# Tag Info

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### Hermitian operators and their physical meaning

If I understand your question correctly, you are asking what is the intuitive meaning of the complex numbers appearing in quantum mechanics, when the classical world appears to deal only in real ...

### How to apply anti-unitary symmetry operators?

You forgot the crucial fact that the observables are selfadjoint and that the relevant matrix elements have the same entries. Only those matrix elements are physically measurable since they are ...

### What is a singular continuous spectrum?

I know it's an old question, but it's a topic I love. ;) I always recommend this paper which goes into details in a very clear way. Let's try to sum up the most important parts for your question. Step ...

### Hermitian operators and their physical meaning

In QM we still want expectation values of our observables to be real numbers, i.e. we want $\langle\psi|\hat A|\psi\rangle$ to be real for observables. To see what this condition imposes on operators ...
1 vote
Accepted

### Prove that every component of angular momentum commutes with $f$

Commuting with $L$ is equivalent to being invariant under rotations. The quantities $r^2$, $r\cdot p$ and $p^2$ are all rotationally invariant, as is any function of them. That is all that is needed.

### What does $\dot x$ mean as an operator in quantum mechanics?

In the Heisenberg picture operators evolve in time, and dot denotes a time derivative $$\dot{\hat{x}}^j~:=~\frac{d\hat{x}^j}{dt}, \qquad \ddot{\hat{x}}^j~:=~\frac{d^2\hat{x}^j}{dt^2},\tag{A}$$ cf. ...
1 vote

### Conditions for the Hamiltonian's spectrum to be discrete

(I propose a second answer because the first was affected by a trivial but devastating mistake). First of all Definition. A selfadjoint operator $A: D(A) \to H$, where $H$ is a complex Hilbert space ...
Accepted

### What does $\dot x$ mean as an operator in quantum mechanics?

$\dot{\hat{x}}$ means the same thing it does in Newtonian mechanics. $$\dot{\hat{x}} = \frac{d\hat{x}}{dt}.$$ The trick is this is happening in the Heisenberg picture, not the Schrodinger picture that ...
1 vote

### Why is Dirac's Phase Operator Non-Hermitian?

There is an even shorter way than the one pointed out by @ACuriousMind. Suppose that there is a unitary $V$ such that $$a= Vn^{1/2}\:, \quad a^* = n^{1/2}V^*$$ on the corresponding domains. As a ...

Accepted

### Normal ordered exponential of one-body operators

Hints: First try to show it for a single bosonic oscillator (for fermions this was done by the OP already). To this end, define the following functions: \begin{align} f(M)&:=\exp{a^\dagger a M} \...

### Decomposing a coherent state?

You are effectively asking for a change of basis from oscillator number states to position eigenstates $|x\rangle$, essentially the wavefunction of the Schrödinger wavepacket. (Momentum eigenstates ...

### The composition property of the time-evolution operators

Yes, the group property (3) holds for any order of $t_1,t_2,t_3$. The proof follows from $$U(t_2,t_1)^{-1}~=~ U(t_1,t_2) \tag{A}$$ and the time-ordered version of eq. (3). Proof of eq. (A): We may ...

### In what way are eigenfunctions of an observable operator complete?

Here, "wave function" refers to any possible wave function for the system. A wave function that is invalid for the system for which you found those eigenfunctions $f$ cannot, in general, be ...

### Srednicki's QFT: Why $\langle p|\phi(0)|0\rangle$ in the interacting theory is Lorentz invariant?

This is why we use the LIPS (Lorentz Invariant Phase Space) normalization $$\langle {\bf p}|{\bf p}'\rangle = (2\pi)^3 2E_{\bf p}\delta^3({\bf p}-{\bf p}')$$ for the single particle states. Without ...
Accepted

### What is the Hamiltonian operator, and is it unique?

In the axiomatic formulation of quantum theory, there is an axiom that is used to define the time evolution of quantum states in an isolated system, \begin{equation} |\psi(t)\rangle = \hat{U}(t) |\psi(...

1 vote

### Conservation of momentum in quantum mechanics

The potential energy of the particles is only dependend on their relative location to each other and therefore only on $x_{12}=x_1-x_2$. We have: \begin{align*} [P_x,x_{12}] =[P_{x1}+P_{x2},x_1-x_2] =[...
You are not pedantic, you are just wrong! $$\hat H= \hat p^2 /2m+ V(\hat x).$$ Recall  \hat p= -i\hbar \int\! dx ~|x\rangle \partial_x \langle x| ,\\ \hat x= \int\! dx ~|x\rangle x \langle x|~, \...