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To show: $$\boxed{\left<0,0|\left[L_n,L_{-n}\right]|0,0\right>=<0,0|n^2L_0+\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left|0,0\right>=<0,0|\frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left|0,0\right>= \frac{D}{2}\sum_{m=1}^{n-1}m(n-m)\left<0,0|0,0\right> = \frac{D}{12}n(n^2-1).}$$ Why is the first term vanishing? To see that ...

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I henceforth assume $\hbar =1$. There is no reason to introduce Dirac deltas here, everything is elementary. Moreover as the function $\psi$ is not differentiable, one cannot use the form of the momentum operator $P$ as derivative which is valid only on smooth functions. Forcing this way would introduce unnecessary difficulties as the derivative must be ...

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That's the covariance of X and Y. It tells you, in a way, "how much" X and Y are correlated, with covariance 0 meaning uncorrelated (Not to be confused with independent). Because $\langle A\rangle$ and $\langle B\rangle$ can be quite large, it is customary to define the correlation coefficient: ($cov(A,B)$ is the covariance of A and B) ...

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I) One problem is that the momentum operator $\hat{p}$ is an unbounded operator, which means that it is only defined on a domain $D(\hat{p}) \subsetneq {\cal H}$ of the Hilbert space ${\cal H}=L^2(\mathbb{R})$. When we apply the differentiation operator $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ to OP's wave function $$\tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), ... 4 Well, you can conclude that something is wrong by the following logic: momentum is an observable, which means its allowed values must be things that you could read off a measuring device (assuming you had one that measures momentum). These are necessarily real values, and since the expectation value is some linear combination of possible measurements, it ... 12 The wavefunction has a discontinuity at x=-a, which gives a term -2aA i \hbar \delta(x+a) when you act with p. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ... 2 We usually say that if two operators, \hat{A} and \hat{B} commute, then they have a simultaneous set of eigenstates. Saying that the eigenstates are the same isn't really correct. For example, let operator \hat{A} be hermitian and act on elements of the Hilbert Space \mathcal{H}_A and let operator \hat{B} also be hermitian and act on elements ... 4 Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, ... 0 No, the only thing you can conclude is that \langle\psi|[H,A]|\psi \rangle =0. Example, for some real constants a,b and for a particle described in L^2(\mathbb R^3) A= aL_x, H=bL_z,$$|\psi\rangle = |\phi(r)\rangle \otimes|l=0,m_z=0\rangle\:.$$In this case [H,A] \neq 0 but \langle \psi(t)|A|\psi(t) \rangle =0 for every t \in \mathbb R since ... 2 Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ... 0 This is one of the open questions in Physics. J.S. Bell felt there was a fundamental clash in orientation between ordinary QM and relativity. I will try to explain his feeling. The whole fundamental orientation of Quantum Mechanics is non-relativistic. Even though, obviously, QM can be made relativistic, it goes against the grain to do so, because the ... 4 Comments to the question (v1): The operator f(\hat{p}) and the identity operator {\bf 1}= \int\!\frac{dp}{2\pi}|p\rangle\langle p| commute. The operator f(\hat{p}) and the integration \int\!\frac{dp}{2\pi} are independent of each other. The q-ket and the q-bra are independent of the integration \int\!\frac{dp}{2\pi}. Of course, if one ... 3 The ladder operators satisfy: \bf{a^{\dagger}}$$|n>=\sqrt{n+1}|n+1>\bf{a}$$|n>=\sqrt{n}|n-1> Taking into account <n|m>=\delta_{n,m} , you get the answer. 0 The momentum operator is the generator of shifts. In 3 dimensions (\hbar =1 ) \begin{equation*} (\exp [i\mathbf{a\cdot p}]f)(\mathbf{x})=f(\mathbf{x+a}) \end{equation*} Expanding in \mathbf{a} \begin{equation*} i\mathbf{a\cdot p}f(\mathbf{x})=\mathbf{a\cdot }\partial _{\mathbf{x}}f(% \mathbf{x)} \end{equation*} or \begin{equation*} ... 2 The momentum operator is not -i\partial_x, rather, that is the representation of the momentum operator on the position basis: namely$$ \langle x|\hat{p}|\psi\rangle = -i\frac{\partial}{\partial x}\psi(x). $$Otherwise, the momentum operator is just defined by action on its eigenstates as \hat{p}|p\rangle = p|p\rangle. I understand the complex ... 1 Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane? In a sense neither the real part nor the imaginary part have physical significance, as these quantities do not directly appear in observables. One way to see this is that any solution \left|\psi \right\rangle to ... 1 Hint. ( I give absolutely no guarantee about not making calculational mistakes!) \begin{eqnarray*} L(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\left( 1-\frac{1}{ \sqrt{\pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{ \sigma ^{2}}]\right) e^{-iPy}\rho e^{+iPy}=L_{1}(\tau )-L_{2}(\tau ) \\ L_{1}(\tau ) ... 0 I think [A,C]=[B,C]=0 with C=[A,B] is an assumption, because there exist counterexamples: for A=\sigma_x are B=\sigma_y Pauli matrix along x, y directions respectively. then C=2i\sigma_z is Pauli matrix along z direction. Obviously [A,C]\neq 0 and [B,C]\neq 0. 2 First of all, there are a few problems with your question: J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of \epsilon is a bit weird, too. What you mean is$$ J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ...

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For the transformation of field \begin{align} \phi_{ R \times S^{D-1}}(t, \Phi) \rightarrow \phi_{R^D}(r, \Phi) = r^{-\Delta}\phi_{ R \times S^{D-1}}(t, \Phi) \end{align} where $\Delta$ is scaling dimension.

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It seems like your answer sidestepped the whole question. When you do the same for $\hat p$ you'll find that its derivative depends on $\hat x$, and on $\lambda$. But these coupled equations can then be solved as a second order equation for the terms individually, which should be what you are looking for.

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Basically, you do need to treat it as a perturbation and no correction is necessary... Calculating the commutator, $$\left[\hat{x}, \hat{H}\right] = \left[ \hat{x}, \frac{\hat{p}^2}{2m} - \frac{k}{2}\hat{x}^2 + \frac{\lambda}{4}\hat{x}^4 \right]$$ but as ...

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It is just an integration by parts considering that boundary terms vanish.

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Generally people define these operators so they follow these rules. It's a requirement for them being number operators and hence of any use.

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The situation is impossible and therefore doesn't happen. When you claim to have two states $\psi_1$ and $\psi_2$ then I'll assume they are linearly independent, otherwise they aren't really two different states. Then you claim you have an Operator $O$ such that $POP^{-1} = \epsilon_3 P$ where $P$ is the parity operator and further that $P \psi_1 = ... 1 This is not a state-operator correspondence map. The 2D state-operator correspondence is given by a map between the Hilbert space of states$\mathcal{H}$with a$\mathrm{PSL}(2,\mathbb{C})$-invariant vacuum$\Omega$and fields$\phi: \mathbb{C}\to\mathrm{U}(\mathcal{H})$explicitly given by $$\{\text{fields}\}\to\mathcal{H},\ \phi \mapsto \lim_{z\to ... 0 For a real scalar field I think what you have written is correct..But if you want to describe a complex scalar field then we need to distinguish between \phi and \phi^{\dagger}... 0 I think the first two things that is transformation of position and momentum operators are defined from definition...because by parity transformation sign of the position coordinates changes and time coordinate remain unchanged...so accordingly we get change in sign for position and momentum...once you do that then angular momentum should remain unchanged ... 1 Under conformal mapping z=>w(z) and \bar{z}=>\bar{w}(\bar{z}) a field of conformal dimension(h,\bar{h}) transforms as \tilde{\phi}(w,\bar{w})=(\frac{\partial w}{\partial{z}})^{-h}(\frac{\partial \bar{w}}{\partial\bar{z}})^{-\bar{h}}\phi(z,\bar{z}).. 1 If you want to reduce the "spaghetti of algebra" you can reorient the coordinates. If \hat{z}'=\hat{m} then in spherical coordinates you have$$ \sigma_\hat{m}= \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \\ \end{array} \right] $$where \cos(\theta)=\hat{n} \cdot \hat{m}. It's easier to find the ... 0 Well, I think that first of all, you should understand, that the reason why you can decompose free field into plain waves is that equations of motions are linear. In case of interaction, equations are motions are nonlinear, so any linear combination of its solutions is no longer a solution. In the second place, about your second question: it depends on ... 0 Your last integral is$$ J = \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial}{\partial r}(r\psi_0(r))\right]} =\\ = 4\pi \int_0^\infty{dr\; r^2 \psi_0(r) \left[-2\pi\frac{\delta(r)}{2\pi r^2}Res(\psi_0(0)) + \sqrt{2}\pi a\frac{\delta(r)}{2\pi r^2}\frac{\partial}{\partial r}(r\psi_0(r))\right]} \\ = ... 1 I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate.${\bf A}$is a vector of operators. For example $${\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3}$$ We can denote this collectively as$A_i$. ... 0 In the video Prof. G. Rangarajan is considering the expectation value $$\mathbb{R}~\ni~ \langle \psi |\underbrace{\hat{L}_z}_{\text{self-adj.}} |\psi \rangle ~=~ \int\! d^3r ~\underbrace{\overline{\psi({\bf r})}}_{\text{real}} \underbrace{ (-i\hbar) \frac{\partial}{\partial \varphi}}_{\text{imaginary}} \underbrace{\psi({\bf r})}_{\text{real}} ... 3 The realness of the interorbital coupling t can indeed tell us something about the symmetries of the system. Here, it entails at least one symmetry: spinless time-reversal symmetry. Consider the Hamiltonian of non-interacting spinless particles that live on a periodic chain with two orbitals in each unit cell as in OP's example. In this case, the ... 0 Answer to this question should start from why we want the physical observables to be represented by linear operators. Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers ... 0 Yes it is wrong because multiplication of matrices, you know it, gives matrices and I don't think it makes sense to put a matrice inside a ket or even a bra vector. Actually even with a constant (complex number) if you have k |v\rangle it does not make any sense to put it inside the ket vector like |kv \rangle. However if you have a constant k in the ... 6 Dirac notation is ill-suited for non-self-adjoint operators. Here's why: Let (-,-) be the inner product on our Hilbert space. The expectation value of AB is then$$ \langle AB \rangle_\psi = (\psi,AB\psi)$$by definition, and Dirac notation writes \langle \psi \vert AB \vert \psi \rangle. for this. But, in this notation, it is no longer clear to which ... 0 \langle\psi|AB|\psi\rangle is a complex number (as opposed to a matrix), so taking its transpose gives you back the same thing, i.e.$$ \langle\psi|AB|\psi\rangle^{\dagger} = \langle\psi|AB|\psi\rangle^*, $$and therefore$$ = \langle\psi|B^\dagger A^\dagger|\psi\rangle . $$EDIT I just realised that you then equated this to \langle\psi|B^\dagger ... 0 You want to use$$ \hat x= i\hbar\frac{\partial}{\partial p} $$in the momentum basis. This means that$$ <p|\hat x|\psi>= i\hbar\frac{\partial}{\partial p} <p|\psi> $$Thus, by hermiticity of \hat x, we evaluate$$ <x|\hat x|p> = (<p|\hat x|x>)^*  =(i\hbar\frac{\partial}{\partial p} <p|x>)^*  ... 0 Even in one dimension the operator$p_r=-i\partial_r$on the half line$r>0$has deficiency indices$(0,1)$. There is thus no way to define it it as a self-adjoint operator. In practical terms this abstract mathematical statement means that there is no set of boundary conditions thta we can impose on the wavefunction$\psi(r)$that lead to a ... 0 Background You already seem to know this stuff but it's worth going over again. So, the adjoint of an operator is the equivalent effect of the operator on the other side of the wavefunction inner product: $$\langle \Phi | \hat A | \Psi\rangle = \int_{-\infty}^{\infty} dx~\Phi^*(x) ~ A[\Psi](x) = \int_{-\infty}^{\infty} dx~{A^\dagger[\Phi]}^*(x) ~ \Psi(x) ... 0 Intuitively, shifting then reflecting is not the same as reflecting then shifting. Consider the case of first shifting 1 unit to the right from 0, then reflecting: you end up at x=-1. If you reflect first, it does nothing, and then shifting to the right by 1 means you end up at x=1. The problem is that$$\hat{T}(-a)\hat{R}\psi(x) =\hat{T}(-a)\psi(-x) ... 0 Actually, if the energy of particles is high enough to take relativity into consideration, the concept of particles in quantum mechanics is no longer as valid. For example, the uncertainty relationship$\Delta E \cdot \Delta t \approx \hbar$and energy-mass relation$E=mc^2$suggest that there will be new particles created and annihilated in those cases. So ... 0 I don't know the term "imaginary operator". I take this to be an antihermitian operator, which eigenvalues are purely imaginary. Then the statement is clearly not true. Take as a counter example any hermitian operator$\hat{A}$and real wavefunction$\psi$with$\langle \psi | \hat{A} |\psi\rangle = A_\psi \neq 0$.$A_\psi$is of course real. Take now ... 5 Comments to the question (v5): In this quantum case the overline/bar notation$\bar{A}=\langle A\rangle$is borrowed from statistics and it denotes a quantum expectation value of a quantity$A$. See also Ehrenfest theorem. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator $$\tag{A} H~=~\frac{p^2}{2m} ... 1 The following does not need the hint, but gives an easy way to retrieve the exact evolved state and prove the conservation of the average number of particles. The idea is to make use of the coherent state expression you had in a previous question. For t=0 let$$ |\alpha(0)\rangle = e^{\alpha_0 \, \hat a^\dagger - \alpha^*_0 \, \hat a} |0\rangle. $$Use ... 0 \langle n| n\rangle = \frac{1}{n!} \langle0|(a)^n (a^{\dagger})^n |0\rangle Because (a^{\dagger})^n |0\rangle = \sqrt{n!} |n\rangle and (a)^n |n\rangle =\sqrt{n!} |0\rangle So we have \langle n| n\rangle = \langle 0| 0\rangle =1 In fact, I think the reasonable process is to assume the normalized \langle n| n\rangle =1 first, then we get the ... 0 In order to show that it is correctly normalised, you need to show that | \left <n|n \right>|^2=1 where in this case \left <n \right|=\frac{1}{\sqrt{n!}} \left <0 \right| \hat a^n. And here I suppose \hat a is a ladder operator. Go take a look on this link to know more about ladder operators https://en.wikipedia.org/wiki/Ladder_operator. ... 3 The quantization prescription$$ [\hat{x},\hat{y}] := \mathrm{i}\hbar\widehat{\{x,y\}}\tag{1}$$for$x,y\$ two classical phase space coordinates does have its subtleties. In particular, as the answer in the linked question says, it leads to inconsistent results when applied to e.g. polar coordinates. The reason for this is two-fold: For the radial ...

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