Time-ordered product of real scalar fields

[Dixanadu]

Hi guys,

So I've got a real scalar field which is the sum of the positive frequency part and negative frequency part:

$\phi(x)=\phi^{(+)}(x)+\phi^{(-)}(y)$

and I'm looking at the time-ordered product:

$T(\phi(x)\phi(y))=\theta(x^{0}-y^{0})\phi(x)\phi(y)+\theta(y^{0}-x^{0})\phi(y)\phi(x)$

for the two cases where $x^{0}>y^{0}$ and $x^{0}<y^{0}$. So in the first case:

$T(\phi(x)\phi(y))=:\phi(x)\phi(y):+[\phi^{(+)}(x),\phi^{(-)}(y)]$, that's all good.

However for the second case, here is what my lecturer has written:

$T(\phi(x)\phi(y))=:\phi(x)\phi(y):+\theta(x^{0}-y^{0})<0|\phi(x)\phi(y)|0>+\theta(y^{0}-x^{0})<0|\phi(y)\phi(x)|0>$.

I have no idea how/why all of a sudden the expectation value is being taken using the vacuum states. Can someone please explain, why is this so different from the first case?

 

[AndyW]

Hi there,

It seems like your lecturer is using the concept of the vacuum expectation value in order to calculate the time-ordered product in the second case. The vacuum expectation value is a mathematical tool used in quantum field theory to calculate the expectation value of a field operator in the vacuum state.

In the first case, where x^{0}>y^{0}, the time-ordered product only involves the field operators themselves and their commutator. However, in the second case, where x^{0}<y^{0}, the time-ordered product also involves a step function, which can be interpreted as the Heaviside function \theta(x^{0}-y^{0}). This function essentially "turns off" the second term in the time-ordered product when x^{0}<y^{0}, as it becomes zero. This is why your lecturer has introduced the vacuum expectation value in order to calculate the expectation value of the second term, which is now the only term present in the time-ordered product.

I hope this helps to clarify the use of the vacuum expectation value in the second case. Please let me know if you have any further questions or if anything is unclear.