Quantum harmonic oscillator, uncertainty relation

[phys-student]

Homework Statement

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>, <a^†a>, and <a^†a^†>
c) Calculate the uncertainty relation (<(x-<x>)²><(p-<p>)²>)^0.5

Homework Equations

Relevant equations given above

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?

 

[blue_leaf77]

<α|a† = <α|α*
Is this the correct way of expressing the eigenvalue equation given in the question in bra notation?

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

 

[phys-student]

If that is correct then the expectation values are as follows; <a^†>=α*, <a> = α, <aa> = α², <a^†a^†> = α*², and <aa^†> = <a^†a> = 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)(<a^†a^†> + <aa> + <a^†a> + <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

 

[blue_leaf77]

<aa†> = <a†a> = αα*.

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 $\langle a^\dagger a \rangle$.

 

[phys-student]

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

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

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

Is that correct?

 

[blue_leaf77]

Is that correct?

Yes.

 

[phys-student]

Okay thanks