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

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

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

$$<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}( \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1)u_{0}.dx$$

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

$$<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}$$

Cheers!

 

[ozymandias]

$$A^\dagger A$$ is indeed the number operator, as
$$A^\dagger A |n> = A^\dagger \sqrt{n} |n-1> = n |n>$$


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