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2

Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal. Notice that, by the spectral theorem, the hamiltonian can be written as follows: $$H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|$$ Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$. Notice, in particular, ...

7

Starting with: $$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$ If $t_i=0$: $$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$ Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$ $$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$ Since the exponential of an operator is (by ...

0

The spin of a right handed photon points in the direction of its momentum, while in a left handed photon it points in opposite way.

0

No, you won't see interference. The cw and ccw states are orthogonal. You can prove that intuitively in the following way. You could think of the incoming light to be cw polarized, then one waveplate would turn it vertical, the other one horizontal.

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I don't think this is a quantum optics problem. Just look up the van-Cittert-Zernike theorem . The (complex) visibility is the Fourier transform of the mutual coherence function of the source.

7

Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states  |\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ...

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