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Jamister

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- TL;DR Summary
- 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.

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.

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