# Homework Help: Creation and annihilation operators

1. Nov 9, 2012

### Lindsayyyy

Hi all

1. The problem statement, all variables and given/known data
Show:

$$(a^\dagger a)^2=a^\dagger a^\dagger a a +a^\dagger a$$

wheres:
$$a= \lambda x +i \gamma p$$
$$a^\dagger= \lambda x -i \gamma p$$

2. Relevant equations
-

3. The attempt at a solution

Well, I haven't got much.

I just tried to use the stuff given, put it into my equation and solve it, but I don't get to the right side.

I calculated a+a first

$$a^\dagger a ={\lambda}^2x^2 + \frac {1}{2} I + \gamma^2 p^2$$

But when I now try to calculate the square of that term I get lost. If I square it I get to:

$$(a^\dagger a)^2= \lambda^4x^4+\gamma^4p^4 +\lambda^2 \gamma^2 (x^2p^2+p^2x^2)-\lambda^2 x^2 -\gamma^2 p^2 +\frac 1 4 I$$

Can anyone help me with this? I don't know what to do now/ If I'm on the right way.

Thanks for your help

edit: I is the identity matrix

Last edited: Nov 9, 2012
2. Nov 9, 2012

### grzz

(a$^{+}$a)$^{2}$
= a$^{=}$aa$^{+}$a
=a$^{+}$(aa$^{+}$)a.

Now try to get aa$^{+}$ in terms of a$^{+}$a.

3. Nov 9, 2012

### Lindsayyyy

what does your a^= mean? I don't know this sign.

Thanks for your help

edit:

$$aa^\dagger=\lambda^2 x^2 + \frac 1 2 I +\gamma^2 p^2$$

4. Nov 9, 2012

### grzz

I used the plus sign instead of the dagger sign.

You are supposed to see a with + as a superscript wherever you see a with ^ as a superscript.

Sorry for the trouble in the notation used.

5. Nov 9, 2012

### Lindsayyyy

I see a = not a + but I think it's just a typing mistake.

so what you mean

$$...= a^\dagger a a^\dagger a=a^\dagger (aa^\dagger) a$$

is that what you meant?

I calculated my aa+ (see my post) and a+a is the same, though there's a minus infront of the I

6. Nov 9, 2012

### Fightfish

"The same"? That difference in sign makes all the difference. Can you now relate $aa^{\dagger}$ to $a^{\dagger}a$?

7. Nov 9, 2012

### Lindsayyyy

With "the same" I meant the rest of the terms. Unlucky word choice from my side I guess.
What do you mean with "can you now relate aa+ to a+a"
do you mean that I should calculate $$(aa^\dagger) (a^\dagger a)$$

thanks for the help

8. Nov 9, 2012

### Fightfish

No, I meant the relationship between $aa^{\dagger}$ and $a^{\dagger}a$ ie express one of them in terms of the other.

9. Nov 9, 2012

### grzz

Yes that is what I meant.

Allow me to use a+ to mean a with + as a superscript.

So aa+ = a+a + 1.

Now one more step.

10. Nov 9, 2012

### Lindsayyyy

Do you mean something like

$$a^\dagger a = 2\lambda^2 x^2 +2 \gamma^2 x^2 -a a^\dagger$$

edit: nevermind, fail lol.

I guess you meant what grzz just said. I have to eat dinner now. I'll try later or tomorrow and come back when I have further questions (I'm pretty sure I will have some :( )

Thanks so far for the help guys.

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