# Dirac Notation: Am I doing this right?

1. Apr 12, 2012

### jumbogala

1. The problem statement, all variables and given/known data
Find <P>. P = i√(mhw/2)(a†-a). Note a† and a are the ladder operators. P is the momentum operator of the harmonic oscillator.

|ψ > = (1/sqrt(2))[ |1> - i |2>]

The answer should be zero, can someone check my work?

2. Relevant equations
a† |n> = sqrt(n+1)|n+1>

a |n> = sqrt(n)|n-1>

3. The attempt at a solution
I think my first step here might be wrong. I assumed that <ψ | = (1/sqrt(2))[ <1| + i <2|].

<P> = <ψ | P | ψ > = (1/sqrt(2))[ <1| + i <2|] i√(mhw/2)(a†-a) (1/sqrt(2))[ |1> - i |2>]

= (1/2)i√(mhw/2)[ <1| + i <2|] * (a†|1> - i*a†|2> - a|1> + i*a|2>]

= (1/2)i√(mhw/2)[ <1| + i <2|] * [sqrt(2)|2> - i*sqrt(3)|3> - sqrt(1)|0> + i*sqrt(2)|1>]

Since <x|y> = 1 when x = y but 0 otherwise, this reduces to

= (1/2)i√(mhw/2)[ <1| + i <2|] * [sqrt(2)|2> - i*sqrt(3)|3> - sqrt(1)|0> + i*sqrt(2)|1>]

=(1/2)i√(mhw/2) (isqrt(2)<1|1> + isqrt(2)<2|2>) = (1/2)i√(mhw/2) ( 2isqrt(2))

= -sqrt(mhw)

But I think the answer is supposed to be zero. What am I doing wrong here?

I already checked for sign mistakes 3 times, and I can't find any.

2. Apr 15, 2012

### wotanub

No matter how I work it out, your result is correct.

What you'r thinking of is $\left\langle n \right| P \left| n \right\rangle$ = 0,

which is a specific case of the general expression

$\left\langle n' \right| P \left| n \right\rangle= i\sqrt{\frac{mhω}{2}}(-\sqrt{n}δ_{n',n-1}+\sqrt{n+1}δ_{n',n+1})$

Since your wavefunction is a superposition of states, it doesn't turn out to be zero. I can show my work if you like.

I found this in a textbook I have. Quantum Mechanics Concepts and Applications 2nd, Zettili, pg 248.

Last edited: Apr 15, 2012
3. Apr 15, 2012

### jumbogala

Thank you very much! I am a bit confused though, what are n and n prime? I don't really get how what I have is different from <n | P | n> = 0?

Does this mean it only works for pure states?

4. Apr 15, 2012

### wotanub

n and n prime are two different quantum states.

This means $\left\langle 1 \right| P \left| 1 \right\rangle = 0$ but $\left\langle 2 \right| P \left| 1 \right\rangle$ is not equal to zero.

when I worked it out, it turns out that $\left\langle 1 \right| P \left| 2 \right\rangle = -\left\langle 2 \right| P \left| 1 \right\rangle = -i\sqrt{mhω}$

Here is a picture of my messy scratchwork.
http://i.imgur.com/ekIjb.jpg

5. Apr 15, 2012

### jumbogala

Excellent, that was super helpful! Thanks again :) I understand now.

6. Apr 15, 2012

### wotanub

You're welcome! I actually remembered not to forget this rule myself!