# Zwiebach quick calculation 10.6

1. Oct 8, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
I am somewhat new to creation and annhilation operators, but I can reduce the problem to showing that:

$$a_{p_1} ^{\dag} a_{p_2} ^{\dag} = a_{p_1} ^{\dag} + a_{p_2} ^{\dag}$$

where
$$a_{p_1} ^{\dag}$$ is the creation operator for a particle with momentum p1 (if I understand it correctly)

Can someone help prove that?
2. Relevant equations

3. The attempt at a solution

Last edited: Oct 8, 2007
2. Oct 8, 2007

### Jimmy Snyder

This can't be right since if it were, then:
$$a_{p_1}^{\dag} a_{p_2} ^{\dag} a_{p_3} ^{\dag} = a_{p_1}^{\dag}(a_{p_2}^{\dag} + a_{p_3}^{\dag}) = a_{p_1}^{\dag}a_{p_2}^{\dag} + a_{p_1}^{\dag}a_{p_3}^{\dag} = a_{p_1}^{\dag} + a_{p_2}^{\dag} + a_{p_1}^{\dag} + a_{p_3}^{\dag}$$
In order to do the calculation, apply H and P as defined in equations (10.60) and (10.61) directly to the state in (10.65). Use induction on k.

3. Oct 8, 2007

### ehrenfest

That's weird though. How can it not be true if

$$\sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p1}^{\dag} a_{p2}^{\dag} = \sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p1}^{\dag} + \sum_{\vec{j}} \vec{j} a_j ^{\dag} a_j a_{p2}^{\dag}$$

?

4. Oct 8, 2007

### Jimmy Snyder

In general, for operators, AB = AC does not imply B = C even when A is non-zero.

5. Oct 8, 2007

### ehrenfest

Yes. I was wrong.

But then somehow you need to split up that long product of $$a_{p_k}^{\dag}$$ operators. You can commute them by 10.59, but they doesn't really help. Is there another way to split it into two terms?

6. Oct 8, 2007

### Jimmy Snyder

The formula [A,BC] = B[A,C] + [A.B]C will do nicely.

7. Oct 10, 2007

### ehrenfest

Ah, I see. Thanks.

For QC 10.7, I am trying to use

$$\sum_{\vec{p}} a_p^{\dag} a_p | \Omega > = \sum_{\vec{p}} (a_p a_p^{\dag} -1) | \Omega >$$

So, there should probably be a Kronecker delta that has an index that gets summed over or something but I am not really sure where this comes in?

Just to be sure, I want to show that $$\sum_{\vec{p}} a_p^{\dag} a_p | \Omega > = n | \Omega >$$.

where n is the number of particles in the state omega, right?

By the way, I should probably not be using omega because we are using that for the vacuum state and we want an arbitrary state here, right?

Last edited: Oct 10, 2007
8. Oct 10, 2007

### Jimmy Snyder

QC 10.7 is done exactly the same way that QC 10.6 is done.