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

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  • #1
quarky2001
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Find the time dependance of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by [tex] V=\frac{1}{2}kx^2 [/tex]



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



When I apply the position operator, I get:

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

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. [itex] x=x_0 cos(2\pi\nu t+\phi) [/itex]Some help would be much appreciated.
 
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  • #2
vela
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Find the time dependence of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by

[tex] V=\frac{1}{2}kx^2 [/tex]

I'm assuming some wavefunction of the form

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

When I apply the position operator, I get:

[tex] \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[/tex]

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. [itex] x=x_0 \cos(2\pi\nu t+\phi) [/itex]. 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).
 
  • #3
quarky2001
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Ah, thank you vela. I must be too used to latex!
 
  • #4
vela
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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.
 
  • #5
ManyNames
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Find the time dependance of the expectation value <x> in a quantum harmonic oscillator, where the potential is given by [tex] V=\frac{1}{2}kx^2 [/tex]



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



When I apply the position operator, I get:

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

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. [itex] x=x_0 cos(2\pi\nu t+\phi) [/itex]Some help would be much appreciated.


[itex]\hat{H}|\psi>=i\hbar \partial_{t}|\psi>[/itex]

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:

[itex]\hat\{H}|\psi_n>=E_n|\phi_n>[/itex]

and the state [itex]|\phi_n>[/itex] evolves as:

[itex]|\phi_n(t)>=e^{-i \omega_{n}t|\phi_n>[/itex]

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

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

Thus the energy of a Hamiltonian can be expressed further:

[itex]E_n=<\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}>[/itex]
 
  • #6
ManyNames
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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...
 
  • #7
quarky2001
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[itex]\hat{H}|\psi>=i\hbar \partial_{t}|\psi>[/itex]

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:

[itex]\hat\{H}|\psi_n>=E_n|\phi_n>[/itex]

and the state [itex]|\phi_n>[/itex] evolves as:

[itex]|\phi_n(t)>=e^{-i \omega_{n}t|\phi_n>[/itex]

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

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

Thus the energy of a Hamiltonian can be expressed further:

[itex]E_n=<\psi^{0}_{n}|\hat{H}|\psi^{0}_{t}>[/itex]

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

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