737 views

### What is the mistake in calculating such a commutator? [duplicate]

$B$ is an Hermitian operator in Hilbert space, and $|b\rangle$ is the eigenstate of $B$. We can have $[A, B] = 1$ where A is arbitary operator. Then we can calculate as below: \begin{align} &\...
• 141
Suppose we want to calculate the expectation value $$\langle x|[\hat{x},\hat{p}]|x\rangle,$$ where $|x\rangle$ is an position eigenstate, so that $\hat{x}|x\rangle=x|x\rangle$ and $\langle x|\hat{x} = ... • 57 2 votes 2 answers 88 views ### What went wrong in the following calculation of$\langle p'|[x,p]|p'\rangle$? [duplicate] We know that $$[x,p]=i\hbar.$$ Consider now the diagonal element in the momentum representation, $$\langle p'|[x,p]|p'\rangle=i\hbar\langle p'|p'\rangle=i\hbar\delta(0).$$ But the LHS = $$\langle p'|... • 51 4 votes 0 answers 66 views ### Mean value of [x,p] on an eigenstate of x [duplicate] The canonical commutation relation states that$$[x,p] = i \hbar\ \mathbb{I}$$if we imagine to be in a one dimensional space. If we take the mean value of the commutator over an eigenstate of the ... 0 votes 0 answers 79 views ### What is \left<x\right|\hat x\hat p-\hat p\hat x\left|x\right>? [duplicate] This is simple question, but I don't know how to do it. \left<x\right|\hat x\hat p-\hat p\hat x\left|x\right>=?, I can solve it in two ways. One of them is$$\left<x\right|\hat x\hat p-\hat ... • 37 0 votes 0 answers 66 views ### How can I reconcile the apparent all-zeroes diagonal of commutators with the canonical commutation relation? [duplicate] Out of interest I was trying to derive some properties of commutators in quantum mechanics. I found that, in my calculations, will have an all-zeroes main diagonal. I must be making a mistake, because ... 1 vote 0 answers 53 views ### The momentum representation of$x$and$ [x,p]$[duplicate] To deduce the momentum representation of$[x,p]$, we can see one paradom $$<p|[x,p]|p>=iℏ$$ $$<p|[x,p]|p>=<p|xp|p>−<p|px|p>=p<p|x|p>−p<p|x|p>=0$$ Why? If we ... • 13 9 votes 3 answers 18k views ### Matrix elements of momentum operator in position representation I have two related questions on the representation of the momentum operator in the position basis. The action of the momentum operator on a wave function is to derive it: $$\hat{p} \psi(x)=-i\hbar\... • 3,732 4 votes 1 answer 3k views ### Matrix elements of the operator \hat{x} \hat{p} in position and momentum basis I want to calculate the matrix elements of the operator \hat{x} \hat{p} in momentum and position basis, that is the two quantities (p,q - momenta, x,y - positions):$$\langle p|\hat{x} \hat{p}|... • 2,835 8 votes 2 answers 505 views ### Deriving the expectation of$[\hat X,\hat H]$For a free particle of mass$m$, with Hamiltonian $$\hat{H} = \frac {\hat{P}^2} {2m},$$ where $$\hat{P} = -i \hbar \frac{\partial} {\partial x}.$$ The commutative relation is given by $$[\hat{X}, \... • 346 3 votes 1 answer 4k views ### Commutation relation of position and momentum using Dirac notation This is likely a very trivial/silly question, but in following a derivation of the position and momentum commutation relation using the dirac notation, I am having trouble justifying a certain step. ... • 127 2 votes 1 answer 3k views ### How do I solve these integrals of wave function and operator? First integral$$\int \Psi^*({\bf r},t)\hat {\bf p} \Psi({\bf r},t)\, d^3r,$$where the \Psi({\bf r},t)=e^{i({\bf k}\cdot{\bf r}-\omega t)}\,\,\, and \hat {\bf p}=-i\hbar \nabla. Second one$$\... • 21 2 votes 2 answers 192 views ### Theoretical Knowledge for Experimental Particle Physicist [closed] I was told that, as an experimental particle physicist, I need to know Quantum Field Theory and the Standard Model. However, are there situations where an experimental particle physicist needs to ... 4 votes 2 answers 181 views ### Heisenberg EOM for$\langle x \rangle\$ in momentum eigenstate - where is my error?
Equation of motion for expectation value of a quantum particle in a momentum eigenstate: $$\frac{d}{dt} \langle x \rangle = \frac{1}{i h} \langle [x,H] \rangle$$ and since it's in a momentum ...