Expectation values and the Harmonic oscillator

1. Oct 10, 2011

Diomarte

1. The problem statement, all variables and given/known data
For the Harmonic Oscillator, the state |ψ> = (|0> + |1>) / √(2)

Find $\overline{x}$ = <ψ|x|ψ> $\overline{p}$ = <ψ|p|ψ>
$\overline{x^2}$ = <ψ|x$^{2}$|ψ> and $\overline{p^2}$ = <ψ|p$^{2}$|ψ>

and

<ψ| (x - $\overline{x}$)^2 |ψ><ψ| (p - $\overline{p}$)^2 |ψ>

2. Relevant equations

3. The attempt at a solution
I got some help from another student on getting started with the first part of this problem, but in all honesty I'm really not even sure how to start some of these, and how the operators work. This is what I've got so far:

<ψ|x|ψ> = <ψ|√(hbar/2mω) (a+a_)|ψ>
<ψ| = 1/√2 (<0| + <1|)
|ψ> = 1/√2 (|0> + |1>)

giving 1/2 √(hbar/2mω) (<0| + <1|) (a+a_) (|0> + |1>)

from here the raising and lowering operators are operating on the nth states of 0 and 1. I know that a_|0> = 0 and a_|n> = √(n)|n-1> and that a+|n> = √(n+1)|n+1> but if anyone can make a suggestion or show me how to get some results here, I would greatly appreciate it. As well, some direction on how to start the next parts would be amazing too, thank you!

2. Oct 10, 2011

dextercioby

So you essentially got this expression (<0| + <1|) (a_+ + a_) (|0> + |1>). Ok, there's some operator 'sandwiched' between bra's and ket's. Can you expand this expression into a sum of 4 terms and then compute each term separately ?

3. Oct 10, 2011

Diomarte

Meaning have the a+ operator act on the bras and then the a_ operate on the kets?

4. Oct 10, 2011

vela

Staff Emeritus
Just have them act on the ket, though it doesn't make a difference.

5. Oct 10, 2011

Diomarte

"Though it doesn't make a difference." Why do you say this vela? Is it because of the orthogonality and cancellation if you end up multiplying through after the operations?

6. Oct 11, 2011

vela

Staff Emeritus
Because if you do the calculations correctly, it doesn't matter whether you have the operators act on the bra or the ket. Why do you think it does?

7. Oct 11, 2011

dextercioby

Actually, there's a trick to take into account. The raising & the lowering ladder operators are not self-adjoint, but one is the adjoint of the other, so that

$$\langle 0|a_{+}|0 \rangle = \langle a_{-}0, 0\rangle = \langle 0,a_{+} 0\rangle$$.

8. Oct 11, 2011

Diomarte

Alright, so I performed the operations and got <x> = 0 and <p> = 0. Two other students I spoke with said that they got <p> = 0 but not <x>. Am I making another mistake here?

9. Oct 11, 2011

Diomarte

Ok, never mind, I figured out what was going on. Thank you all very much for your help. After following what was going on, I've successfully solved four of my five assignments! I'll post what I did later on tonight when I've got the time, so if other people have a similar question, they have a direction to go.