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:

&lt;x&gt;=\int_{-\infty}^\infty {\psi_0}^2 (x) e^{\-iEt/\hbar}e^{iEt/\hbar}dx
&lt;x&gt;=\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:

&lt;x&gt;=\int_{-\infty}^\infty {\psi_0}^2 (x) e^{\-iEt/\hbar}e^{iEt/\hbar}dx
&lt;x&gt;=\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&gt;=i\hbar \partial_{t}|\psi&gt;

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&gt;=E_n|\phi_n&gt;

and the state |\phi_n&gt; evolves as:

|\phi_n(t)&gt;=e^{-i \omega_{n}t|\phi_n&gt;

These are called time-dependant evolutions of the schrodinger equation which describes the evolution of a system. If we have an expectation value &lt;x&gt; 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)&gt;=\frac{p^2}{2m} |\psi(t)&gt;=i \hbar \frac{d}{dt} |\psi(t)&gt;

Thus the energy of a Hamiltonian can be expressed further:

E_n=&lt;\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}&gt;

 

[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&gt;=i\hbar \partial_{t}|\psi&gt;

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&gt;=E_n|\phi_n&gt;

and the state |\phi_n&gt; evolves as:

|\phi_n(t)&gt;=e^{-i \omega_{n}t|\phi_n&gt;

These are called time-dependant evolutions of the schrodinger equation which describes the evolution of a system. If we have an expectation value &lt;x&gt; 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)&gt;=\frac{p^2}{2m} |\psi(t)&gt;=i \hbar \frac{d}{dt} |\psi(t)&gt;

Thus the energy of a Hamiltonian can be expressed further:

E_n=&lt;\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}&gt;

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