# Time-ordered calculation for equal time

I have a question about how to calculate the following expectation value:

$$\langle0|\mathcal{T}\{{a^{\dagger}}(0,0) a(0,0)\}|0\rangle$$ where $|0\rangle$ is the ground state and $a^{\dagger}(x,t)$ is the creation operator which creates a particle at position $x$ and time $t$. Is the answer just $0$ since time-order operator does nothing in this case?

By definition : $\langle0|\mathcal{T}\{\Phi(x) \Phi(y)\}|0\rangle = \theta(x^0-y^0) \langle0|\Phi(x) \Phi(y) |0\rangle + \theta(y^0-x^0) \langle0|\Phi(y) \Phi(x) |0\rangle$

Now, we may take $\theta(0) = \frac{1}{2}$, so in our special case, we would have :

$\langle0|\mathcal{T}\{{a^{\dagger}}(0,0) a(0,0)\}|0\rangle = \frac{1}{2} \langle0|a^{\dagger}(0,0) a(0,0)|0\rangle + \frac{1}{2} \langle0|a(0,0) a^{\dagger}(0,0)|0\rangle \\ = \frac{1}{2} \langle0|a(0,0) a^{\dagger}(0,0)|0\rangle$

With $a(0,0) = \int \frac{d^3k}{\sqrt{(2\pi)^3 2E_k}} a(\vec k), \quad a^{\dagger}(0,0) = \int \frac{d^3k}{\sqrt{(2\pi)^3 2E_k}} a^{\dagger}(\vec k)$, and $[a(\vec k, a^{\dagger}(\vec k')]= \delta^3(\vec k - \vec k')$, we would have :

$\langle0|\mathcal{T}\{{a^{\dagger}}(0,0) a(0,0)\}|0\rangle = \frac{1}{2} \int \frac{d^3k}{(2\pi)^3 2E_k}$

• It's not clear to me why you passed to momentum representation. The state $|0>$ is apparently in the x-space so it's better to stay with the x representation of $a$ and $a^{\dagger}$. I would say that answer is just $\frac{1}{2}[a,a^{\dagger}]$. – Yair Nov 8 '13 at 21:18
• @Yair : The state $|0\rangle$ represents the vacuum; it is not in "$x$ space" or "$p$ space". A application of the creation operator $a^{\dagger}(\vec k)$ applied to the vacuum creates a one-particle state of momentum $k$. Creation operator at a position $x$ and time $t$ are defined as Fourier transform of the creation operator $a^{\dagger}(\vec k)$. – Trimok Nov 9 '13 at 9:15
• You're right, the representation is not important here. But still, I would just say it's $\frac{1}{2}[a,a^{\dagger}]=\frac{\delta(x)}{2}$ where x denotes a vanishing vector. Anyway, I think both options are equivalent. – Yair Nov 9 '13 at 14:13