# Time-dependence of expectation value <x> in a quantum harmonic oscillator?

Find the time dependance of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by $$V=\frac{1}{2}kx^2$$

I'm assuming some wavefunction of the form $\Psi(x,t)=\psi(x) e^{-iEt/\hbar}$

When I apply the position operator, I get:

$$<x>=\int_{-\infty}^\infty {\psi_0}^2 (x) e^{\-iEt/\hbar}e^{iEt/\hbar}dx$$
$$<x>=\int_{-\infty}^\infty {\psi_0}^2 (x)$$

which is time-independent, and also wrong...

I think it's supposed to be an easy question, and I'm supposed to get the classical result, i.e. $x=x_0 cos(2\pi\nu t+\phi)$Some help would be much appreciated.

Last edited:

vela
Staff Emeritus
Homework Helper
Find the time dependence of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by

$$V=\frac{1}{2}kx^2$$

I'm assuming some wavefunction of the form

$$\Psi(x,t)=\psi(x) e^{-iEt/\hbar}$$

When I apply the position operator, I get:

$$\langle x\rangle=\int_{-\infty}^\infty x{\psi_0}^2 e^{-iEt/\hbar}e^{iEt/\hbar}dx = \int_{-\infty}^\infty x{\psi_0}^2 dx$$

which is time-independent, and also wrong...

I think it's supposed to be an easy question, and I'm supposed to get the classical result, i.e. $x=x_0 \cos(2\pi\nu t+\phi)$. Some help would be much appreciated.

PS - why doesn't my latex appear typeset?
You have TeX on the brain. ;) The closing tag uses a forward slash: [/tex]

P.S. I reformatted your post slightly, fixed a few typos, and corrected the argument of the cosine (2 pi should be in the numerator, not the denominator).

Ah, thank you vela. I must be too used to latex!

vela
Staff Emeritus
Homework Helper
You got the right answer for the wavefunction you chose. That wavefunction is an eigenstate of the Hamiltonian, so the expectation value of an observable doesn't change with time.

Find the time dependance of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by $$V=\frac{1}{2}kx^2$$

I'm assuming some wavefunction of the form $\Psi(x,t)=\psi(x) e^{-iEt/\hbar}$

When I apply the position operator, I get:

$$<x>=\int_{-\infty}^\infty {\psi_0}^2 (x) e^{\-iEt/\hbar}e^{iEt/\hbar}dx$$
$$<x>=\int_{-\infty}^\infty {\psi_0}^2 (x)$$

which is time-independent, and also wrong...

I think it's supposed to be an easy question, and I'm supposed to get the classical result, i.e. $x=x_0 cos(2\pi\nu t+\phi)$Some help would be much appreciated.

$\hat{H}|\psi>=i\hbar \partial_{t}|\psi>$

In the mathematics of quantum mechanics, the Hamiltonian operator is self-adjoint so it's diagonalisable and all its eigenvalues are real. There is always atleast one family of orthogonal states |\phi_n> that span the state space:

$\hat\{H}|\psi_n>=E_n|\phi_n>$

and the state $|\phi_n>$ evolves as:

$|\phi_n(t)>=e^{-i \omega_{n}t|\phi_n>$

These are called time-dependant evolutions of the schrodinger equation which describes the evolution of a system. If we have an expectation value $<x>$ then it means the expectation of energy, which is equivalent to the expression $V=\frac{1}{2}kx^2$. But these are time-dependant, and the Hamiltonian is given as

$\hat{H}|\psi(t)>=\frac{p^2}{2m} |\psi(t)>=i \hbar \frac{d}{dt} |\psi(t)>$

Thus the energy of a Hamiltonian can be expressed further:

$E_n=<\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}>$

By the way, what where the equations you where given? I know of the ones you gave, but the solutions you tried to arise at did not seem familiar...

$\hat{H}|\psi>=i\hbar \partial_{t}|\psi>$

In the mathematics of quantum mechanics, the Hamiltonian operator is self-adjoint so it's diagonalisable and all its eigenvalues are real. There is always atleast one family of orthogonal states |\phi_n> that span the state space:

$\hat\{H}|\psi_n>=E_n|\phi_n>$

and the state $|\phi_n>$ evolves as:

$|\phi_n(t)>=e^{-i \omega_{n}t|\phi_n>$

These are called time-dependant evolutions of the schrodinger equation which describes the evolution of a system. If we have an expectation value $<x>$ then it means the expectation of energy, which is equivalent to the expression $V=\frac{1}{2}kx^2$. But these are time-dependant, and the Hamiltonian is given as

$\hat{H}|\psi(t)>=\frac{p^2}{2m} |\psi(t)>=i \hbar \frac{d}{dt} |\psi(t)>$

Thus the energy of a Hamiltonian can be expressed further:

$E_n=<\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}>$

Ah ha! Thank you. I'm kicking myself for not seeing that, but it seems obvious now!