# Why is QFT insensitive to absolute energies?

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1. Aug 19, 2015

### "Don't panic!"

In the canonical formulation of QFT (to which I've been exposed), it is always argued that only differences in energy are physically observable and so we can deal with the fact that the vacuum energy is infinite by redefining the vacuum such that its energy is zero and we subsequently measure all energies relative to it. Why is this the case? Why are only energy differences physically observable (neglecting gravity, as one does in ordinary QFT). Is it because the vacuum energy is spatially homogeneous (i.e. constant throughout 3D space), it is in a sense "normalised" such that, for intents and purposes, one can only measure a difference in energy relative to it?

2. Aug 19, 2015

### The_Duck

The Schrodinger equation is

$i \hbar \frac{d}{dt} | \psi(t) \rangle = \hat H |\psi(t) \rangle$

Suppose we have some solution $|\psi(t)\rangle$ to this equation. Now let's add an overall constant $\delta$ to the Hamiltonian:

$i \hbar \frac{d}{dt}|\psi'(t)\rangle = (\hat H + \delta) | \psi'(t)\rangle$

You can check that a solution to this new equation is

$|\psi'(t)\rangle = e^{-i\delta t/\hbar} |\psi(t)\rangle$

The new solution differs from the original solution only by an overall phase. Overall phases are unobservable. So a constant shift in the Hamiltonian produces no observable effects on the dynamics of the system.

3. Aug 19, 2015

### "Don't panic!"

So as far as QFT is concerned is the vacuum energy deemed to be unphysical (a relic of the mathematical description)?

4. Aug 19, 2015

### The_Duck

Yes. Since we have the freedom to shift the Hamiltonian without affecting the dynamics, we can choose to shift it such that the energy of the vacuum state is zero, or any other desired value. So the absolute vacuum energy is arbitrary and it's fair to call it unphysical.

5. Aug 19, 2015

### "Don't panic!"

Ok great, thanks for your help.
Just one more question though. When one includes gravity is the reason why the vacuum energy becomes physical because gravity is sensitive to absolute energies - all forms of matter and energy affect geometry and hence gravitate?

6. Aug 19, 2015

### The_Duck

Right. Have a look at the action for classical gravity. Here $R$ is the Ricci scalar, $\Lambda$ is the cosmological constant, and $\mathcal{L}_M$ is the Lagrangian density for matter and fields (excluding the gravitational field). If we change the vacuum energy of some field, this is basically adding a constant to the Lagrangian of that field and hence adding a constant to $\mathcal{L}_M$. In the gravitational action this is the equivalent to shifting the cosmological constant $\Lambda$. So once fields are coupled to gravity, shifting their vacuum energy *does* affect the overall dynamics.

7. Aug 19, 2015

### "Don't panic!"

Right, but even before considering the vacuum energy, gravity is sensitive to absolute energies isn't it? (As energy density, which is absolute, crops up in the stress-energy tensor $T_{\mu\nu}$)
Slightly off topic, but why is the stress-energy tensor defined as $$T_{\mu\nu}:=-\frac{2}{\sqrt{-g}}\frac{\delta (\sqrt{-g}\mathcal{L}_{M})}{g^{\mu\nu}}$$

8. Aug 20, 2015

### The_Duck

Sure. An overall shift in the energy density *does* affect gravity, by effectively shifting the cosmological constant, as I described above. So gravity is sensitive to the absolute energy density.

One way of thinking about it is the following. Consider electromagnetism. In electromagnetism the electromagnetic field couples to the electric current. In fact we can define the electric current as whatever couples to the electromagnetic field, which is to say, whatever multiples the electromagnetic potential in the Lagrangian. Mathematically, we write this as:

$$j^\mu = \frac{\delta \mathcal{L}_{M}}{\delta A_\mu}$$

where $A_\mu$ is the electromagnetic potential and $\mathcal{L}_M$ is the Lagrangian of whatever matter is coupled to the electromagnetic field. If you have seen the Lagrangian for QED you can verify that this gives the right electric current.

The equation you wrote is basically the same but for gravity. Basically it says: "the stress-energy tensor is whatever couples to the metric tensor, which is to say, whatever multiplies $g_{\mu\nu}$ in the Lagrangian."

Last edited: Aug 20, 2015
9. Aug 20, 2015

### "Don't panic!"

So is the definition for the stress-energy tensor in terms of the variation of the matter Lagrangian wrt the metric deduced by analogy to other field theories and also the requirement that a variation of the total action for GR should reproduce Einstein's equation?

10. Aug 20, 2015

### Jano L.

But the shift is infinite in the case of Poynting EM energy. There is a substantial difference between shifting the energy scale by finite number (trivial) and eliminating a highly problematic term from the Hamiltonian (non-trivial, changes the usability of the Hamiltonian).

Note that even if the zero-point terms are removed by hand from the Hamiltonian, the lowest eigenfunctions of harmonic oscillators still give non-zero 2nd moments (variances) for the canonical variables (vacuum fluctuations).