Energy operator in curved space-time

[jfy4]

Hi,

I was wondering what the expression would be for the energy operator in general relativistic quantum mechanics. So I attempted it for strictly diagonal metrics. Here it is:

Assume a stationary observer. Then the expression for a particle is

$$E=-g_{\alpha\beta}p^\alpha u_{obs}^{\beta}.$$

However, the observer is stationary, and must obey the time-like condition of 4-velocity, hence

$$g_{\alpha\beta}u_{obs}^{\alpha}u_{obs}^{\beta}=g_{tt}\left(u_{obs}^{t}\right)^2=-1.$$

Then

$$u_{obs}=\sqrt{-g^{tt}}$$

which implies

$$E=-g_{\alpha\beta}p^\alpha u_{obs}^{\beta}\rightarrow \sqrt{-g_{tt}}p^t.$$

Then substituting the expression for the energy operator in Minkowski space-time

$$\boxed{\hat{E}=i\hbar \sqrt{-g_{tt}}\frac{\partial}{\partial t}.}$$

This is the final form for the energy operator in an arbitrary diagonal metric. Would anyone have an idea of how to construct the momentum operator for a general metric (it can be diagonal too, anything will make me happy). Corrections and comments entirely welcomed!

 

[Matterwave]

By energy operator, do you mean the Hamiltonian?

 

[Demystifier]

By energy operator, do you mean the Hamiltonian?

He obviously doesn't. Instead, his energy operator is a kind of a general-relativistic variant of the energy operator (4) in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

 

[Demystifier]

Would anyone have an idea of how to construct the momentum operator for a general metric (it can be diagonal too, anything will make me happy). Corrections and comments entirely welcomed!

Your energy operator transforms as a scalar, not as a time-component of a vector. How would you expect the appropriate momentum operator to transform? As a 4-vector? As a 3-vector?

I think a better strategy is to construct a 4-vector energy-momentum operator that contains both energy and 3-momentum in one vector quantity. This is indeed very easy to construct, because it is given by Eq. (2) in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]
In curved spacetime the ordinary derivative should be replaced by the covariant one, but they coincide as long as they act on a wave function which transforms as a scalar (which is the case for spin 0).

Another benefit of Eq. (2) is the fact that it is observer-independent, in the sense that you don't need the vector u_obs.

 

[jfy4]

Your energy operator transforms as a scalar, not as a time-component of a vector. How would you expect the appropriate momentum operator to transform? As a 4-vector? As a 3-vector?

Good point...

I think a better strategy is to construct a 4-vector energy-momentum operator that contains both energy and 3-momentum in one vector quantity. This is indeed very easy to construct, because it is given by Eq. (2) in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

Clearly the generalization to flat-space relativistic quantum mechanics is (2). I had no problem with that from the very start, in fact I used the $p^0$ component to finish my derivation. I was hoping to construct the $p^0$ component of that 4-vector for an arbitrary Einstein metric.

In curved spacetime the ordinary derivative should be replaced by the covariant one, but they coincide as long as they act on a wave function which transforms as a scalar (which is the case for spin 0).

You are suggesting the energy-momentum operator should be expressed as

$$\hat{p}_{\alpha}\psi=\nabla_{\alpha}\psi$$

which as you also pointed out turns into

$$\hat{p}_{\alpha}\psi=\partial_{\alpha}\psi$$

if $\psi$ is a scalar. Let me pose a question back, if I may. Since we know that the shape of space (like the Schwarzschild geometry) affects the energy of a particle (the photon), would you say then that the operators can be the same, but instead it is the expectation value which varies with the geometry of space-time? Then it is the wave function that changes with the geometry of space-time, not the operators?

Another benefit of Eq. (2) is the fact that it is observer-independent, in the sense that you don't need the vector u_obs.

This would be a fine trait to have.

 

[jfy4]

Ok,

after a bit of reading I have come up with the following: As I said in my earlier post, if the operators on scalars are going to be the same regardless of the metric, then the wave functions must not be the same between the cases of curved vs flat space-time. The wave functions would have to satisfy

$$-\hbar^2 g^{\alpha\beta}\nabla_{\alpha}\nabla_{\beta}\psi+m_0c^2\psi=0$$

whereas for flat space-time it would need to satisfy

$$-\hbar^2 \eta^{\alpha\beta}\partial_{\alpha}\partial_{\beta}\psi+m_0c^2\psi=0.$$

Point being that in the first equation that the covariant derivatives may not strictly reduce down to partials because of the sum between the covariant derivatives. Is this right?

 

[Demystifier]

Ok,

after a bit of reading I have come up with the following: As I said in my earlier post, if the operators on scalars are going to be the same regardless of the metric, then the wave functions must not be the same between the cases of curved vs flat space-time. The wave functions would have to satisfy

$$-\hbar^2 g^{\alpha\beta}\nabla_{\alpha}\nabla_{\beta}\psi+m_0c^2\psi=0$$

whereas for flat space-time it would need to satisfy

$$-\hbar^2 \eta^{\alpha\beta}\partial_{\alpha}\partial_{\beta}\psi+m_0c^2\psi=0.$$

Point being that in the first equation that the covariant derivatives may not strictly reduce down to partials because of the sum between the covariant derivatives. Is this right?

Yes.

 

[Demystifier]

Let me pose a question back, if I may. Since we know that the shape of space (like the Schwarzschild geometry) affects the energy of a particle (the photon), would you say then that the operators can be the same, but instead it is the expectation value which varies with the geometry of space-time? Then it is the wave function that changes with the geometry of space-time, not the operators?

Yes.