# Ladder operators in Klein -Gordon canonical quantisation

1. Oct 18, 2015

### spaghetti3451

The quantum Klein-Gordon field $\phi({\bf{x}})$ and its momentum density $\pi({\bf{x}})$ are given in Fourier space by

$\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)$ and

$\pi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} (-i) \sqrt{\frac{ \omega_{{\bf{p}}}}{2}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} - a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)$.

[These are equations (2.25) and (2.26) from the Peskin and Schroeder.]

Now, I used the substitution ${\bf{p}} \rightarrow {\bf{-p}}$ in the expression for $\phi({\bf{x}})$ and obtained

$a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}$ and $a_{-{\bf{p}}} = a^{\dagger}_{{\bf{p}}}$.

On the other hand, I used the same substitution ${\bf{p}} \rightarrow {\bf{-p}}$ in the expression for $\phi({\bf{x}})$ and obtained

$a_{{\bf{p}}} = - a^{\dagger}_{-{\bf{p}}}$ and $a_{-{\bf{p}}} = - a^{\dagger}_{{\bf{p}}}$.

Can someone explain what's going on?

Last edited: Oct 19, 2015
2. Oct 19, 2015

### vanhees71

There is a - missing in the second term of $\Pi(\vec{p})$. That's seen in the Heisenberg picture and the relation $\Pi=\dot{\phi}$.

3. Oct 19, 2015

### spaghetti3451

Sorry, that's my own mistake. Peskin and Schroeder has the minus sign. I've made the edit now.

4. Oct 19, 2015

### spaghetti3451

Actually, with the minus sign in the second term of $\pi({\bf{x}})$, I get the inconsistency in the ladder operators.

5. Oct 19, 2015

### vanhees71

I don't understand, how you get these equations in the first place. The $\hat{a}_{\vec{p}}$ are an independent set of annihilation operators. There shouldn't be a relation between $\hat{a}_{\vec{p}}$ and $\hat{a}_{-\vec{p}}^{\dagger}$.

6. Oct 19, 2015

### spaghetti3451

Well, under the substitution ${\bf{p}} \rightarrow {\bf{-p}}$,

$\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)$

becomes

$\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{-\bf{p}}}}} \big( a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{-{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} \big)$,

and then I compared the coefficients of the exponential $e^{-i{\bf{p}} \cdot {\bf{x}}}$ on both sides to obtain $a^{\dagger}_{{\bf{p}}} = a_{-{\bf{p}}}$ and the coefficients of the exponential $e^{i{\bf{p}} \cdot {\bf{x}}}$ on both sides to obtain $a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}$.

But now that you mention it, I'm beginning to think my argument is wrong. The creation and annihilation operators are, in fact, independent of each other, i.e. for instance, $a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}}$ is the same as $a_{-{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}}$ because the integration is over the entire range of momenta from $-\infty$ to $\infty$.

Am I correct?