Expectation value, harmonic oscillator

[broegger]

Hi,

I have to find the expectation values of xp and px for nth energy eigenstate in the 1-d harmonic oscillator. If I know <xp> I can immediately find <px>since [x,p]=ih. I use the ladder operators a_{\pm}=\tfrac1{\sqrt{2\hslash m\omega}}(\mp ip+m\omega x) to find <xp>, but I get a complex value, <xp>=ih/2. It doesn't seem right in the context of the rest of the exercise...

 

[dextercioby]

Why not...?Why should it be real...?

Daniel.

 

[Galileo]

What's your expression for xp in terms of the ladder operators?

 

[broegger]

Well, xp is not hermitian, I see your point, dexter. My expression for xp is i\hslash/2({a_+}^2 + {a_-}^2 + 1).

 

[dextercioby]

Voilà.

Daniel.

 

[dreamspy]

Could I be so bold as to ask how you got that expression for xp ?

 

[dreamspy]

I got it. I convert p to ladder operators using this formula:

\hat a_\pm=\frac{1}{\sqrt{2mh\omega}}(m\omega \hat x \mp i\hat p)

Just isolate p. Then we use this formula for x in ladder operators:

\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat a_++\hat a_-)

Then we simply multiply x and p.

But what i get is a little bit different than above:

\hat x \hat p=\frac{i\hbar}{2}(\hat a_+^2-\nf\hat a_-^2+1)

(note the - in the last formula)

But the final result is the same
&lt;\hat x \hat p&gt; = \frac{i\hbar}{2}