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Yes, you just have to check that $e^{-i\beta P/\hbar}|q\rangle$ is an eigenket of $Q$ (uppercase represents operators) with eigenvalue $q+\beta$. We evaluate $$Q \big ( e^{-i\beta P/\hbar}|q\rangle \big )$$ by using the first identity (which is actually a consequence of the canonical commutation relation): $$Q e^{-i\beta P/\hbar} = e^{-i\beta P/\hbar}Q + ... 6 I am gonna give a much shorter answer than @Cyro. Yes it is also true for the other components of angular momentum. It is simply due to vector addition. The total angular momentum of the system (were it classical) would be \mathbf{J} = \sum_i\mathbf{J}_i. For quantum, it's the same thing, but you just quantise the observable into an operator. The "catch" ... 5 I think this is a great question. It also puzzled me for a while. The key here is irreducible representations of the rotation group. You start with one quantum particle, the state of this quantum partile is |\psi\rangle_1 which is a vector in some Hilbert space \mathcal{H}_1. You also have a set of operators \exp\left(iJ_{1x}\theta_x\right),\, \exp\... 4 Obviously, no, for a simple counting argument: The entropy is only a single number, while states for which \rho=U\sigma U^\dagger must have identical spectrum, which is specified (up to trace) by d-1 numbers. So if the Hilbert space has dimension d>2, there are counter-examples. (Just take any two diagonal states with different entries but same ... 4 Are -|0\rangle, i|0\rangle, \frac{1+i}{\sqrt2}|0\rangle valid (that is, physically possible) kets? How are they different from |0\rangle? Yes, they are valid, but describe the same physical state. A state can be characterized by its expectation values, and a phase factor e^{i\varphi} cancels with the complex-conjugate factor from the bra. ... 4 The point is that a physical observable corresponds to a basis of state vectors, which are assigned definite values of that observable. For example, position is a physical observable, and a state |x \rangle whose wavefunction is concentrated at x has a definite value of position, namely x. Now, a Hermitian operator happens to have a basis of ... 4 It measures the (square of the) magnitude of the spin angular momentum, ignoring the spin’s direction. 3 I may have misunderstood your question, but bear with me. I don't think your calculations are right at all. I agree with you up to$$\hat{A} = \sum_n \sum_m \left<\phi_n\right|\hat{A}\left|\phi_m\right> \left| \phi_n\right> \left<\phi_m\right|$$Now, using your initial basis, \begin{pmatrix}1 \\ 0\end{pmatrix} and \begin{pmatrix}0 \\ 1\end{... 3 The reason is that, in my view, this representation is not the representation of any observable: there is no observable which is multiplicative in this representation. Instead, the creation and annihilation operators (which are not observables) take a simple form. Therefore this representation is useful in contexts where those operators play an important ... 3 Without any further context, there isn't any difference because you can always redefine |1 \rangle by a sign to go from one to the other. But once you fix the definitions of |0\rangle and |1 \rangle, there is a difference. For example, if |0 \rangle represents photon polarization along the north direction and |1 \rangle is along the east ... 3 A Fock space is just one special construction of a Hilbert space. The basic idea is that the Fock space allows you to superpose tensor products of distinct degree. In other words, it allows you to make sense of expressions of the form$$|a\rangle+|b\rangle\otimes |c\rangle.where |a\rangle,|b\rangle,|c\rangle are one-particle states. From the quantum ... 2 \newcommand{ket}[1]{\left|#1\right>}Marcus has given the mathematical answer using the commutation rules of ladder operators a_p and a_q^\dagger. I want to give a bit the intuition behind the operator P^\mu and how you can almost guess the answer. Note that this intuition is solidified by the calculation that Marcus gives and in no way replaces it.... 2 Use the canonical commutation relation [a_p, a^{\dagger}_q] = (2\pi)^3\delta^3(p-q): We have that a_{p_2}p^{\mu}a^{\dagger}_pa_pa^{\dagger}_{p_1} =p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}a_p + p^{\mu}a_{p_2}a^{\dagger}_p[a_p, a^{\dagger}_{p_1}] The first term is ignored, because when considering <0|p^{\mu}a_{p_2}a^{\dagger}_pa^{\dagger}_{p_1}... 2 I think there's a misprint in your question so I'll go with what I think is right. Using the coupled basis (which is already permutation symmetric) \begin{align} \vert 11\rangle = \vert +\rangle_1\vert +\rangle_2\, ,\qquad \vert 10\rangle &= \frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2 +\vert -\rangle_1\vert +\rangle_2\right)\, ,\quad \vert ... 2 They are used extensively in so-called coherent state (and vector coherent state) representations, albeit the form of the inner product that you give is bypassed for explicit calculations. Indeed the primary interest of this is the simplicity of representing creation and destruction operators in terms of differential operators and multiplication by Bargmann ... 2 There is no need to symmetrize or antisymmetrize the tensor product of the spin spaces. Each spin is tied to a particlular site in the lattice, so different spins are distinguishable. It would be different if the objects with spin were able to hop/tunnel from site to site. Then their statistics would matter. 2 This is yet more of why I favor a subjective and information-based understanding of quantum theory. First off, I should point out that what you have come up with is really more of taking the experiment a bit too literally or strictly, when what it really is is a conceptual issue with quantum theory that, while as you've pointed out runs into some troubles ... 2 OP asks good questions. Let us try to sketch the logic of the LSZ reduction formula. In the Heisenberg picture, the real field \varphi(x) has a Fourier expansion \varphi(x)~=~\int \widetilde{dk}\left[a({\bf k})e^{ik\cdot x}+ a^{\dagger}({\bf k})e^{-ik\cdot x} \right], \tag{3.19} $$where k\cdot x= {\bf k}\cdot {\bf x}-\omega_{\bf k}t and \omega_{... 2 If you want X have eigenvalue x_0 at t_0 you should take either t_0=0 or take U(t)= \exp\{-iH(t-t_0)\} so that U(t)={\rm Id} at t=t_0. 2 Yes this is the right way to look at this except you need U(s) instead of U(s-\epsilon) in your formula. If you have a map s\mapsto U(s) into the set of bounded operators on a Hilbert (e.g., unitary operators), then you have two main approaches for defining the derivative$$ \frac{d}{ds}U(s)=\lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}(U(s+\epsilon)-...

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In order to have dissipation you need somewhere for the energy to go, and this somewhere has to be included in the quantization process. In addition to absorbing the energy, the extra system causes quantum decoherence and so makes the problem quite tricky. The damped harmonic oscillator is one of the problems that were attacked and solved by Caldeira and ...

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First, the one-particle state localized in the point, e.g. $\phi(x)|0\rangle$ is very singular. What behaves more nicely insteas is a smeared state $\int_V d^D x \eta(x) \phi(x)|0\rangle$. However if we look at this "particle at a point" we will see, $$\phi(0)=\int \frac{d^{D-1}p}{(2\pi)^D\sqrt{2 E_p}} a^\dagger(\vec{p})|0\rangle$$ I.e. to produce ideally ...

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If we have a rotating body, it's classical rotational energy is $$E_{rot} = \frac{L^2}{2\theta}$$ Therefore, if we include a rotational particle in quantum mechanics it seams "natural" to use the same term and replace $L^2$ by it's operator -- the mathematical argument involves some group theory and so called generators [see e.g. the book of Sakurai for a ...

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In a photonics context, a standard method to generate entangled photons is SPDC. This generates a pair of photons whose polarisation are entangled. This is "practical" enough to be standard in quantum optics laboratories. Note that the direction in which the photons are emitted can be controlled, and thus the photons used for whatever protocol one is trying ...

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Let's be more careful: $$K(q,t|q^\prime, t^\prime)=\int_{q(t)=q}^{q(t^\prime)=q^\prime} Dq(t) \exp\Big[i\int_0^Tdt(\frac{1}{2}m\dot{q}^2-V(q)\Big].$$ In this integral you need integrate over all continuous trajectories with boundary conditions at $t$ and $t^\prime$: $q(t)=q$ and $q^\prime(t^\prime)=q^\prime$. Using $K$ you easily can calculate evaluation ...

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