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Yes, if it's in a pure momentum eigenstate, which is the only way you would obtain a single eigenvalue of the momentum operator. There are complications due to the fact that you can't really have a system in a pure momentum state (though you can get arbitrarily close). But conceptually, that's what it means to have a single eigenvalue. Note: applying the ...

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Under a unitary $U$, for which $U^\dagger U = UU^\dagger = \mathbb I$, any arbitrary operator $A$ transforms as $$A \mapsto A' = U^\dagger AU.$$ You're being asked to show that if $C=[A,B]$, then $C'=[A',B']$, or in other words, that $$U^\dagger[A,B]U = [U^\dagger AU, U^\dagger BU].$$ The proof is simple but it's for you to work out.

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No. Mathematically multiplying your state vector by an operator is not how you determine the outcome of a measurement. First, that would assume you can get a deterministic value, which we know isn't true of measurements of quantum systems. Second, if your system is (most likely) in a superposition of momentum states, then this operation will not even give ...

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The number operator does not commute with the position operator. We have $$\hat{H}=\frac{\hat{P}^2}{2m}+\frac{1}{2}m\omega^2\hat{X}^2=\left(a^\dagger a+\frac{1}{2}\right)\hbar\omega=\left(\hat{N}+\frac{1}{2}\right)\hbar\omega$$ Then, $$\hat{N}=\frac{\hat{H}}{\hbar\omega}-\frac{1}{2}$$ where $\hat{H}$ is the Hamiltonian for the harmonic oscillator. Using $[\... 2 Each pure State$|\psi_n\rangle$of the quantum system is a multi-particle state describing all particles of the system. So your pure states are all possible states of the system. Since it is not prepared in a state$|\psi\rangle$there are plenty of possibilites of states in which your systems could be. Let us say you prepared your system in$|\psi\rangle$... 0 You're almost there. Note that $$T(\delta x) \delta x = \delta x + \mathcal O(\delta x^2)$$ so if you stick to first order (i.e. consider infinitesimal$\delta x$), your result and the more general result from wikipedia agree. 3 Question (1) Yes,$A$is in a pure state. By definition, a state$\rho$is pure if and only if it is a projector, that is$\rho^2 = \rho$. This is equivalent to$\mathrm{Tr}(\rho^2) = 1$. From this, it is clear that$\rho_A$is a pure state. Question (2) I suppose you want an explanation of the definitions behind the notion of entanglement. If$\rho$is ... 2 The requested lower limit is zero already for$X$and$P$as I am going to prove. Let us consider the Fourier-Plancherel transform$F: L^2(\mathbb{R},dx)\to L^2(\mathbb{R},dx)$, formally for integrable functions (otherwise a further extension is necessary) $$(F\psi)(x) = \frac{1}{(2\pi)^{1/2}} \int_{\mathbb R} e^{ixy} \psi(y) dx$$ It is clear that if$\psi$... 0 The Hamiltonian operator in quantum field theory is constructed by using the fields in question. For example, the Hamiltonian for a free scalar field reads $$\notag H = \frac{1}{2} \int_V d^3x \left( \pi^2 + ( \partial_i \phi )^2 + m^2 \phi^2 \right) \, .$$ The key observation is now that a given field evaluated at different ... 2 As mentioned in the comments, are there any operators where I cannot simplify$\langle \mathbf{r}|\hat O|\mathbf{r}'\rangle$to$O(\mathbf{r})\delta(\mathbf{r}-\mathbf{r}')$and obtain $$\hat{O}=\int d^3r\,\hat\psi^\dagger(\mathbf{r})O(\mathbf{r})\hat\psi(\mathbf{r})?$$ Yes, there are ─ the momentum operator is the trivial example, and it cannot be ... 1 Basically there is nothing that prevents you from writing such a term. The question is whether such non-local one particle operator will occur in a physical system. On the face of it, this operator 'teleports' a particle from${\bf r}$to${\bf r}'$instantly. Therefore, normally we would suspect any such single particle operator as unphysical. However, in ... 1 Yes. You are correct. We can write operator$A$as $$A = \left[\lambda_1 |\phi_1 \rangle \langle \phi_1| + \lambda_2\sum |\phi_n\rangle \langle \phi_n| \right] \otimes I_b$$ as it acts as the identity operator on subsystem$b$. Consequently, it doesn't matter what is the basis in which we write subsystem$b$, as the identity operator is the same in every ... 3 It may not be an eigenstate of$\hat L_z$but, if the system is in a pure state, it will be an eigenstate of$\hat L_{\hat n}=\hat n\cdot \vec L$, i.e. it will be an eigenvector of the projection of angular momentum in some direction$\hat n$. It may be tricky to find this direction but one way might be to make a beam that travels through weak magnetic ... 6 There seems to be some sort of misunderstanding here. Making a measurement of observable$A$of a system in the state$|\psi\rangle$does not mean we need to get a number from the calculation$A|\psi\rangle$. The issue here is that$A|\psi\rangle$is still a vector. If you are expecting the measurement to give a value of$a$for$A|\psi\rangle=a|\psi\rangle$... 5 We can always move to the eigenbasis given a hermitian operator like$L_z, H, L^2, \cdots.$There is a theorem called Spectral theorem, which states that there exists an eigenbasis given hermitian operator, and this applies to finite dimensional, infinite dimensional, including Hilbert spaces. Thus, since we have an eigenbasis, we can expand$|\psi\rangle=...

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Briefly, the action (Lagrangian$^1$ density) in the path integral are functionals (functions), respectively, as opposed to operators. This is a consequence of how the path-integral formalism is derived from the operator formulation (by inserting infinitely many completeness relations). The action & Lagrangian density usually don't depend on Planck's ...

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The main point is that the 2-pt function $\langle {\cal O}_1{\cal O}_2\rangle$ acts a non-degenerate bilinear form on the infinite-dimensional vector space of linear operators (which consists of primary operators and descendants thereof), i.e. $$[\forall {\cal O}_2:~~ \langle {\cal O}_1{\cal O}_2\rangle~=~0]\quad\Rightarrow\quad {\cal O}_1~=~0.$$ It ...

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My quantum mechanics is a little rusty but I will attempt an answer. The problem is that the Hamiltonian is not Hermitean. A Hermitean operator in this case would have two eigenvectors. You need to add the hermitean conjugate of the S_{-}S_{+} term to the hamiltonian. From what I remember this is common practice in condensed matter physics.

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