Covariant Uncertaintly Principle

[jfy4]

Covariant Uncertainty Principle

Hi,

is it possible to write the uncertainty principle as a dot product like:

$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq\hbar$$

or even to generalize it as

$$g^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq\hbar$$

?

 

[vanhees71]

That's not so clear. What's meant by $$\Delta x$$ and $$\Delta p$$ here? Note that it is not trivial to define, e.g., position operators for massless particles with spin (e.g., there's no position operator for photons in the strict sense).

You find a good review on the implications of the uncertainty relations in the context of relativistic quantum theory in the beginning of Landau-Lifgarbages vol. IV. This goes back to a famous paper by Bohr.

 

[kof9595995]

@vanhees71: I'm curious why there's no position operator for photons in the strict sense? Could you explain?

 

[Demystifier]

Hi,

is it possible to write the uncertainty principle as a dot product like:

$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq\hbar$$

Yes, provided that you work in an extended Hilbert space as in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

But you are missing a factor of D/2=2 for D=4 spacetime dimensions. The correct relation for D=4 is
$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq 2\hbar$$

or even to generalize it as

$$g^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq\hbar$$

Yes (with the same correction).

 

[jfy4]

Yes, provided that you work in an extended Hilbert space as in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

But you are missing a factor of D/2=2 for D=4 spacetime dimensions. The correct relation for D=4 is
$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}\geq 2\hbar$$


Yes (with the same correction).

That's curious. When I first wrote it out, I had the 2, but then I thought with the lorentzian signature that the temporal component would subtract one of the the $\hbar/2$s.

Would it not contract as follows?

$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}=-\Delta t\Delta E+\Delta x\Delta p_x +\Delta y\Delta p_y +\Delta z\Delta p_z\geq -\frac{\hbar}{2}+\frac{\hbar}{2}+\frac{\hbar}{2}+\frac{\hbar}{2}=\hbar$$

 

[Demystifier]

That's curious. When I first wrote it out, I had the 2, but then I thought with the lorentzian signature that the temporal component would subtract one of the the $\hbar/2$s.

Would it not contract as follows?

$$\eta^{\alpha\beta}\Delta x_{\alpha}\Delta p_{\beta}=-\Delta t\Delta E+\Delta x\Delta p_x +\Delta y\Delta p_y +\Delta z\Delta p_z\geq -\frac{\hbar}{2}+\frac{\hbar}{2}+\frac{\hbar}{2}+\frac{\hbar}{2}=\hbar$$

No!

You should start from the canonical uncontracted uncertainty relation
$$\Delta x^{\alpha}\Delta p_{\beta}=\frac{\hbar}{2} g^{\alpha}_{\beta}$$
and observe that, in any spacetime,
$$g^{\alpha}_{\beta}=\delta^{\alpha}_{\beta}$$

 

[jfy4]

No!

You should start from the canonical uncontracted uncertainty relation
$$\Delta x^{\alpha}\Delta p_{\beta}=\frac{\hbar}{2} g^{\alpha}_{\beta}$$
and observe that, in any spacetime,
$$g^{\alpha}_{\beta}=\delta^{\alpha}_{\beta}$$

Interesting.

Where did you find that/How did you know to use that equation (inequality?)? Else, what's wrong with my contraction above?

 

[Demystifier]

Interesting.

Where did you find that/How did you know to use that equation (inequality?)? Else, what's wrong with my contraction above?

Well, the uncertainty relation above is a consequence of

1) The canonical commutation relations
$$[x^{\alpha},p_{\beta}]=-i\hbar g^{\alpha}_{\beta}$$
which is Eq. (3) in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

and

2) The fact that the canonical momentum associated with $$x^{\alpha}$$ is $$p_{\alpha}$$ (not $$p^{\alpha}$$ as one might naively think).

Your contraction is wrong because it starts from a wrong uncontracted relation. This is related to 2) above.

 

[A. Neumaier]

@vanhees71: I'm curious why there's no position operator for photons in the strict sense? Could you explain?

See the entry ''Particle positions and the position operator" in my theoretical physcs FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#position

 

[kof9595995]

See the entry ''Particle positions and the position operator" in my theoretical physcs FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#position

Thanks, though I can't really understand the proof.

 

[A. Neumaier]

Thanks, though I can't really understand the proof.

In a nutshell, the essence of the argument is the following.

A position operator for a spin s particle needs a 2s+1 dimensional spin space.
(Why this is necessary is the most difficult part of the argument.)
This is the case for massive particles. But massless particles only have a 1- or 2-dimensional spin space, depending on whether or not they are chiral.

Photons are not chiral and have spin 1, hence they have only 2 instead of the 3 spin components needed. The missing component would be ''longitudinal'' photons, which do not exist (and cannot for a massless particle).

 

[kof9595995]

In a nutshell, the essence of the argument is the following.

A position operator for a spin s particle needs a 2s+1 dimensional spin space.
(Why this is necessary is the most difficult part of the argument.)
This is the case for massive particles. But massless particles only have a 1- or 2-dimensional spin space, depending on whether or not they are chiral.

Photons are not chiral and have spin 1, hence they have only 2 instead of the 3 spin components needed. The missing component would be ''longitudinal'' photons, which do not exist (and cannot for a massless particle).

I'm curious what's the physical consequence, does it simply mean we can never measure (or even define?) the position of a photon?

 

[A. Neumaier]

I'm curious what's the physical consequence, does it simply mean we can never measure (or even define?) the position of a photon?

Not according to the textbook view of measurements. But there are more approximate notions of measurable position, defined in terms of POVMs rather than operators.
http://en.wikipedia.org/wiki/POVM

 

[jfy4]

Well, the uncertainty relation above is a consequence of

1) The canonical commutation relations
$$[x^{\alpha},p_{\beta}]=-i\hbar g^{\alpha}_{\beta}$$
which is Eq. (3) in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

and

2) The fact that the canonical momentum associated with $$x^{\alpha}$$ is $$p_{\alpha}$$ (not $$p^{\alpha}$$ as one might naively think).

Your contraction is wrong because it starts from a wrong uncontracted relation. This is related to 2) above.

Thanks,

I'll spend some time pondering it over.