1
$\begingroup$

When we write inner product in QM for example $\langle\psi\rvert x \psi\rangle$it means (in position space) $\int\psi^*(x,t)x\psi(x,t)dx$. But when we write, in QFT, $\langle0\rvert0\rangle=1$ what does it mean? As $\rvert0\rangle$ stands for vacuum state and since there is no particle as of now (since no creation operator is operated on it yet), why are we choosing it to be one?

$\endgroup$
  • 1
    $\begingroup$ The vacuum state still has to be normalised ... $\endgroup$ – John Rennie Nov 23 '18 at 9:34
  • $\begingroup$ @JohnRennie yeah it has to but if I think about vacuum it reminds me of just empty space with zeroth energy level so idea of normalization sounds a bit weird to me. Like in QM we back the normalization by giving a physical argument that particle has to be found in between $-\infty$ and $+\infty$ but what to say when we are dealing with the vacuum. $\endgroup$ – aitfel Nov 23 '18 at 9:42
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/8602/2451 and links therein. $\endgroup$ – Qmechanic Nov 23 '18 at 10:18
3
$\begingroup$

In the standard treatment of quantum field theory, states are treated as abstract vectors in a large Hilbert space (specifically, a Fock space), and no "wavefunction" interpretation is given to them like it is in 1D quantum mechanics.

For applications to black hole physics (and other areas of quantum gravity), however, one often refers to the "wave functional" (the probability amplitude for the field to be in a specific configuration) $\Psi[\phi(\textbf{x})]$, for which a matrix element is written as a path integral

$$\langle\Psi|\mathcal{O}|\Phi\rangle=\int\mathcal{D}\phi(\textbf{x})\,\Psi[\phi(\textbf{x})]^*\mathcal{O}[\phi]\Phi[\phi(\textbf{x})],$$

which is calculated in perfect analogy to expectation values in 1D quantum mechanics.

When we write $\langle 0|0\rangle=1$, what we're saying is that the vacuum wave functional $\Psi_0[\phi(\textbf{x})]$ is normalized with respect to this (somewhat ill defined) inner product.

$\endgroup$
1
$\begingroup$

There is no physically-meaningful relationship between the norm of a state-vector and the number of particles it contains. QFT is not constructed in terms of particles anyway; it is constructed in terms of quantum fields. Particles are phenomena that the theory predicts. The vacuum state is a physical state defined to be the state of lowest energy. Particles are defined relative to the vacuum state, and the vacuum state has no particles by definition. A typical state in QFT does not have any well-defined number of particles.

(By the way, the meaning of "particle" becomes even more blurred in QFT in curved spacetime, where there is no longer any obvious way to choose which state to call the "vacuum".)

Only non-zero vectors in the Hilbert space can be used to represent physical states, and a non-zero vector has a non-zero norm. Just like any other physical state, the vacuum state is represented by a non-zero vector in the Hilbert space.

As Bob Knighton's answer highlighted, we can use a representation in which a state-vector is a function of the field variables, and the field operators are differential operators acting on those functions. A simple but explicit example of a calculation using such a representation can be found here: https://physics.stackexchange.com/a/439381/206691. A famous published example can be found here:

To help emphasize that the vacuum state should be regarded as a physical state (and therefore should be represented by a non-zero vector in the Hilbert space), remember that even "empty space" always includes a non-zero metric field. Flat spacetime is no exception. We usually treat the metric field as a prescribed background in QFT, not as one of the quantum fields, but it is still an essential part of the model. Even though "empty space" might not have any particles, it always has at least one entity (the metric field) that is "physical" in the sense that it would interact with other fields and particles if they were present — at least it would in the real world.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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