Time evolution of state

  • Thread starter KFC
  • Start date
  • #1
KFC
488
4

Main Question or Discussion Point

Suppose I know an initial state [tex]\Phi(x)= \exp(-x^2)[/tex], the Hamiltonian is

[tex]H = p^2/2m + x^2/2[/tex]

where p is the mometum operator. If I want to find the time evolution of the state [tex]\Phi(x)[/tex], should I write it as the following?

[tex]\Psi(x, t) = \exp(-i H t/\hbar)\Phi(x)[/tex]

However, since [tex]H[/tex] contains an operator, I don't know how to find the close form of the time-dependent state. Should I expand it as a series and then operate it on [tex]\Phi(x)[/tex] term by term? But in this way, it seems not easy to combine the result to get the close form!?
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
The H in the exponential should be E - as in the energy of that eigenstate. I think what you have there is an Eigenstate (it's been a while since I did Quantum harmonic oscillators, and the forms I solved for were somewhat different so I'm not sure), if it isn't then you can't do this. You need to normalize the wave function, in any case.
 
  • #3
KFC
488
4
The H in the exponential should be E - as in the energy of that eigenstate. I think what you have there is an Eigenstate (it's been a while since I did Quantum harmonic oscillators, and the forms I solved for were somewhat different so I'm not sure), if it isn't then you can't do this. You need to normalize the wave function, in any case.
Thank you. I read a text in which the author use the fourier transformation to carry out the time-dependent state.

[tex]\Phi(k) = \frac{1}{\sqrt{2\pi}}\int dx \exp(-ikx) \Phi(x)[/tex]

and then the time-dependent state is the inverse transformation of [tex]\exp(-iEt/\hbar)\Phi(k)[/tex]

[tex]\Psi(x, t) = \frac{1}{\sqrt{2\pi}}\int dk \exp(ikx)\Phi(k) \exp(-iEt/\hbar)[/tex]

For harmonic oscillator, if [tex]H = p^2/2m[/tex], then [tex]E=\hbar^2k^2/2m[/tex], plug this into the inverse fourier transformation will give the result directly. But if H also includes [tex]x^2/2[/tex], what does [tex]E[/tex] (in terms of k) look like?

By the way, if we consider two coupling hamonic oscillator such that

[tex]H = p_1^2/2m + x_1^2/2 + p_2^2/2m + x_2^2/2[/tex] and we apply the Fourier transformation again to find the time-dependent state, how to write the Fourier transformation? Should it be 2D fourier transformation?
 
  • #4
Matterwave
Science Advisor
Gold Member
3,965
326
Uhm...I don't think you need to worry about Fourier transforms for this problem (all you did was transform the function over to wave number space and then transform it back... tagging on the exponential term does nothing with that integral since it's over dk). The problem can be solved directly from the Schroedinger's equation using ladder operators or using Legendre polynomials.

The energies of the quantum harmonic oscillator is:

[tex]E=(n+\frac{1}{2})\hbar\omega[/tex]

I believe the wave function you provided corresponds to n=1 state...but there are some missing coefficients in front of the x squared term...so I'm not sure if it indeed is... You may want to check that.

Where [tex]\omega = \sqrt{\frac{k}{m}}[/tex]

k is the spring constant as usual, and in your case it is 1.

You can obtain this by using ladder operators. This problem has been completely solved so I don't think I need to redo everything here. You can find information about this problem in the wiki: http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

I don't understand your coupled oscillator, why is there a [tex]p_2[/tex] and [tex]x_2[/tex] term? Are there 2 particles in there?
 
  • #5
KFC
488
4
Uhm...I don't think you need to worry about Fourier transforms for this problem (all you did was transform the function over to wave number space and then transform it back... tagging on the exponential term does nothing with that integral since it's over dk). The problem can be solved directly from the Schroedinger's equation using ladder operators or using Legendre polynomials.

The energies of the quantum harmonic oscillator is:

[tex]E=(n+\frac{1}{2})\hbar\omega[/tex]

I believe the wave function you provided corresponds to n=1 state...but there are some missing coefficients in front of the x squared term...so I'm not sure if it indeed is... You may want to check that.

Where [tex]\omega = \sqrt{\frac{k}{m}}[/tex]

k is the spring constant as usual, and in your case it is 1.

You can obtain this by using ladder operators. This problem has been completely solved so I don't think I need to redo everything here. You can find information about this problem in the wiki: http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

I don't understand your coupled oscillator, why is there a [tex]p_2[/tex] and [tex]x_2[/tex] term? Are there 2 particles in there?
Yes. There are two particles.
 
  • #6
Matterwave
Science Advisor
Gold Member
3,965
326
Hmmm, sorry but I can't help you there. I haven't studied two particle systems yet.
 

Related Threads for: Time evolution of state

  • Last Post
Replies
1
Views
1K
Replies
2
Views
393
Replies
3
Views
811
Replies
18
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
1K
Replies
8
Views
1K
  • Last Post
Replies
1
Views
3K
Top