How to compute normalization of one-particle states for Klein-Gordon field quantization

Question

I am reading through Dr. Schwartz's book on quantum field theory; in section 2.3.1, he writes the following relation: $$\langle\mathbf{p}|\mathbf{k}\rangle=2\omega_p(2\pi)^3\delta^3(\mathbf{p}-\mathbf{k})$$ where $\omega_p=|\mathbf{p}|$. Do note his conventions: $[\hat{a}_p,\hat{a}_k^\dagger]=(2\pi)^3\delta(\mathbf{p}-\mathbf{k})$ and $\sqrt{2\omega_p}\hat{a}_p^\dagger|0\rangle=|\mathbf{p}\rangle$. However, when I do the (fairly trivial) calculation myself, I get $$\begin{align} \langle\mathbf{p}|\mathbf{k}\rangle&=2\sqrt{\omega_p\omega_k}\langle0|\hat{a}_p\hat{a}_k^\dagger|0\rangle \\ &=2\sqrt{\omega_p\omega_k}\left((2\pi)^3\delta(\mathbf{p}-\mathbf{k})\langle0|0\rangle+\langle 0|\hat{a}_k^\dagger\hat{a}_p|0\rangle\right) \\ &=2\sqrt{\omega_k\omega_p}(2\pi)^3\delta(\mathbf{p}-\mathbf{k}) \end{align}$$ I fail to see how $\sqrt{\omega_p\omega_k}=\omega_p$! Do I have some fundamental misunderstanding of the situation? There doesn't seem to be anything wrong with the calculation. Any help is much appreciated.

Answer 1

Since $$\omega_p=\sqrt{\mathbf{p}^2+m^2}$$ and $$f(x)\delta(x-y)=f(y)\delta(x-y) ,$$ in the sense of integration, you have $$\sqrt{\omega_p}\delta(\mathbf{p}-\mathbf{k})=\sqrt{\omega_k}\delta(\mathbf{p}-\mathbf{k}) .$$

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As pointed out by John Dumancic and kaylimekay in the comments below, the identities for the $\delta$ function are only meaningful when they are utilized in integration. To be specific, one can perform substitution in the following expression $$ \int {d\mathbf{p}}\rho(\mathbf{k})\sqrt{\omega_p}\delta(\mathbf{p}-\mathbf{k})$$ to get $$ \int {d\mathbf{p}}\rho(\mathbf{k})\sqrt{\omega_k}\delta(\mathbf{p}-\mathbf{k})=\rho(\mathbf{k})\sqrt{\omega_k}\int {d\mathbf{p}}\delta(\mathbf{p}-\mathbf{k})=\rho(\mathbf{k})\sqrt{\omega_k} .$$ But without the integral $\int d\mathbf{p}$, the identity/equality is not rigorously defined. You may try to verify in your favorite textbook whether the identity is always utilized in the above context.