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10

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 ...

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 ...

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 $$\tag{A} H~=~\frac{p^2}{2m} ... 4 Comments to the question (v1): The operator f(\hat{p}) and the identity operator {\bf 1}= \int\!\frac{dp}{2\pi}|p\rangle\langle p| commute. The operator f(\hat{p}) and the integration \int\!\frac{dp}{2\pi} are independent of each other. The q-ket and the q-bra are independent of the integration \int\!\frac{dp}{2\pi}. Of course, if one ... 4 Assumptions: I will be talking about Hermitian (more generally self-adjoint) operators only. This means that I will assume that the operators in question have a set of eigenvectors that span the Hilbert space. As mentioned by tomasz in a comment, this is not exactly necessary, since more general statements can be made, but since we are dealing with basic QM, ... 4 I) One problem is that the momentum operator \hat{p} is an unbounded operator, which means that it is only defined on a domain D(\hat{p}) \subsetneq {\cal H} of the Hilbert space {\cal H}=L^2(\mathbb{R}). When we apply the differentiation operator \hat{p}=\frac{\hbar}{i}\frac{d}{dx} to OP's wave function$$\tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), ...

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.

3

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 ...

3

(1) Yes, take ${\cal H} = L^2(\mathbb R, dx)\oplus L^2(\mathbb R, dx)$ and thereon $\left(X (\psi, \phi)\right)(x,y) := (x\psi(x),y\phi(y))$. We have $\sigma(X)=\sigma_c(X)$ and the degeneracy is just $2$. (2) Yes, use the example (1) with a countably infinite copies of $L^2(\mathbb R, dx)$ and use the Hilbertian direct sum of Hilbert spaces. (There are ...

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 ...

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 ...

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 $$J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ... 2 Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ... 2 The momentum operator is not -i\partial_x, rather, that is the representation of the momentum operator on the position basis: namely$$ \langle x|\hat{p}|\psi\rangle = -i\frac{\partial}{\partial x}\psi(x). $$Otherwise, the momentum operator is just defined by action on its eigenstates as \hat{p}|p\rangle = p|p\rangle. I understand the complex ... 2 We usually say that if two operators, \hat{A} and \hat{B} commute, then they have a simultaneous set of eigenstates. Saying that the eigenstates are the same isn't really correct. For example, let operator \hat{A} be hermitian and act on elements of the Hilbert Space \mathcal{H}_A and let operator \hat{B} also be hermitian and act on elements ... 2 I henceforth assume \hbar =1. There is no reason to introduce Dirac deltas here, everything is elementary. Moreover as the function \psi is not differentiable, one cannot use the form of the momentum operator P as derivative which is valid only on smooth functions. Forcing this way would introduce unnecessary difficulties as the derivative must be ... 1 Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane? In a sense neither the real part nor the imaginary part have physical significance, as these quantities do not directly appear in observables. One way to see this is that any solution \left|\psi \right\rangle to ... 1 The following does not need the hint, but gives an easy way to retrieve the exact evolved state and prove the conservation of the average number of particles. The idea is to make use of the coherent state expression you had in a previous question. For t=0 let$$ |\alpha(0)\rangle = e^{\alpha_0 \, \hat a^\dagger - \alpha^*_0 \, \hat a} |0\rangle. $$Use ... 1 I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate. {\bf A} is a vector of operators. For example$$ {\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3} $$We can denote this collectively as A_i. ... 1 If you want to reduce the "spaghetti of algebra" you can reorient the coordinates. If \hat{z}'=\hat{m} then in spherical coordinates you have$$ \sigma_\hat{m}= \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \\ \end{array} \right] $$where \cos(\theta)=\hat{n} \cdot \hat{m}. It's easier to find the ... 1 Under conformal mapping z=>w(z) and \bar{z}=>\bar{w}(\bar{z}) a field of conformal dimension(h,\bar{h}) transforms as \tilde{\phi}(w,\bar{w})=(\frac{\partial w}{\partial{z}})^{-h}(\frac{\partial \bar{w}}{\partial\bar{z}})^{-\bar{h}}\phi(z,\bar{z}).. 1 This is not a state-operator correspondence map. The 2D state-operator correspondence is given by a map between the Hilbert space of states \mathcal{H} with a \mathrm{PSL}(2,\mathbb{C})-invariant vacuum \Omega and fields \phi: \mathbb{C}\to\mathrm{U}(\mathcal{H}) explicitly given by$$ \{\text{fields}\}\to\mathcal{H},\ \phi \mapsto \lim_{z\to ...

1

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 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 ... 1 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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