# Momentum Operator of Klein-Gordon field

I am studying the second chapter of Peskin and Schroeder's QFT text. In equation 2.27 and 2.28, the book defines the field operators:

$$\phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 w_p}} (a_p + a^\dagger_{-p}) \, e^{ipx} \\ \pi(x) = \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx}$$

The momentum operator is then calculated in equation 2.33. However, my own derivation gives a different answer. I am reproducing my steps hoping that someone will be able to find where I went wrong. Starting with the definition of the momentum (conserved charge of spatial translations):

$$\mathbf{P} = -\int d^3x \, \pi(x) \nabla \phi(x) \\ \mathbf{P} = -\int d^3x \, \Bigg[ \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx} \Bigg]\Bigg[\int \frac{d^3p'}{(2\pi)^3} \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \, \nabla e^{ip'x} \Bigg] \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \int d^3x \, (-i^2) \, \mathbf{p}' \, e^{i(p+p')x} \, \Bigg[\sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \Bigg]\Bigg[\frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \Bigg] \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \mathbf{p}' (2\pi)^3 \delta(p+p') \, \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \\ \mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{-p}}} (a_{-p} + a^\dagger_{p}) \\$$ Since $w_{p} = w_{-p} = |p|^2 + m^2$, we get $$\mathbf{P} = -\int \frac{d^3p}{2(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{w_{p}}} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} + a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] \\ \mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] + \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg] \\$$

The first integral is odd with respect to p, and vanishes. For the second term, we can formally prove that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$, but we can also argue that from noting that this operator pair creates a particle but then destroys it, with any possible constants only depending on the magnitude of \mathbf{p}. This line of reasoning gives us:

$$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{p} a_{p}\bigg) = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \, [ a_p,a^\dagger_{p}] \\$$

The commutator here is proportional to the delta function, and hence this expression doesn't match what Peskin & Schroeder, and other QFT books have, i.e., $$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p} \, a^\dagger_{p} a_p$$

UPDATE: I realized later that my assumption that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$ was wrong. When I was trying to prove this using the expansion of the ladder operators in terms of $\phi(x)$ and $\pi(x)$ I was making an algebra error.

• Have you tried going the other direction (i.e. start from the last line you expect, and work back to the starting point)? This relationship should work both ways, so working backwards should yield some insight. – Sean E. Lake Dec 21 '17 at 1:01
• I know that if the negative sign in the line before the commutator was a positive, it could work out. But I don’t see any way of making that happen. – Sidd Dec 21 '17 at 1:03
• Next recommendation: what do the equal time canonical commutation relations, $\left[\phi(\mathbf{x}),\,\phi(\mathbf{x}')\right] = \left[\pi(\mathbf{x}),\,\pi(\mathbf{x}')\right] = 0$ and $\left[\phi(\mathbf{x}),\,\pi(\mathbf{x}')\right] = i\delta(\mathbf{x}-\mathbf{x}')$, imply for $\left[a(\mathbf{p}), \,a^\dagger(\mathbf{p}')\right]$ and $\left[a(\mathbf{p}),\, a(\mathbf{p}')\right]$? – Sean E. Lake Dec 21 '17 at 2:13
• The commutation relations are $\left[ a(\mathbf{p}),a^\dagger(\mathbf{p'}) \right] = (2\pi)^3 \delta(\mathbf{p}-\mathbf{p'}), \left[a(\mathbf{p}),a(\mathbf{p'}) \right] = 0, and \left[a^\dagger(\mathbf{p}),a^\dagger(\mathbf{p'}) \right] = 0$. This however doesn't seem to solve the problem. – Sidd Dec 21 '17 at 2:19

From $$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg)$$ you actually have $$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{\mathbf{p}}{2}a^\dagger_{-p} a_{-p}\right) = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{-\mathbf{p}}{2}a^\dagger_{p} a_{p}\right) = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a^\dagger_{p} a_p + \frac{\mathbf{p}}{2}a^\dagger_{p} a_{p}\right) + \mbox{renormalization term.}\\$$ Dropping the infinite renormalization term due to $\delta({\bf 0})$ $$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p}\: a^\dagger_{p} a_p \:.$$
• Thanks for the answer! Shouldn't we use the symmetrized momentum operator since $\hat\phi(x)$ and $\hat\pi(x)$ don't commute? In particular, for example. in the 1 D version, shouldn't we use $\hat{P}=-\dfrac{1}{2}\displaystyle\int dx \bigg(\hat{\pi}{(x)}\partial_x\hat{\phi}{(x)}+\partial_x\hat{\phi}{(x)}\hat{\pi}{(x)}\bigg)$? I checked and it reproduces the same result but is there a reason behind it or is it a coincidence? If the latter, which form should one be using in general? Thanks! :) – Dvij Mankad Dec 4 '18 at 3:07