The vanishing of vacuum expectation value

Question

I have some difficulty understanding why the vacuum expectation value vanishes. As illustrated in my notes, we can split the field into two parts: $$ \phi(x) = \phi^+(x) + \phi^-(x), $$ where $\phi^+(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\omega({\vec{p}})} a(\vec{p })e^{-ipx}$, and $\phi^-$ the conjugate of $\phi^+$.

Then, consider the product of fields: $$ \begin{align} \phi(x) \phi(y) &= (\phi^+(x) + \phi^- (x))\cdot (\phi^+(y) + \phi^- (y)) \\ & = \phi^+(x) \phi^+(y) + \phi^+(x) \phi^-(y) + \phi^-(x) \phi^+(y) + \phi^-(x) \phi^-(y), \end{align} $$ while the normal ordering of this product is: $$ :\phi(x) \phi(y): = \phi^+(x) \phi^+(y) + \phi^-(x) \phi^+(y) + \phi^-(x) \phi^+(y) + \phi^-(x) \phi^-(y). $$

It is clear to me that the first three terms in the normal ordering would be zero as the annihilation operator is on the right and give zero value. But why is the last term giving zero value? It only involves two creation operators. Any help would be appreciated.

Answer 1

$\phi^+(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\omega({\vec{p}})} a(\vec{p })e^{-ipx}$, and $\phi^-$ the conjugate of $\phi^+$.

It might help to write $\phi^-$ more explicitly: $$ \phi^-(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\omega({\vec{p}})} a^\dagger(\vec{p })e^{ipx} $$

$$ :\phi(x) \phi(y): = \phi^+(x) \phi^+(y) + \phi^-(x) \phi^+(y) + \phi^-(x) \phi^+(y) + \phi^-(x) \phi^-(y). $$ ...But why is the last term giving zero value?

Because $$ a(\vec p)|0\rangle = 0 $$ implies that $$ \langle 0|a(\vec p)^\dagger = 0\;. $$

So $$ \langle 0|\phi^-(x)\phi^-(y)|0\rangle =\langle 0 | \int \frac{d^3p}{(2\pi)^3}\frac{1}{\omega({\vec{p}})} a^\dagger(\vec{p })e^{ipx} \int \frac{d^3k}{(2\pi)^3}\frac{1}{\omega({\vec{k}})} a^\dagger(\vec{k })e^{iky}|0\rangle $$ $$ = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\omega({\vec{p}})} e^{ipx} \int \frac{d^3k}{(2\pi)^3}\frac{1}{\omega({\vec{k}})}e^{iky} \underbrace{\langle 0 |a^\dagger(\vec{p })}_0a^\dagger(\vec{k })|0\rangle $$ $$ =0 $$

Answer 2

The expectation has the vacuum on both sides: $<0|\phi(x)\phi(y)|0>$. When you hit the left side with $\phi^{-}$ it gets annihilated.