# Quantum operators and commutation relations

1. Apr 4, 2016

### CAF123

1. The problem statement, all variables and given/known data
Given the mode expansion of the quantum field $\phi$ and the conjugate field one can derive $$\mathbf P = \int \frac{d^3 \mathbf p}{(2\pi)^3 2 \omega(\mathbf p)} \mathbf p a(\mathbf p)^{\dagger} a(\mathbf p)$$ By writing $$e^X = \text{lim}_{n \rightarrow \infty} \left(1+\frac{X}{n}\right)^n$$ calculate $[\mathbf P, a(\mathbf q)^{\dagger}]$ and hence evaluate $$\exp(-i \mathbf P \cdot \mathbf z) a(\mathbf q)^{\dagger} \exp(i \mathbf P \cdot \mathbf z),$$ with $\mathbf z$ a constant vector.

2. Relevant equations
$[AB,C] = A[B,C] + [A,C]B$

Commutation relations for the creation and annihilation operators.

3. The attempt at a solution
I think I can show that $[\mathbf P, a(\mathbf q)^{\dagger}] = \mathbf q a(\mathbf q)^{\dagger}$ but why in particular is this definition of the exponential useful? I can use it to rewrite the exponential factors appearing in that expression but that gives me $$\text{lim}_{n \rightarrow \infty} \text{lim}_{r \rightarrow \infty} \left(1- i \frac{\mathbf P \cdot \mathbf z}{n}\right)^n a(\mathbf q)^{\dagger} \left(1+ i \frac{\mathbf P \cdot \mathbf z}{r}\right)^r$$ but I can't see an easy way to proceed.

It seems to me that I could make more progress using the Baker Campbell Hausdorff formula for non commuting exponentials

2. Apr 4, 2016

### strangerep

Last edited: Apr 4, 2016
3. Apr 5, 2016

### CAF123

Hi strangerep,
Is this to help me in computing $[\mathbf P,a(q)^{\dagger}]?$ I think I managed to do this using the integral representation of the spatial momentum in terms of c/a operators. My result was $qa(q)^{\dagger}$. So this tells me that $a(q)^{\dagger}$ and $\mathbf P$ do not commute and so the expression $\exp(-iP \cdot z) a(q)^{\dagger} \exp(i P \cdot z)$ can't be simplified trivially. i thought about using the BCH formula for $e^{-X} A e^X$ and operators X and A but I am instructed to use the limit definition of the exponential and I can't see how to progress using such a definition.
Yup, I should have said $\phi$ is a real scalar field.

4. Apr 5, 2016

### TSny

Can you use your commutation relation for $[\mathbf P,a(q)^{\dagger}]$ to move $a(q)^{\dagger}$ all the way to the left in the expression $\left( 1-i P \cdot z \right)a(q)^{\dagger}$?

Repeat for $\left( 1-i \Large{\frac{P \cdot z}{2}}\right)^2a(q)^{\dagger}$.

5. Apr 5, 2016

### CAF123

Hi TSny,

I suppose you are maybe hinting at me trying a few cases and seeing if there is a generalisation? $$(1-iP \cdot z) a(q)^{\dagger} = ((a(q)^{\dagger} - i((a(q)^{\dagger} P \cdot z + q \cdot z a(q)^{\dagger}))$$ Then $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = (1-iP \cdot z) a(q)^{\dagger} - \frac{1}{4} (P \cdot z)^2 a(q)^{\dagger}$$ It seems like the first term will be common but the second term is not?

Thanks!

6. Apr 5, 2016

### TSny

Yes.
Simplify by factoring $a(q)^{\dagger}$ out to the left on the RHS.
Write $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = \left(1-i \frac{P \cdot z}{2}\right) \left(1-i \frac{P \cdot z}{2}\right)a(q)^{\dagger}$$ Try moving $a(q)^{\dagger}$ all the way to the left (one set of parentheses at a time).

7. Apr 5, 2016

### CAF123

$$a(q)^{\dagger}(1-i(P+q)\cdot z)$$
I get that $$a(q)^{\dagger}\left(1-\frac{i}{2}(P+q)\cdot z \right)^2$$

So, we end up with $$\text{lim}_{n,r \rightarrow \infty} a(q)^{\dagger} \left(1-\frac{i}{n}((P+q)\cdot z)\right)^n \left(1+\frac{i}{r} P \cdot z \right)^r = a(q)^{\dagger} \exp(-i(P+q)\cdot z) \exp(i P \cdot z) = a(q)^{\dagger} \exp(-iq \cdot z)$$

Similarly, I can compute e.g $$I = \exp(-iP \cdot z) \phi(x) \exp(i P \cdot z)$$ where $$\phi(x) = \int \frac{d^3 q}{(2\pi)^3 2 w(q)} (a(q) e^{-iq \cdot x} + a(q)^{\dagger} e^{iq \cdot x})$$ and using the results above gives the result that $I$ is $\phi(t, \mathbf x - \mathbf z)$ so that $I$ is a space translation operator for the field $\phi$ in the $-z$ direction. The last result makes sense to me, so I guess my answer to the first part was also correct?

Last edited: Apr 5, 2016
8. Apr 5, 2016

### TSny

That looks good. You've shown that the momentum operator is the generator of translations in space. I think most people would refer to $\exp(-iP \cdot z)$ as the translation operator rather than $I$.

Yes, I believe you worked the first part correctly.

9. Apr 5, 2016

### strangerep

No. That part's easy.

I'm not sure whether you need any more hints from me -- maybe TSny's hints are enough?