How do I find the time evolution of a state with an operator in the Hamiltonian?

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Discussion Overview

The discussion revolves around finding the time evolution of a quantum state given a Hamiltonian operator, specifically in the context of the quantum harmonic oscillator. Participants explore various methods, including the use of Fourier transforms and ladder operators, while addressing the complexities introduced by the Hamiltonian's structure.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using the expression \(\Psi(x, t) = \exp(-i H t/\hbar)\Phi(x)\) to find the time evolution of the state, but expresses uncertainty about how to handle the operator in the Hamiltonian.
  • Another participant suggests that the Hamiltonian in the exponential should represent the energy \(E\) of the eigenstate, questioning whether the initial state is indeed an eigenstate and emphasizing the need for normalization.
  • A different participant introduces the idea of using Fourier transformation to derive the time-dependent state, but raises a question about how to express energy \(E\) in terms of wave number \(k\) when the Hamiltonian includes both kinetic and potential energy terms.
  • Some participants argue against the necessity of Fourier transforms, suggesting that the problem can be approached directly using the Schrödinger equation and ladder operators, while noting the energies of the quantum harmonic oscillator.
  • There is a discussion about the coupled harmonic oscillator, with one participant questioning the presence of multiple momentum and position terms, seeking clarification on the system's configuration.
  • One participant admits a lack of knowledge regarding two-particle systems, indicating a limitation in their ability to contribute further to that aspect of the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and appropriateness of using Fourier transforms versus direct methods involving the Schrödinger equation. There is no consensus on the best approach to take, and several questions remain unresolved regarding the specifics of the Hamiltonian and the nature of the initial state.

Contextual Notes

Participants note potential missing coefficients in the wave function and the need for normalization, indicating that assumptions about the state may not be fully addressed. The discussion also highlights the complexity of dealing with multi-particle systems without fully resolving the implications of the Hamiltonian's structure.

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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!?
 
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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.
 
Matterwave said:
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?
 
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?
 
Matterwave said:
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.
 
Hmmm, sorry but I can't help you there. I haven't studied two particle systems yet.
 

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