3

The spin is undergoing Larmor precession about the magnetic field, so yes the $x$ and $y$ components will change while the component parallel to the field remains fixed..


2

I think he is implicitly assuming that the filled Fermi sea $\vert{\phi_0}\rangle$ is a good approximation for the ground state of the weakly interacting Hamiltonian. Therefore he is assuming that under time evolution, this state only picks up a phase. In terms of Green's functions in terms of Feynman diagrams, we can say that he is approximating the ...


2

The Ehrenfest theorem uses the expected value of the commutator. i.e. for some operator $\hat{A}$ the Ehrenfest theorem says $$\frac{d}{dt} \langle\hat{A}\rangle = \frac{1}{i\hbar} \langle[\hat{A}, \hat{H}]\rangle + \langle\frac{\partial\hat{A}}{\partial t}\rangle.$$ In your case while $[\hat{S}_x, \hat{H}] \neq 0,$ its expectation value will still be zero ...


2

If H depends on $\lambda$ then so to do its its eigenvalues. That is $E = E(λ)$. And since $E$ is the eigenvalue corresponding to the state $|E\rangle$ then $E(\lambda) |E\rangle = H(\lambda)|E\rangle$ or $E(\lambda)\langle E|E\rangle = \langle E|H(\lambda)|E\rangle$ or $E(\lambda) = \langle E|H(\lambda)|E\rangle$ due to orthonormal condition (we could have ...


2

It is true in this specific case that, if $\hat H=\frac{1}{2m}\mathbf{p}^2$ is the Hamiltonian for the free particle, then $U^\dagger\hat{\mathbf{p}}^2 U=\hat{\mathbf{p}}^2$ for $U$ a rotation. Maybe the 2d case is sufficient to illustrate the point. We have \begin{align} R^\dagger_z(\theta)\hat{p}_xR(\theta)&=\hat{p}_x^\prime=\hat{p}_x\cos\theta+\hat{p}...


2

If the operator $U$ is indeed the unitary operator, then $U U^{\dagger}= U^{\dagger} U = 1$ then expression you have (which appears to be applied incorrectly) above will reduce back to its original form $\frac{1}{2m} U^{\dagger} p^2 U = U^{\dagger} H U = H$


1

The energy function is not always equivalent to the energy of the system. Consider the case when it happens to be: Suppose we have lagrangian given by $$\mathcal{L}= \mathcal{L}_0+\mathcal{L}_1+\mathcal{L}_2 $$ where $$\mathcal{L}_0=\mathcal{L}_0(q,t)$$, $$\mathcal{L}_1=\sum_{i}b_i\dot{q}_i$$ and $$\mathcal{L}_2=\sum_{i,j}c_{ij}\dot{q}_i\dot{q}_j$$ In this ...


1

The statement in question is made in a specific context - that of electrons in condensed matter, which interact via Coulomb interaction, which is quartic in fermion operators. Physically, a quadratic Hamiltonian does not necessarily correspond to non-interacting particles: e.g., bosonization approach reduces Hamiltonians to interactions to quadratic ones. ...


1

Sketch C is drawn correctly for a glass of water under horizontal acceleration. As it is accelerating gravity is not the only force acting on it, so it would not remain horizontally level. Tilting the tray would be a way to keep the glass from spilling or sliding on the tray while accelerating. However, I doubt that very many waiters would accelerate quickly ...


1

Of course they commute, in general, manifestly, $$ P_{12}P_{34} = I\otimes \sigma_1 ; \qquad P_{13}P_{24} = \sigma_1\otimes I, $$ so $$ P_{12}P_{34} ~ P_{13}P_{24}= P_{13}P_{24}~P_{12}P_{34} . $$ Of course, $$\sigma_1=\begin{bmatrix} 0& 1\\1 &0\end{bmatrix}$$ permutes the two entries of 2-vectors/spinors it acts on.


1

It won't affect the measurement, due to the fact, that identity operator commutes with everything. In Schroedinger picture: $$ \langle \psi(t) | A | \psi(t) \rangle = \langle \psi | e^{-i (H - \lambda I) t} A e^{i (H - \lambda I) t} | \psi \rangle = $$ $$ = e^{-i \lambda t }\langle \psi | e^{-i H t} A e^{i H t} | \psi \rangle e^{i \lambda t} = \langle \...


1

Let $|\Psi \rangle$ be some state in which the system is. Then the energy expectation value is: $$\langle \Psi|(H-\lambda I)|\Psi \rangle = \langle \Psi|H|\Psi \rangle - \lambda.$$ Note that $|\Psi \rangle$ is arbitrary so we see that this just corresponds to an overall shift of the energy. For comparing energy levels this does not matter but for energy ...


1

The reason is that $H_0$ as an observable is hermitian $H_0 = H_0^\dagger$. That means you can use $$\langle\Psi_b|H_0 r|\Psi_a\rangle = (\langle\Psi_a|r^\dagger H_0^\dagger|\Psi_b\rangle)^\ast = (\langle\Psi_a|r^\dagger H_0|\Psi_b\rangle)^\ast$$ You might also just say that it is part of the definition of the bra-ket notation, that operators $A$ inside act ...


1

I don't fully understand what you are after, but from the comment I understand that you assume something like $$ H_1=\operatorname{diag} (2,2,-2,-2), \qquad H_2=\operatorname{diag} (2,-2,2,-2), $$ in a space parameterized by eigenvectors $\psi_{1,2,3,4}$, that is, $H_1|\psi_1\rangle = H_2|\psi_1\rangle = 2|\psi_1\rangle$; $H_1|\psi_2\rangle = -H_2|\psi_2\...


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