Exploring the Solution to a Harmonic Oscillator Problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
rhysticlight
Messages
3
Reaction score
0
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!
 
Physics news on Phys.org


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!