Discretizing a 1D quantum harmonic oscillator, finding eigenvalues

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  • #1
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Homework Statement:
Consider the harmonic oscillator defined by the hamiltonian
$$\hat{H} = \frac{\hat{p}^{2}}{2} + \frac{\omega^{2}}{2}\hat{x}^2$$

Compute the first ##n## eigenvalues ##E_n##, and eigenvectors ##\ket{\psi_n}##
Relevant Equations:
The harmonic oscillator is governed by the time-independent SE:
$$\hat{H}\ket{\psi_n} = E_n\ket{\psi_n}$$.

Written explicitly (with ##\hbar = 1##):

$$ \frac{\psi_n^{''}}{2} + \frac{\omega^{2}}{2}x^2\psi_n = E_n\psi_n $$
##x## can be discretized as ##x \rightarrow x_k ## such that ##x_{k + 1} = x_k + dx## with a positive integer ##k##. Throughout we may assume that ##dx## is finite, albeit tiny.

By applying the Taylor expansion of the wavefunction ##\psi_n(x_{k+1})## and ##\psi_n(x_{k-1})##, we can quickly approximate ##\psi_n(x_k)^{''}## as:

$$\psi_n(x_k)^{''} = \frac{\psi_n(x_{k + 1}) - 2\psi_n(x_k) + \psi_n(x_{k-1})}{dx^2}$$

Which in turn allows us to write the TISE as

$$ \frac{\psi_n(x_{k + 1}) - 2\psi_n(x_k) + \psi_n(x_{k-1})}{2dx^2} + \frac{\omega^{2}}{2}x^2\psi_n(x_k) = E_n\psi_n(x_k)$$

With this, we are capable of writing a tri-diagonal Hamiltonian, with diagonal terms

$$H_{ii} = \frac{1}{dx^2} + \omega^{2}x_{i}^2$$

with ##i = 1, 2, ... , k##
with the 2nd (upper, lower) diagonal terms all being ## 1/(2dx^2) ##.

This entire derivation is fine. However, I am having trouble interpreting it.

The diagonalization of this Hamiltonian gives me ##k## eigenvalues which do not correspond to the energy eigenvalues ##E_n##. That is because the equation written is for a single state ##n##. So what do these ##k## eigenvalues correspond to?

At this point, I imagine that a system with ##n## states discretized like this would have a block-diagonal Hamiltonian, with each entry corresponding to the tri-diagonal matrix I wrote above, one for each state of energy. But this sounds incorrect and I have no idea how to put that in coherent mathematical terms.

To find the first ##n## states, there should be ##kn## eigenvalues, because each state will have ##k## of those? And if that is the case, what is ##E_n## in terms of these ##k## eigenvalues?
 
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Answers and Replies

  • #2
DrClaude
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The diagonalization of this Hamiltonian gives me ##k## eigenvalues which do not correspond to the energy eigenvalues ##E_n##.
They should be the same.

Note that ##n## doesn't enter anywhere in the Hamiltonian matrix, so it is incorrect to say that the equation is written for a single state ##n##.
 
  • #3
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They should be the same.

Note that ##n## doesn't enter anywhere in the Hamiltonian matrix, so it is incorrect to say that the equation is written for a single state ##n##.
Thanks for the response.

I see what you're saying. And I suppose that we did not change the Hamiltonian anywhere, so throughout the algebra naturally the system did not become different. Meaning that I am expecting the results for the usual oscillator.

That said, I think the discretization is throwing me off. Why is it that the ##k## eigenvalues correspond to the ##n## energy states?

More specifically: We know the energy eigenstates are orthogonal. It seems to me that what I am doing is computing the energy eigenstates for each position ##x_k##. If that is the case, then this means that the set of positions ##\ket{x_k}## I am considering forms an orthogonal basis which I can use to express the matrix of the Hamiltonian (,##\bra{x_i}\hat{H}\ket{x_j}##). I am struggling to come to terms with this, because why do the eigenvalues corresponding to the different positions also describe the different energy states?
 
  • #4
DrClaude
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That said, I think the discretization is throwing me off. Why is it that the ##k## eigenvalues correspond to the ##n## energy states?
Because you are solving for the eigenvectors of the matrix H. That matrix represents the (discretized) Hamiltonian of the harmonic oscillator, hence the eigenvectors will be the eigenstates of the Hamiltonian.


More specifically: We know the energy eigenstates are orthogonal. It seems to me that what I am doing is computing the energy eigenstates for each position ##x_k##. If that is the case, then this means that the set of positions ##\ket{x_k}## I am considering forms an orthogonal basis which I can use to express the matrix of the Hamiltonian (,##\bra{x_i}\hat{H}\ket{x_j}##). I am struggling to come to terms with this, because why do the eigenvalues corresponding to the different positions also describe the different energy states?
There are two things going on here. The first is the discretization of the h.o. Hamiltonian over a finite range. This represents of course an approximation, as the Hamiltonian would normally have an infinite number of eigenstates. Also, the results will be discretized approximations to the actual eigenfunctions.

Within that discrete approximation, you have a set of basis kets, ##\{ \ket{x_i} \}##, which are eigenstates of the ##\hat x## operator. The matrix representation of the Hamiltonian is then, as you mention,
$$
H_{ij} = \braket{x_i | \hat{H} | x_j}
$$

Any wave function ##\psi## is then represented as a column vector, the elements of which are
$$
\begin{pmatrix}
\braket{ x_0 | \psi} \\ \braket{ x_1 | \psi} \\ \vdots \\ \braket{ x_n | \psi}
\end{pmatrix}
$$

When you solve the eigenvalue problem for the matrix H, what you get are eigenvalues corresponding to the energies ##E_k##, with eigenvectors containing the components as above, a discretized approximation of the eigenfunctions ##\psi_k##.
 
  • #5
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Because you are solving for the eigenvectors of the matrix H. That matrix represents the (discretized) Hamiltonian of the harmonic oscillator, hence the eigenvectors will be the eigenstates of the Hamiltonian.

Of course, this makes a lot of sense. It's been a minute since I've done linear algebra for real, so I forget the eigenvectors pertain to how the matrix is structured.

Any wave function ψ is then represented as a column vector, the elements of which are
(⟨x0|ψ⟩⟨x1|ψ⟩⋮⟨xn|ψ⟩)

And this was the one thing I was missing. For some reason I was thinking each energy state had a different wave function, and not of different projections of the wave function itself. (Is this correct?)

Anyway, thank you so much for your answers, they really helped me understand how to get this done!
 
  • #6
DrClaude
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And this was the one thing I was missing. For some reason I was thinking each energy state had a different wave function, and not of different projections of the wave function itself. (Is this correct?)
Each energy state has its own wave function. The vector I wrote is the generic case, replace ##\psi## with the specific solution for a given energy eigenstate.
 

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