Harmonic oscillator superposition amplitude evaluation

  • Thread starter Trance
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  • #1
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Hi all

Homework Statement


I have the first three states of the harmonic oscillator, and I need to know the amplitudes for the states after the potential is dropped.


Homework Equations


[itex]u_{0}=(\frac{1}{\pi a^{2}})^{\frac{1}{4}} e^{{\frac{-x^2}{2a^2}}}[/itex]

[itex]u_{1}=(\frac{4}{\pi})^{\frac{1}{4}} (\frac{1}{a^2})^\frac{3}{4} x e^{({\frac{-x^2}{2a^2}})}[/itex]

[itex]u_{2}=(\frac{1}{4\pi a^2})^{\frac{1}{4}} (\frac{2x^2}{a^2}-1) e^{{\frac{-x^2}{2a^2}}}[/itex]

Then, the potential is lowered suddenly, by setting [itex]\omega ' = \frac{\omega}{2}[/itex].

The system is then modelled with the expansion postulate, as a superposition of these three new states, with [itex]\omega'[/itex] in them.

The particle was originally in the ground state before the potential was lowered, so I have that:

[itex]C_{i}=\int u_{i}^* \psi dx[/itex]

Where [itex]\psi[/itex] represents the ground state function, as stated above, with [itex]\omega[/itex]. The states are real though, so the complex conjugate is unchanged from the normal state.


The Attempt at a Solution


When I try to evaluate these integrals, I find that I can't actually do it. The problem is that the different symbols keep tripping me up... The ω and ω' values don't seem to gel.

Am I doing the physics wrong, or am I simple falling short on the maths?
Thanks, all.
 

Answers and Replies

  • #2
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Ah! Solved the maths problems...

I have one more question, though. For a one-dimensional quantum harmonic oscillator, when determining the amplitudes,[itex]C_{i}[/itex], should I integrate from 0 to infinity, or from -infinity to infinity? I'm guessing it's the latter.
 
  • #3
vela
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You're right, the latter.
 
  • #4
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You're right, the latter.
Thanks!

To that end, I have a standard integral that gives the answer for the limits 0 and infinity. Is it simply a matter of multiplying by two or have I missed something?
 
  • #5
vela
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If the integrand is an even function, you can simply multiply by 2. If it's odd, the integral will vanish.
 
  • #6
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If the integrand is an even function, you can simply multiply by 2. If it's odd, the integral will vanish.


It's an exponential, similar in form to:

[itex]\int e^{-ax^2} dx = 0.5[\frac{\pi}{a}]^\frac{1}{2}[/itex]

Where the limits are taken over 0 and infinity for this identity.

So, am I okay to multiply by two?
 
  • #7
vela
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You should be able to answer that on your own. A function f(x) is even if f(x)=f(-x) and odd if f(x)=-f(x). Which one do you have on your hands here?
 
  • #8
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I've been having a duh moment here...(I blame late night maths). it's clearly even, because the x has an even exponent. So multiplying by two is permitted.
 
  • #9
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Okay, so now I have a superposition of the 3 states, complete with amplitudes. I've introduced a time dependence by setting
[itex]\psi = \Sigma C_{i} U_{i} e^{\frac {-iE_{i}t} {h}} [/itex]

Which gives me a superposition state with 3 terms. Now, I want to find the probability of finding this particle in an area -a to a, so I've multiplied the above by its complex conjugate, which gives an expression with 9 terms. If I now integrate over -a to a, am I right in thinking that the terms that contain two of the original eigenstates (eg U0 and U1) will reduce to the kronecker delta, or is that only valid for an integral over all space?
 
  • #10
vela
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To rely on orthogonality, you have to integrate over all space.
 

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