Stress-energy tensor in quantum field theories is defined in terms of quantum field's derivatives and Lagrangian, which itself is defined in terms of quantum fields.

This suggests that stress-energy tensor in quantum field theories must be some sort of expected (statistical) tensor rather than physical (realized) tensor. Is this correct understanding, or should we understand stress-energy tensor as physical (realized) tensor, with quantum field's derivatives understood relative to physical path taken for the quantum field, in terms of Feynman's path formalism?


Just like any observable in Quantum Mechanics, stress-energy tensor is an operator acting on the Hilbert space of the quantum field.

For example, consider the Scalar field with Lagrangian $$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \, \partial^{\mu} \phi - \frac{m}{2} \phi^2. $$

Its classical stress-energy tensor is given by $$ T_{\mu \nu} = \partial_{\mu}\phi \, \partial_{\nu} \phi - \frac{1}{2} \eta_{\mu \nu} \partial_{\sigma}\phi \, \partial^{\sigma} \phi + \frac{m}{2} \phi^2 \eta_{\mu \nu}. $$

In the Klein-Gordon QFT, the corresponding object is the stress-energy tensor operator:

$$ \hat{T}_{\mu \nu} = :\partial_{\mu}\hat{\phi} \, \partial_{\nu} \hat{\phi}: - \frac{1}{2} \eta_{\mu \nu} :\partial_{\sigma}\hat{\phi} \, \partial^{\sigma} \hat{\phi}: + \frac{m}{2} \eta_{\mu \nu} :\hat{\phi}^2:\;. $$

Here $:\dots:$ stands for normal-ordering the operator string. This is just reordering the creation and annihilation operators of which the contents in the colons are made such that the result has zero value when acting on the vacuum state: $$ :\dots:\,\left|0\right> = 0, $$ by definition.

Note that normal-ordering is a purely quantum-mechanical phenomenon. All commutators are proportional to $\hbar$ and thus vanish in the classical limit. For classical physics, the order of multipliers in the product doesn't matter.

The time derivatives in this formula can be eliminated by substituting the Heisenberg evolution equation for operators. This is what one usually does when she expands the field operators in terms of creation and annihilation operators.

Finally, by evaluating the energy-momentum 4-vector operator $\hat{P}_{\mu} = \intop d^3 x \hat{T}_{\mu 0}$ on the 1-particle states of the Fock space, you realize that these particles carry energy and momentum.

P.S. your last sentence makes little sense to me. You're doing QFT, so I assume you are familiar with Quantum Mechanics. This tensor is not "realized" or "statistical"; just like any other observable in Quantum Mechanics, it is "quantum". Which means that it is an operator used to label arbitrary states with probabilities of having values of the physical observable via the Born rule.


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.