Exploring the Solution to a Harmonic Oscillator Problem

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SUMMARY

The discussion centers on the harmonic oscillator problem involving a particle in a potential state represented as |psi(x,0)>=1/5 * [3|0> + 4|1>]. The user seeks clarification on the application of the raising operator a+ to the state |0>*exp(-i*E0*t/h), questioning why the energy term remains E0 instead of changing to E1. The response confirms that the energy terms E1 and E0 are constants, specifically multiples of \hbar\omega, and advises substituting these values before applying the ladder operators.

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  • Familiarity with quantum state notation and ladder operators (a+ and a-)
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rhysticlight
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I am not really asking how to solve the problem but just for explanation of what I know to be true from the problems solution. Basically the original problem statement is this:

A particle in a harmonic oscillator potential starts out in the state
|psi(x,0)>=1/5 * [3|0> + 4|1>] and it asks to find the expectation value of position <x>.

Now the way I approached the problem was to first find |psi(x,t)> by simply "tacking on" the time dependent exponential terms and then expressing x through the ladder operators a+ and a-.

What I am wondering is when I, for example, apply the raising operator a+ to the state |0>*exp(-i*E0*t/h) does the function become |1>*exp(-i*E0*t/h) rather than |1>*exp(-i*E1*t/h) (i.e. why does the energy term in the time dependent part not change?)

Thanks!
 
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rhysticlight said:
I am not really asking how to solve the problem but just for explanation of what I know to be true from the problems solution. Basically the original problem statement is this:

A particle in a harmonic oscillator potential starts out in the state
|psi(x,0)>=1/5 * [3|0> + 4|1>] and it asks to find the expectation value of position <x>.

Now the way I approached the problem was to first find |psi(x,t)> by simply "tacking on" the time dependent exponential terms and then expressing x through the ladder operators a+ and a-.

What I am wondering is when I, for example, apply the raising operator a+ to the state |0>*exp(-i*E0*t/h) does the function become |1>*exp(-i*E0*t/h) rather than |1>*exp(-i*E1*t/h) (i.e. why does the energy term in the time dependent part not change?)

Thanks!

The energy terms E1 and E0 are constants, namely multiples of [tex]\hbar[/tex][tex]\omega[/tex]. Put in these values, if you wish, before you operate with the ladder operators and see what happens.
 


Ah o.k. I see now, thanks!
 

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