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.


1 Answer 1


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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.