Recent content by Jamister

I made a new post explaining what I mean.

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|...

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

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...

great! I find a way and proved it! thank you!

I want to say that the matrix is tridiagonal

I want to edit the post. why It seems like I can't do it...

yes 2x2 is easy I can find. the eigenvector $$(1,-1)$$

I don't know...

I added another condition to the question.

I tried many things... 1. using the characteristic polynomial to show that there are negative roots 2. using Sylvester's criterion to show the matrix is not a positive definite 3. using the inequalities ##\left|m_{i j}\right| \leq \sqrt{m_{i i} m_{j j}} \quad \forall i, j## (and it turns out...

I need to prove the following: A symmetric real matrix ##A## with positive elements ##a_{i,j}\geq 0## can’t be definite positive matrix (i.e. with only positive eigenvalues) if the following condition holds: $$\sum_{i=1}^{N-1}a_{i,i+1}>\frac{1}{2}\sum_{i=1}^{N}a_{i,i}=\frac{1}{2}\text{Tr}(A)$$...

In balanced homodyne detection, it is claimed that one can do state tomography. I understand most of the derivation except one part. Here is a figure describing homodyne detection. the operator that is being measured is $$ R=N_{1}-N_{2}=a^{\dagger} b+b^{\dagger} a $$. taking the mode b to be...

why in QFT 4-momentum is conserved? how can it be derived from basic principles of the Hamiltonian formalism? Is it conserved because of the golden rule?

I don't think it's obvious