Expectation value using ladder operators

[T-7]

I wonder if someone could examine my argument for the following problem.

Homework Statement

Using the relation

\widehat{x}^{2} = \frac{\hbar}{2m\omega}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )

and properties of the ladder operators, determine the expectation value <\widehat{x}^{2}> for the ground state of the simple harmonic well.

The Attempt at a Solution

<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )u_{0}.dx

I then argue

<br /> \left[ \widehat{A},\widehat{A}^{+}\right] = \widehat{A}\widehat{A}^{+} - \widehat{A}^{+}\widehat{A} = 1<br />
<br /> \Rightarrow \widehat{A}\widehat{A}^{+} = 1 + \widehat{A}^{+}\widehat{A}<br /> \Rightarrow \widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2\widehat{A}^{+}\widehat{A} + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2n + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1<br />

I am not quite clear if it is right that \widehat{A}^{+}\widehat{A} is equal to n (the eigenfunction number), which is zero here. Could someone comment on that, and on whether or not my treatment above is ok?

Well, if this is true, then

&lt;\widehat{x}^{2}&gt; = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}( \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1)u_{0}.dx

and by exploiting the orthonormal properties, I argue that the first two integrals are zero (you have one eigenfunction multiplied by another and integrated over infinity), but the third integral is 1, and then

&lt;\widehat{x}^{2}&gt; = \frac{\hbar}{2m\omega}

Cheers!

 

[ozymandias]

A^\dagger A is indeed the number operator, as
A^\dagger A |n&gt; = A^\dagger \sqrt{n} |n-1&gt; = n |n&gt;


------
Assaf
http://www.physicallyincorrect.com/"