(x-x'), \quad \int_0^L \mathrm{d} x |x \rangle \langle x|=\hat{1},$$

You are saying that basis for this problem appears to be the position eigenvectors, which are delta functions. I am not sure why you say they are living in the dual of the domain of the position operator.

and you have to use that in the position representation (i.e., matrix mechanics) the Hamiltonian is given as the differential operator
$$\hat{H}=-\frac{\hbar}{2m} \partial_x^2.$$
It's living on a proper subset of the square-integrable functions, which represent the states $\psi(x)=\langle x|\psi \rangle$. The Hilbert space is further specified (and that's what you can do only in the position representation) by $\psi(0)=\psi(L)=0$. With this you can solve the eigenvalue problem for $\hat{H}$, and then use the eigenvectors of $\hat{H}$ for other matrix-mechanics calculations.

I think that what you are saying is that we have to use the position basis to solve this problem but the position basis is not easy to work with because there are an infinite number of eigenvalues and the eigenvectors are delta functions. But we should be able to calculate the elements of $\hat{A}$ using the formula from above.

 

[mike1000]

Here's an R-Code that forms the Hamiltonian matrix for the discretized square well problem and solves the eigenstate n=3 (the ground state is n=1), plotting the probability density as an output:

L <- 1                                               # Length of the domain
N <- 100                            # Number of discrete points
dx <- L/N
A = matrix(nrow = N, ncol = N)                # Hamiltonian matrix
n_state <- 3                        # Number of the eigenstate to be calculated

for(m in c(1:N))                    # Fill the Hamiltonian matrix with elements appropriate for an infinite square well problem
{
for(n in c(1:N))
{
A[m,n]=0
if(m == n) A[m,n] = 2/dx^2
if(m == n-1 || m == n+1) A[m,n]=-1/dx^2
}
}

v = eigen(A)                       # Solve the eigensystem
vec = v$vectors

soln = c(1:N)
xaxis = c(1:N)*L/N

for(m in c(1:N))
{
soln[m] = vec[m,n_state]               # Fill the vector "soln" with the wavefunction values
}

jpeg(file = "plot.jpg")                   # Plot the probability density
plot(xaxis,abs(soln)^2)
lines(xaxis,abs(soln)^2)
dev.off()

The plot that this code produces looks just like you'd expect from a square of a sine function.

Thank you for posting the link to this article. http://www.dartmouth.edu/~pawan/final%20project.pdf

The thing that jumps out at me from both the paper and the code you provided is that the solution is completely independent of the Energy! From the code it is obvious that the matrix is composed entirely of spatial coordinates (because the potential, V(x), is zero). The code makes it very clear that the eigenvalues are the eigenvalues of the operator, which, in this case, is simply the second derivative operator.

Having said that, one thing I noticed in the code was there is no dependence, at all, on the Energy. There should have been an energy term on the diagonal of the matrix but there is none. So I went back to look at the paper and I found the following quote. from page 5 of the paper

Note that E is simply a scalar, and equation (4) may be alternatively written as
$$H_E~u = 0 $$where $H_E$is simply H with E subtracted from its diagonal entries.

What justification is there for removing E from the diagonal entries? I do not understand that. What do the eigenvalues represent if we subtracted E from the diagonal elements?

 

[hilbert2]

In that code I have made the problem nondimensionalized by setting the particle mass to value $m=1/2$ and Planck's constant to value $\hbar = 1$. This just defines the system of units that is used for energy, length and so on.

The eigenvalues (energies) of the discretized H operator are stored in the vector "v$values" after the "v = eigen(A)" command but are not printed by the code above.

 

[mike1000]

In that code I have made the problem nondimensionalized by setting the particle mass to value $m=1/2$ and Planck's constant to value $\hbar = 1$. This just defines the system of units that is used for energy, length and so on.

The eigenvalues (energies) of the discretized H operator are stored in the vector "v$values" after the "v = eigen(A)" command but are not printed by the code above.

I do not see that in the code. There are no values defined for the Energy in the code(unless I missed it). Your matrix A does not reference Energy anywhere!

 

[hilbert2]

You don't know the energies before you've solved the eigenvalue problem $\mathbf{H}\psi = E\psi$. The solution consists of a set of eigenvectors and the set of eigenvalues that correspond to the vectors.

The discretized system gives you an approximation for the N lowest energy eigenstates and eigenvalues, where N is the number of discrete points.

 

[PeterDonis]

I suppose you are trying to say that the rows represent positions in some way.

I am saying that there is one row for each possible position. More precisely, a "state vector" in this formalism is an infinite column vector, with one entry for each possible position (giving the amplitude to be at that position). An operator in this formalism is then an infinite by infinite matrix that multiplies an infinite column vector to give another infinite column vector.

 

[mike1000]

You don't know the energies before you've solved the eigenvalue problem $\mathbf{H}\psi = E\psi$. The solution consists of a set of eigenvectors and the set of eigenvalues that correspond to the vectors.

The discretized system gives you an approximation for the N lowest energy eigenstates and eigenvalues, where N is the number of discrete points.

But we have not solved the eigenvalue problem have we? The energies were subtracted from the diagonal before the eigenvalues and eigenvectors were calculated.

What do the eigenvalues and eigenvectors really represent? And how do we get the energies?

 

[hilbert2]

If the domain $x \in [0,L]$ consisted of only six points with spacing $\Delta x$, the matrix-vector equation would be

$\frac{1}{(\Delta x)^2}\begin{bmatrix} 2 & -1 & 0 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0 & 0\\ 0 & -1 & 2 & -1 & 0 & 0\\ 0 & 0 & -1 & 2 & -1 & 0\\ 0 & 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & 0 & -1 & 2 \end{bmatrix}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix} = \frac{2mE}{\hbar^2}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix}$

and here the numbers $\psi_n$ correspond to the wavefunction values $\psi (n \Delta x)$. The values -1 and 2 in the matrix come from the coefficients of $\psi_{n-1}$, $\psi_n$ and $\psi_{n+1}$ in the discretized second derivative operator.

 

[mike1000]

If the domain $x \in [0,L]$ consisted of only six points with spacing $\Delta x$, the matrix-vector equation would be
$\frac{1}{(\Delta x)^2}\begin{bmatrix} 2 & -1 & 0 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0 & 0\\ 0 & -1 & 2 & -1 & 0 & 0\\ 0 & 0 & -1 & 2 & -1 & 0\\ 0 & 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & 0 & -1 & 2 \end{bmatrix}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix} = \frac{2mE}{\hbar^2}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix}$

and here the numbers $\psi_n$ correspond to the wavefunction values $\psi (n \Delta x)$. The values -1 and 2 in the matrix come from the coefficients of $\psi_{n-1}$, $\psi_n$ and $\psi_{n+1}$ in the discretized second derivative operator.

Yes but that is not the eigenvalue matrix. You have to subtract the vector on the right hand side from the diagonal and set the resulting matrix equal to zero. The paper recognizes this, but then they say that they can subtract the Energy(E) from the diagonal entry because it is just a scalar. That is what I do not understand.

Isnt this the matrix for which we need the eigenvalues?
$\def \X{\frac{2mE}{\hbar^2}}$
$\frac{1}{(\Delta x)^2}\begin{bmatrix} 2 - \X& -1 & 0 & 0 & 0 & 0\\ -1 & 2-\X & -1 & 0 & 0 & 0\\ 0 & -1 & 2-\X & -1 & 0 & 0\\ 0 & 0 & -1 & 2-\X & -1 & 0\\ 0 & 0 & 0 & -1 & 2-\X & -1\\ 0 & 0 & 0 & 0 & -1 & 2-\X \end{bmatrix}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix} = 0$

 

[mike1000]

Yes but that is not the eigenvalue matrix. You have to subtract the vector on the right hand side from the diagonal and set the resulting matrix equal to zero. The paper recognizes this, but then they say that they can subtract the Energy(E) from the diagonal entry because it is just a scalar. That is what I do not understand.

Isnt this the matrix for which we need the eigenvalues?
$\def \X{\frac{2mE}{\hbar^2}}$
$\frac{1}{(\Delta x)^2}\begin{bmatrix} 2 - \X& -1 & 0 & 0 & 0 & 0\\ -1 & 2-\X & -1 & 0 & 0 & 0\\ 0 & -1 & 2-\X & -1 & 0 & 0\\ 0 & 0 & -1 & 2-\X & -1 & 0\\ 0 & 0 & 0 & -1 & 2-\X & -1\\ 0 & 0 & 0 & 0 & -1 & 2-\X \end{bmatrix}\begin{bmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\\ \psi_5\\ \psi_6 \end{bmatrix} = 0$

I made a big mistake. And it took your analysis to help me see it. The matrix that we use to determing the eigenvalues is not what I wrote above. It is your matrix. We calculate the determinant of the following matrix and get a characteristic polynomial for $\lambda$. The roots of that polynomial are all the eigenvalues of the matrix.
$\def \X{\lambda}$
$\frac{1}{(\Delta x)^2}\begin{vmatrix} 2 - \X& -1 & 0 & 0 & 0 & 0\\ -1 & 2-\X & -1 & 0 & 0 & 0\\ 0 & -1 & 2-\X & -1 & 0 & 0\\ 0 & 0 & -1 & 2-\X & -1 & 0\\ 0 & 0 & 0 & -1 & 2-\X & -1\\ 0 & 0 & 0 & 0 & -1 & 2-\X \end{vmatrix} = 0$
$\def \X{\frac{2mE}{\hbar^2}}$
Guess what? This has been my problem all along. I forgot how to calculate ALL eigenvalues of a matrix. We do not need to know the values of the energy( $\X$) to calculate ALL the possible values for the Energy( $\X$) . So you are right. The eigenvalues of the following matrix are what we need to know, which is/are the eigenvalues of the gradient operator. I am sorry I messed up your posts.

$\frac{1}{(\Delta x)^2}\begin{bmatrix} 2 & -1 & 0 & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0 & 0\\ 0 & -1 & 2 & -1 & 0 & 0\\ 0 & 0 & -1 & 2 & -1 & 0\\ 0 & 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & 0 & -1 & 2 \end{bmatrix}$

We do not need to know specific eigenvalues of the operator in order to calculate ALL the eigenvalues of that operator. (Everything is falling into place now until someone tells me I still got it wrong! )

 

[DrClaude]

The thing that jumps out at me from both the paper and the code you provided is that the solution is completely independent of the Energy! From the code it is obvious that the matrix is composed entirely of spatial coordinates (because the potential, V(x), is zero). The code makes it very clear that the eigenvalues are the eigenvalues of the operator, which, in this case, is simply the second derivative operator.

Note that this is based on an approximation, and doesn't correspond exactly to what @vanhees71 was describing. The wave function is considered only at a discrete number of points, so $x$ is no longer continuous. This correspond to a particular choice of an approximate basis set of vectors- This approach is also known as the Discrete Variable Representation.

Having said that, one thing I noticed in the code was there is no dependence, at all, on the Energy. There should have been an energy term on the diagonal of the matrix but there is none. So I went back to look at the paper and I found the following quote. from page 5 of the paper

The potential of the square well is uniquely 0 on the finite spatial grid. Otherwise, a term $V(x)$ would appear on the diagonal.

What justification is there for removing E from the diagonal entries? I do not understand that. What do the eigenvalues represent if we subtracted E from the diagonal elements?

It is simply
$$
\mathbf{H} \mathbf{u} = E \mathbf{u} = E \mathbf{I} \mathbf{u} \\
\mathbf{H} \mathbf{u} - E \mathbf{I} \mathbf{u} = 0 \\
\left( \mathbf{H} - E \mathbf{I} \right) \mathbf{u} = 0 \\
\mathbf{H}_E \mathbf{u} = 0
$$
where $\mathbf{I}$ is the indentity matrix and $\mathbf{H}_E \equiv \mathbf{H} - E \mathbf{I}$. This requires the eigenenergies $E$ to be known.

 

[vanhees71]

The observables are represented by operators which are square matrices. The eigenvalues of the matrices(observables) are the allowed values for the observable. The associated eigenvector is the quantum state associated with that eigenvalue.

The observables are represented by self-adjoint operators. These act on abstract Hilbert space as linear operators with some additional properties, i.e., for any two Hilbert-space vectors in the domain of the operator you have
$$\langle \phi|\hat{A} \psi \rangle=\langle \hat{A} \phi|\psi \rangle,$$
which means the operator is Hermitean and its domain $D$ (i.e., the subspace where the operator is well defined) is the same as its codomain (i.e., it maps any vector in $D$ linearly to a vector in $D$).

I think I know what basis means, but I am not sure I know what appropriate basis means.

Of course, in principle you can choose any basis you want, but what I mean by "appropriate basis" is that you want a basis, where you can evaluate the matrix elements of the operators you like to solve the eigenvalue problem for in practice.

You are saying that basis for this problem appears to be the position eigenvectors, which are delta functions. I am not sure why you say they are living in the dual of the domain of the position operator.

The dual of a vector space is the space of linear functionals. For a Hilbert space there's a one-to-one mapping between the linear functionals and vectors, because one an show that any linear functional $L: H \rightarrow \mathbb{C}$ on a Hilbert space $H$ uniquely defines a vector $|L \rangle \in H$ such that $L(\psi)=\langle L|\psi \rangle$. In this sense you can identify the dual of $H$ with $H$ itself, i.e., you have $H^*=H$. For the position operator the domain is a smaller but dense subspace of $H$, and thus the Dual is "larger" than $H$, i.e., it contains distributions (in the sense of generalized functions). E.g. the momentum operator $\hat{p}$ in the position representation is $\hat{p}=-\mathrm{i} \hbar \partial_x$ and thus the eigenvectors are plane waves in this representation, i.e.,
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp \left (\frac{\mathrm{i} x p}{\hbar} \right ).$$
It's obviously not square integrable but you have
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{2 \pi \hbar}\exp \left (\frac{\mathrm{i} x (p'-p)}{\hbar} \right )=\delta(p-p'),$$
which clearly shows that it's to be understood as a distribution (defined on the domain of $\hat{p}$ and $\hat{x}$ as "test functions").

I think that what you are saying is that we have to use the position basis to solve this problem but the position basis is not easy to work with because there are an infinite number of eigenvalues and the eigenvectors are delta functions. But we should be able to calculate the elements of $\hat{A}$ using the formula from above.

In this case the position basis is in fact very easy to work with, because you just solve a simple boundary value differential-equation problem to get the energy eigenvalues and eigenstates.

For a gentle introduction on the modern formulation of QT in terms of the "rigged Hilbert space" formalism, see the excellent textbook by Ballentine, Quantum Mechanics, Addison Wesley.

 

[hilbert2]

The potential of the square well is uniquely 0 on the finite spatial grid. Otherwise, a term $V(x)$ would appear on the diagonal.

Yes, if for instance we have a harmonic oscillator potential $V(x) \propto (x-x_0 )^2$, the discretized potential $V_n = V(n\Delta x)$ appears on the diagonal elements. My code that calculates the three lowest energy states of this system is below:

L <- 6.0                             # Length of the domain
N <- 150                    # Number of discrete points
dx <- L/N
A = matrix(nrow = N, ncol = N)        # Hamiltonian matrix
V = c(1:N)

for(m in c(1:N))
{
V[m] = 3.0*(m*dx - 3.0)*(m*dx - 3.0)       # define a harmonic oscillator potential with spring constant k = 6
}

for(m in c(1:N))             # Fill the Hamiltonian matrix with elements appropriate for a harmonic oscillator system
{
for(n in c(1:N))
{
A[m,n]=0
if(m == n) A[m,n] = 2/dx^2 - V[m]
if(m == n-1 || m == n+1) A[m,n]=-1/dx^2
}
}

v = eigen(A)                       # Solve the eigensystem
vec = v$vectors

psi1 = c(1:N)
psi2 = c(1:N)
psi3 = c(1:N)
xaxis = c(1:N)*L/N

for(m in c(1:N))
{
psi1[m] = vec[m,1]               # Fill the psi-vectors with the eigenfunction values
psi2[m] = vec[m,2]
psi3[m] = vec[m,3]
}

jpeg(file = "plot.jpg")             # Plot the probability densities for the ground state and two excited states above it
plot(xaxis, 0.01*V, ylim=c(0,0.04))
lines(xaxis, 0.01*V)
lines(xaxis,abs(psi1)^2)
lines(xaxis,abs(psi2)^2)
lines(xaxis,abs(psi3)^2)
dev.off()

A plot of V(x) and the approximate probability densities for quantum numbers n=0, n=1 and n=2 looks like this:

 

[vanhees71]

One should emphasize that these nice numerical calculations are not what's known as "matrix mechanics". It's solving an approximate eigenvalue problem by discretizing space. It's a kind of "lattice calculation" for quantum theory.

 

[mike1000]

One should emphasize that these nice numerical calculations are not what's known as "matrix mechanics". It's solving an approximate eigenvalue problem by discretizing space. It's a kind of "lattice calculation" for quantum theory.

If the discretization was fine enough, such that the numerical solution approached the analytic solution what would you say about the finite difference matrix? If the matrix used in the finite difference solution gave the same eigenvalues and the same eigenvectors as the analytic solution wouldn't the finite difference matrix equal the unknown, operator matrix?

I guess what I am asking if two matrices have the same eigenvalues and the same eigenvectors are they equivalent?

 

[hilbert2]

I'm not sure if someone has actually published a proof that the ground state of the discretized system approaches the exact gaussian harmonic oscillator ground state when $L \rightarrow \infty$ and $\Delta x \rightarrow 0$.

 

[hilbert2]

I guess what I am asking if two matrices have the same eigenvalues and the same eigenvectors are they equivalent?

If they are hermitian and have same eigenvalues and eigenvectors, they have to be the same matrix as far as I know. If they only have the same eigenvalues and same dimension and are hermitian, then they can differ by a unitary transformation.