# Total momentum operator of the Klein-Gordon field (before limit to the continuum)

I'm following K. Huang's QFT: From Operators to Path Integrals book. In the second chapter, he introduces the Klein-Gordon equation (KGE), and its scalar field $$\phi(x)$$, which satisfies this equation. He then gives an Ansatz for this scalar field: $$\phi(x) = \frac{1}{\sqrt{\Omega}}\sum_{k}q_{k}(t) \, \text{e}^{i\mathbf{k}\cdot\mathbf{r}},$$ which, because of its real nature, it follows that $$q_{k}^{*}=q_{-k}$$. Then, using this solution in the KGE, another equation for the $$q_{k}$$ is found: $$\ddot{q}_{k}+\omega_{k}^{2}q_{k}=0,$$ where $$\omega_{k}^{2}=k^{2}+m^2$$. Then, the field is quantized by imposing the canonical commutation relations $$i\left[\dot{q}_{k}^{\dagger}(t),q_{k^{\prime}}(t)\right]=\delta_{kk^{\prime}}$$ $$\left[q_{k}(t),q_{k^{\prime}}(t)\right]=0$$ which are equivalent to $$i\left[\dot{\phi}(\mathbf{r},t),\phi(\mathbf{r}^{\prime},t)\right]=\delta^{3}(\mathbf{r}-\mathbf{r}^{\prime})$$ $$\left[\phi(\mathbf{r},t),\phi(\mathbf{r}^{\prime},t)\right]=0.$$ Lastly, he then finds a solution for the $$q_{k}$$ coefficients, from the previous differential equation, which is $$q_{k}(t)=\frac{1}{\sqrt{2\omega_{k}}}\left(a_{k} \, \text{e}^{-i\omega_{k}t} + a_{-k}^{\dagger} \, \text{e}^{i\omega_{k}t}\right).$$ Thus, the coefficients from $$a_{k}$$ also follow a commutation relation, and are operators. These are the commutation relations from the creation and annihilation operators of the quantum harmonic oscillator. Later, he proceeds to find the Hamiltonian and the Momentum operators for the Klein-Gordon field, which is where I have my doubt.

As I understand, and as it is mentioned in the first chapter, the operators in the quantum field's Fock space are obtained as $$\hat{O}=\int\text{d}^{3}r \, \psi^{\dagger}(\mathbf{r})O\psi(\mathbf{r}),$$ where $$\psi(\mathbf{r})$$ is the field now promoted to operator. Then, as I understand, if I want to calculate the total momentum operator, eq. (2.26) in the book, I would need to work out \begin{align*} \hat{P}=& \ \int\text{d}^{3}r\phi^{*}(\mathbf{r})(-i\nabla)\phi(\mathbf{r}) \\ =& \ \sum_{k,k^{\prime}}\mathbf{k} \, q_{-k^{\prime}}q_{k} \, \int\text{d}^{3}r\frac{\text{e}^{i(\mathbf{k} - \mathbf{k}^{\prime})\cdot\mathbf{r}}}{\Omega} \\ =& \ \sum_{k}\mathbf{k} \, q_{-k}q_{k} \\ =& \ \sum_{k}\frac{\mathbf{k}}{2\omega_{k}}\left(a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} + a_{-k}a_{-k}^{\dagger} + a_{k}^{\dagger}a_{k} + a_{k}^{\dagger}a_{-k}^{\dagger} \, \text{e}^{2i\omega_{k}t}\right). \end{align*} Now, noting that, by considering the change of index $$k\to-k$$ \begin{align*} \sum_{k=-\infty}^{\infty}\frac{\mathbf{k}}{2\omega_{k}}a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} =& \ \frac{0}{2\omega_{0}}a_{0}a_{0} \, \text{e}^{-2i\omega_{0}t} + \sum_{k=1}^{\infty}\frac{\mathbf{k}}{2\omega_{k}}a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} + \sum_{k=-\infty}^{-1}\frac{\mathbf{k}}{2\omega_{k}}a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} \\ =& \ \sum_{k=1}^{\infty}\frac{\mathbf{k}}{2\omega_{k}}a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} - \sum_{k=1}^{\infty}\frac{\mathbf{k}}{2\omega_{k}}a_{-k}a_{k} \, \text{e}^{-2i\omega_{k}t} \\ =& \ 0. \end{align*} Then, all that is left is \begin{align*} \hat{P} =& \ \sum_{k}\frac{\mathbf{k}}{2\omega_{k}}\left(a_{k}a_{k}^{\dagger} + a_{k}^{\dagger}a_{k}\right) \\ =& \ \sum_{k}\frac{\mathbf{k}}{2\omega_{k}}\left(2a_{k}^{\dagger}a_{k} + 1\right) \\ =& \ \sum_{k}\frac{\mathbf{k}}{\omega_{k}}a_{k}^{\dagger}a_{k}, \end{align*} where the term $$\sum_{k}\mathbf{k}/2\omega_{k}$$ is also evaluated to zero by being odd on $$k$$. However, this is not the expression in the book, which reads $$\begin{equation*} \hat{P}=\sum_{k}\mathbf{k} \, a_{k}^{\dagger}a_{k}, \end{equation*}$$ which must also be right, as all it does is counting the momentum $$\hbar\mathbf{k}$$ of the number of particles of each $$k$$. Where is it that my procedure is wrong? Where is this factor of $$1/\omega_{k}$$ eliminated?

• The momentum along the spatial direction $i$ is the spatial integral of $T^{0i}(x)$. Performing computations, a factor $\omega_k$ takes place removing the denominator you found…Your general formula for $\hat{O}$ is only valid for the non relativistic case. Commented Mar 29 at 6:56

## 1 Answer

When you related the quadrature operator $$q$$ to the ladder operators $$a$$'s in the "solution," there is an implied commutation relation for the ladder operators that you can derive from the commutation relation for the quadrature operators. You should use this commutation relation when you do your computation at the end which should then be consistent.

• Hi! Thanks for your answer! Yes, this relation is $\left[a_{k},a_{k^{\prime}}^{\dagger}\right]=\delta_{kk^{\prime}}$, which is what I used to get from $a_{k}^{\dagger}a_{k} + a_{k}a_{k}^{\dagger}$ to $2a_{k}^{\dagger}a_{k} + 1$. Is this what you meant? Commented Mar 28 at 10:27
• Yes, you should check whether it is consistent with the expression that you gave for $q$ in terms of these $a$'s. I suspect not. Commented Mar 29 at 4:03