Measurement of the phase of coherent states

[Jamister]

TL;DR

It is commonly said that phase can't be measured, just the relative phase. why?

It is commonly said that the phase of coherent states can't be measured, just the relative phase between two coherent states.

A qubit example: define the states
$$|\phi\rangle=[|0\rangle+\exp (\mathrm{i} \phi)|1\rangle] / \sqrt{2}$$
and the measurement operators
$$\hat{M}_{\phi}=|\phi\rangle\langle\phi| / \sqrt{\pi}$$
and
$$\hat{E}_{\phi}=\hat{M}_{\phi}^\dagger \hat{M}_{\phi}=\frac{1}{\pi}|\phi\rangle\langle\phi|$$
The completeness relation holds
$$\int_{0}^{2 \pi} \mathrm{d} \phi \hat{E}_{\phi}=|0\rangle\langle 0|+| 1\rangle\langle 1|=\hat{1}$$
Therfore under the measurement:
$$\rho\rightarrow\int\hat{M}_{\phi}\rho\hat{M}_{\phi}^{\dagger}\text{d}\phi$$
I expect to measure the phase (with some error).

The coherent states hold a similar completeness relation relation
$$\frac{1}{\pi} \int|v\rangle\langle v| \mathrm{d}^{2} v=\hat{1}$$
and therefore I expect that it is also possible to measure the phase of coherent states.

 

[vanhees71]

It's not clear to me what you want to achieve physics wise, but what you "measure" here is a relative phase, since in $|\phi \rangle$ it's the phase difference between $|0 \rangle$ and $|1 \rangle$.

An overall phase of the state ket is unobservable, because in quantum theory it's not the phase ket itself that represents the corresponding pure state, but the statistical operator $\hat{\rho}_{\psi} = |\psi \rangle \langle \psi|$, for which any overall phase factor cancels. It's important to keep in mind that the statistical operator (or equivalently the unit ray defined by the state ket) represents the state, not the state ket itself.

 

[Jamister]

It's not clear to me what you want to achieve physics wise, but what you "measure" here is a relative phase, since in $|\phi \rangle$ it's the phase difference between $|0 \rangle$ and $|1 \rangle$.

An overall phase of the state ket is unobservable, because in quantum theory it's not the phase ket itself that represents the corresponding pure state, but the statistical operator $\hat{\rho}_{\psi} = |\psi \rangle \langle \psi|$, for which any overall phase factor cancels. It's important to keep in mind that the statistical operator (or equivalently the unit ray defined by the state ket) represents the state, not the state ket itself.

Sorry, I was not clear in my statements, that's not what I meant. What I meant is what you called the relative phase. For coherent states I know for sure that there is no meaning in measuring this phase. I edited the post above

 

[Jamister]

It is commonly said that the phase of coherent states can't be measured, just the relative phase between two coherent states. but I show here that there are measurement operators that can measure phase. so what is wrong?

Define the measurement operators
$$\hat{M}_{\phi}=|\phi\rangle\langle\phi| / \sqrt{\pi}$$
and
$$\hat{E}_{\phi}=\hat{M}_{\phi}^\dagger \hat{M}_{\phi}=\frac{1}{\pi}|\phi\rangle\langle\phi|$$
The completeness relation holds
$$\frac{1}{\pi} \int|v\rangle\langle v| \mathrm{d}^{2} v=1$$
Therefore under the measurement:
$$\rho\rightarrow\int\hat{M}_{\phi}\rho\hat{M}_{\phi}^{\dagger}\text{d}\phi$$
I expect to measure the phase (with some error).

 

[Jamister]

I made a new post explaining what I mean.

 

[Dale]

It is commonly said

Please provide a reference for this

 

[Twigg]

I'm going to second the request for a reference, since I don't understand why that statement would be specific to coherent states.

However, I can respond to this:

but I show here that there are measurement operators that can measure phase. so what is wrong?

Just because a quantity $Q$ has a corresponding observable operator $\hat{Q}$ does not mean that it's physically meaningful. For example, potential energy is meaningless unless you have a point of reference (setting potential energy to 0 as $r\rightarrow \infty$, for example). Phase works the same way. You can measure phase differences, as in interferometry (e.g., double slit experiment or Michelson interferometer); however an absolute phase just doesn't hold any meaning until you compare it to another reference phase.

 

[Cthugha]

The completeness relation holds
$$\frac{1}{\pi} \int|v\rangle\langle v| \mathrm{d}^{2} v=1$$

What is your integration range? If you allow the range outside of the range between 0 and 2 pi, the states will not be orthogonal. If you limit the range to the range between 0 and 2 pi, you will get a photon number spectrum that must necessarily extend towards minus infinity which is obviously wrong.

Details are given, e.g., in Carruthers and Nieto, Phase and Angle Variables in Quantum Mechanics, Rev. Mod. Phys. 40, 411 (1968). Online version of the paper
The relevant discussion starts on the right side of 428.

A meaningful phase measurement is typically done via two orthogonal quadrature measurements - which in turn needs some relative phase to define the quadrature axes.

 

[vanhees71]

You also find some basic thoughts in the parallel thread:

https://www.physicsforums.com/threads/photon-interference-and-beamforming.1016452/post-6648164

 

[Demystifier]

It is commonly said that the phase of coherent states can't be measured, just the relative phase between two coherent states. but I show here that there are measurement operators that can measure phase. so what is wrong?

Define the measurement operators
$$\hat{M}_{\phi}=|\phi\rangle\langle\phi| / \sqrt{\pi}$$
and
$$\hat{E}_{\phi}=\hat{M}_{\phi}^\dagger \hat{M}_{\phi}=\frac{1}{\pi}|\phi\rangle\langle\phi|$$
The completeness relation holds
$$\frac{1}{\pi} \int|v\rangle\langle v| \mathrm{d}^{2} v=1$$
Therefore under the measurement:
$$\rho\rightarrow\int\hat{M}_{\phi}\rho\hat{M}_{\phi}^{\dagger}\text{d}\phi$$
I expect to measure the phase (with some error).

The problem is that your $\hat{E}_{\phi}$ operators are not complete, i.e.
$$\int d\phi \,\hat{E}_{\phi} \neq 1$$