- #1

JTFreitas

- 18

- 3

- 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?

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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