# Time evolution of state

1. Apr 13, 2009

### KFC

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

$$H = p^2/2m + x^2/2$$

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

$$\Psi(x, t) = \exp(-i H t/\hbar)\Phi(x)$$

However, since $$H$$ 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 $$\Phi(x)$$ term by term? But in this way, it seems not easy to combine the result to get the close form!?

2. Apr 13, 2009

### Matterwave

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. Apr 13, 2009

### KFC

Thank you. I read a text in which the author use the fourier transformation to carry out the time-dependent state.

$$\Phi(k) = \frac{1}{\sqrt{2\pi}}\int dx \exp(-ikx) \Phi(x)$$

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

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

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

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

$$H = p_1^2/2m + x_1^2/2 + p_2^2/2m + x_2^2/2$$ 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. Apr 14, 2009

### Matterwave

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:

$$E=(n+\frac{1}{2})\hbar\omega$$

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 $$\omega = \sqrt{\frac{k}{m}}$$

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 $$p_2$$ and $$x_2$$ term? Are there 2 particles in there?

5. Apr 14, 2009

### KFC

Yes. There are two particles.

6. Apr 14, 2009

### Matterwave

Hmmm, sorry but I can't help you there. I haven't studied two particle systems yet.