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The mode expansion of a real massive real field is given by:

$$A^{\mu} (x)= \sum_r \int dp [ c_r(p)\epsilon^\mu_r(p) e^{-ip\cdot x} + c_r^\dagger(p)\epsilon^\mu_r(p) e^{ip\cdot x} ]$$

where $dp$ is the Lorentz invariant integration measure, $c^\dagger$ and $c$ are the creation and annihilation operators, $\epsilon^\mu_r$ are the polarisation vectors.

Now I'm asked to show the following:

$$\langle 0 \lvert A^\mu(x) \lvert p,r\rangle = \epsilon^\mu_r(p) e^{-ip\cdot x}$$

where $\vert p,r\rangle = c^\dagger_r(p) \vert 0 \rangle$. $A^\mu$ satisifies the Proca equation.

My idea to do this was to compute

$$A^{\dagger \mu} \vert 0 \rangle = \left( \sum_r \int dp (c_r^\dagger(p)\epsilon^\mu_r(p) e^{ip\cdot x} + (c_r(p)\epsilon^\mu_r(p) e^{-ip\cdot x} \right) \vert 0 \rangle = \sum_r \int dp \ \epsilon^\mu_r(p) e^{ip \cdot x } \vert p,r \rangle$$

Since I thought that the annihilation operator would give zero when acting on the vacuum and we have $c^\dagger_r |0 \rangle = |p,r\rangle$.

Taking the conjuagate I got:

$$\langle 0 | A^\mu = \langle p,r | \sum_r \int dp \ \epsilon^\mu_r(p) e^{-ip \cdot x } $$

I'm not sure if this is correct but then I would "stack" $c^\dagger_r(p) | 0 \rangle$ to the right of this expression but that doesn't seem to give me the right answer.

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In order to compute your time contraction (because it is what it is), you just have to remember that $\langle 0|a^\dagger=0$. So you have the following: \begin{align*} \langle 0 | A^\mu |p,r\rangle & =\langle 0 | A^{-,\mu} c^\dagger_r (p)|0\rangle \\ &=\int dp' e^{-ip'_\mu x^\mu} \sum_{s=1}^3 \epsilon^\mu_s \langle 0| c_s (p') c^\dagger_r(p) |0 \rangle \\ &=\int dp' e^{-ip'_\mu x^\mu} \sum_{s=1}^3 \epsilon^\mu_s \langle 0| c^\dagger_r(p) c_r (p') + \delta_{rs}\delta^{(3)}(p-p')|0 \rangle \\ &=\epsilon^\mu_r e^{-ip_\mu x^\mu} \end{align*}

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