Quantum harmonic oscillator, uncertainty relation

1. Mar 28, 2016

phys-student

1. The problem statement, all variables and given/known data
Consider a particle with mass m oscillates in a simple harmonic potential with frequency ω. The position, x, and momentum operator, p, of the particle can be expressed in terms of the annihilation and creation operator (a and a respectively):
x = (ħ/2mω)^0.5 * (a + a)
p = i(ħmω/2)^0.5 * (a - a)
The annihilation and creation operators satisfy the following commutation relations:
[a,a] = 1
The eigenvalue equation of a is found as:
a|α> = α|α>
Where |α> is a coherent state
a) Find the expectation value <a> in coherent state |α>
b) Find the expectation values of <aa>, <aa>, and <aa>
c) Calculate the uncertainty relation (<(x-<x>)2><(p-<p>)2>)0.5
2. Relevant equations
Relevant equations given above

3. The attempt at a solution
I've worked through most of the problem and only noticed that I may have done something wrong when I got to part c, where the expression that I get for <x> causes the expectation value <(x-<x>)2> to be equal to 0, which also means the uncertainty relation is equal to 0. I think the error may be when I wrote the eigenvalue equation for a in bra notation like this:
<α|a = <α|α*
Is this the correct way of expressing the eigenvalue equation given in the question in bra notation?

2. Mar 28, 2016

blue_leaf77

That's correct. But I don't know where you went wrong unless you provide your calculation.

3. Mar 28, 2016

phys-student

If that is correct then the expectation values are as follows; <a>=α*, <a> = α, <aa> = α2, <aa> = α*2, and <aa> = <aa> = αα*.

This means that the expectation value <x> is: <x> = (ħ/2mω)0.5(<a> + <a>) = (ħ/2mω)0.5(α* + α)

So the expression (x-<x>)2 = x2 - 2x(ħ/2mω)0.5(α* + α) + (ħ/2mω)(α* + α)2

So the expectation value from the uncertainty relation is: <(x-<x>)2> = <x2> - 2*(ħ/2mω)0.5(α*+α)<x> + (ħ/2mω)(α* + α)2

The expectation value <x2> is:
<x2> = (ħ/2mω)(<aa> + <aa> + <aa> + <aa>) = (ħ/2mω)(α* + α)2

If you sub the expression for <x> and <x2> back into the expression for <(x-<x>)2> it ends up reducing to 0

4. Mar 28, 2016

blue_leaf77

That's where you go wrong. Your $\langle a^\dagger a \rangle$ is correct, but your $\langle aa^\dagger \rangle$ is not. To calculate the latter, you could have used the commutation between the annihilation and raising operators and $\langle a^\dagger a \rangle$.

5. Mar 28, 2016

phys-student

Okay so you are saying that <aa> = αα* is correct, then using the commutation relation:

[a,a] = aa - aa = 1, therefore: aa = 1 + aa

So then the expectation value of aa is: <aa> = 1 + <aa> = 1 + αα*

Is that correct?

6. Mar 28, 2016

blue_leaf77

Yes.

7. Mar 28, 2016

Okay thanks