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

[quarky2001]

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:

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

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.

 

[vela]

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).

 

[quarky2001]

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

 

[vela]

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.

 

[ManyNames]

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:

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

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-dependent evolutions of the Schrödinger 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-dependent, 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:

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

 

[ManyNames]

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...

 

[quarky2001]

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

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-dependent evolutions of the Schrödinger 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-dependent, 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!