I'm trying to prove that the position-space representation of a single particle wave function of the state $|\textbf{p}⟩$ in Quantum Field theory is given by $e^{i \textbf{p}\cdot\textbf{x}}$

i.e. $⟨0|\phi(\textbf{x},t)|\textbf{p}⟩=e^{i \textbf{p}\cdot\textbf{x}}$, where $|\textbf{p}⟩=\sqrt{2E_{\textbf{p}}}a_{\textbf{p}}^\dagger |0⟩$ is the normalised single particle state.

Here is my attempt to prove it. $$ \begin{aligned} ⟨0|\phi(\textbf{x},t)|\textbf{p}⟩ &= ⟨0|\int\frac{d^3p^\prime}{(2\pi)^3}\frac{1}{\sqrt{2E_{\textbf{p}^{\prime}}}}\Big(a_{\textbf{p}^{\prime}}e^{i\textbf{p}^{\prime}\cdot\textbf{x}}+a_{\textbf{p}^{\prime}}^\dagger e^{-i\textbf{p}^{\prime}\cdot\textbf{x}}\Big)\sqrt{2E_{\textbf{p}}}a_{\textbf{p}}^\dagger|0⟩ \\ &= \int\frac{d^3p^\prime}{(2\pi)^3}\sqrt{\frac{E_{\textbf{p}}}{E_{\textbf{p}^{\prime}}}}\Big(⟨0|a_{\textbf{p}^{\prime}} a_{\textbf{p}}^\dagger |0⟩e^{i\textbf{p}^{\prime}\cdot\textbf{x}}+⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩ e^{-i\textbf{p}^{\prime}\cdot\textbf{x}}\Big) \end{aligned} $$ Focusing on $⟨0|a_{\textbf{p}^{\prime}} a_{\textbf{p}}^\dagger |0⟩$ and $⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩$, I get that the first term is equal to $$ ⟨0|a_{\textbf{p}^{\prime}} a_{\textbf{p}}^\dagger |0⟩=⟨\textbf{p}^\prime|\textbf{p}⟩=(2\pi)^3\delta^{(3)}(\textbf{p}^\prime-\textbf{p}) $$

For the second term I'm not as confident to what it should equal to? I'm trying to use the following trick and write $⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩$ as $(a_{\textbf{p}^{\prime}}|0⟩)^\dagger a_{\textbf{p}}^\dagger|0⟩$ and use the fact that $a_{\textbf{p}^{\prime}}|0⟩=0$, but I don't know if this is the proper way to do it.

If $⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩$ is zero I arrive at the desired result.

$$ \begin{aligned} ⟨0|\phi(\textbf{x},t)|\textbf{p}⟩ &= \int\frac{d^3p^\prime}{(2\pi)^3}\sqrt{\frac{E_{\textbf{p}}}{E_{\textbf{p}^{\prime}}}}(2\pi)^3\delta^{(3)}(\textbf{p}^\prime-\textbf{p})e^{i\textbf{p}^{\prime}\cdot\textbf{x}}\\ &=e^{i \textbf{p}\cdot\textbf{x}} \end{aligned} $$ Basically I need someone to confirm if my proof of $⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩=0$ is correct or if there's any other property that makes $⟨0|a_{\textbf{p}^{\prime}}^\dagger a_{\textbf{p}}^\dagger|0⟩$ zero.

  • 1
    $\begingroup$ That is correct and it is exactly the expected reasoning. $\endgroup$ Oct 5 '17 at 12:28
  • 2
    $\begingroup$ You should however be a bit mot careful in derivations. You first use $| \mathbf{p} \rangle = \sqrt{2 E_\mathbf{p}} a^\dagger_\mathbf{p} | 0 \rangle$, and then $\langle 0 | a_{\mathbf{p}'} a_\mathbf{p}^\dagger | 0 \rangle = \langle \mathbf{p}' | \mathbf{p} \rangle$, which is inconsistent. $\endgroup$
    – Darkseid
    Oct 5 '17 at 14:07
  • $\begingroup$ Good note @Darkseid $\endgroup$ Oct 5 '17 at 14:08

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