# Tag Info

## New answers tagged commutator

1

I try to measure the energy and the position of the system simultaneously The states with definite energy are not states with definite position so there is no particle state of both definite energy and definite position.

1

At the first look the question seemed very interesting, but later I found the mistake. You said you are measuring the position of the particle precisely. But how? You can tell that the particle is inside the well but you can not know the exact position of the particle. For more info read ...

2

If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $H = \frac{p^2}{2m} + V(x)$ where $V(x)$ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $\frac{p^2}{2m}$ does not commute with $x$, how are you saying that ...

4

I'd like to expand a little bit on the interpretation of commutators as a measure of disturbance (related to incompatibility, as touched on in the other answers). My interpretation of the commutator is that $[A,B]$ quantifies the extent to which the action of $B$ changes the value of the dynamical variable $A$, and vice versa. Let's assume that $A$ is a ...

1

Let's start with the Schrödinger equation: $$\mathrm i\hbar\frac{\partial}{\partial t}\left|\psi\right> = H\left|\psi\right>$$ Since $H$ is self-adjunct, this also implies $$\mathrm -i\hbar\frac{\partial}{\partial t}\left<\psi\right| = \left<\psi\right|H$$ Now consider the most general quantum state, expressed by a density matrix $$\rho = \sum_k ... 1 It may be helpful to assign the students following HW problem : Suppose A and B be two observables i) What is the necessary condition that A and B can be simultaneously measured in an experiment without any uncertainty ? ii) Write down all the second degree polynomials in A and B which are again observables. iii) Suppose ... 4 At a basic level : 1) if [A,B]=0, and if A and B are infinitesimal generators of a symmetry (so also conserved quantities), this means that both A is invariant by B, and B is invariant by A. For instance, [H,J_z]=0, means that the angular momentum is conserved during time evolution, and that the hamiltonian is invariant by rotation. As ... 9 Self adjoint operators enter QM, described in complex Hilbert spaces, through two logically distinct ways. This leads to a corresponding pair of meanings of the commutator. The former way is in common with the two other possible Hilbert space formulations (real and quaternionic one): Self-adjoint operators describe observables. Two observables can be ... 3 Short intro to ladders As you say, they're ladder operators. Let's get rid of the annoying \hbar by setting it to one, and call them more systematically L_{-1},L_0,L_1 instead of L_-,L_z,L_+. Then, the commutation relations take the uniform form$$[L_n,L_m] = (m-n)L_{m+n}$$If we had countably many of these, we'd have a Witt algebra, if there was ... 3 (Disclaimer: The more rigourously inclined individual may be better suited by looking at the Stone-von Neumann theorem, as Qmechanic notes) One can deduce that the momentum operator takes the form \hat p = -\mathrm{i}\hbar\partial_x in the position representation from the fact that the momentum operator generates the infinitesimal translations as ... 3 I) Comment to the question (v1): The Schrödinger position representation$$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{\partial }{\partial x^k}, \qquad \hat{x}^j ~=~x^j,$$correctly reproduces the canonical commutation relations$$ [\hat{x}^j,\hat{p}_k ]~=~i\hbar ~\delta^j_k ~{\bf 1}, $$while the proposal$$\hat{p}_k ~=~ \frac{\hbar}{i} \frac{1}{x^k}, ...

1

The answer to both questions is that D act on Hilbert space states. I'll answer them in reverse order. what exactly do we mean by the eigenvectors of D? Are they fields in space-time? No, in this context, eigenvectors of D are states living in the Hilbert space of the field theory. Because it is only in this sense that the commutation relations between ...

4

The commutation relations $$[D,P_{\mu}] = +i P_{\mu} , \qquad [D,K_{\mu}] = -i K_{\mu}$$ show that $P_{\mu}$ and $K_{\mu}$ raise and lower the conformal dimension of a state. In other words, if you have a state $|\phi\rangle$ of dimensions $\Delta$, so that $D\, |\phi\rangle = i\Delta |\phi\rangle$, then $$D \, P_{\mu} \, |\phi\rangle = [D,P_{\mu}]\, ... 1 In addition to Qmechanic's interesting answer, you might want to explicitly see that \mathrm{SU(2)} only has three independent variables. Let us start by writing the U \in \mathrm{SU(2)} as: $$U=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ with a,b,c,d \in \mathbb{C}. Since we are looking at the special unitary ... 2 I) Quantum mechanically, the Lie group associated with rotational symmetry is$$G~:=~Spin(3)~\cong~ SU(2),$$which is a double cover of SO(3), and has a 3-dimensional real Lie algebra$$L~:=~so(3)~\cong~ su(2)$$with generators J_i satisfying$$[J_i,J_j]~=~i\hbar\sum_{k=1}^3\epsilon_{ijk} J_k. II) In quantum mechanics, we are interested in Lie ...

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