New answers tagged operators
0
votes
Action of Symmetry Operators on Hamiltonian
There is a typo in your equations that was not present in the book print that you posted. The correct expression is
$$
\mathcal{CH^*C^{-1}=-H}.
$$
The presence of conjugation in l.f.s. is important, ...
- 3,335
0
votes
How to handle a delta-function divergence from an infinite-dimensional trace of a quantum operator with a general continuous parameter?
When we have a operator that is diagonalizable in a continuous basis, its trace is rather pathological. We can express such an operator as
$$ \hat{B} = \int |\lambda\rangle B(\lambda) \langle\lambda|\ ...
- 15.8k
0
votes
Accepted
Computing descendants of primary operator in 2D CFT
For the descendant obtained by hitting $\partial \phi(w)$ with $L_{-2}$, we obtain
$$[L_{-2}, \partial \phi(w)] = \frac{1}{2\pi i} \oint_{c_w} dz \; z^{-1} T(z) \partial \phi(w) = - \frac{\partial \...
- 283
1
vote
Commutator of angular momentum operator $[\hat{l}_a, F_b(\hat{x})]$ with a position-dependent vector-function $\vec{F}(\vec{x})$?
When you write
$$
\hat{l}_a = -i\epsilon_{abc}x_b \partial_c
$$
Actually what you're doing is to write it in position basis $| x\rangle$. The expression
$$
\hat l_a F_d(\hat x) = -i\epsilon_{abc}x_b {\...
- 3,335
2
votes
Does Haag's theorem lack dynamics?
You're mixing up different things.
First, there's the interaction picture, which means the following: if you write your Hamiltonian as $H=H_1+H_2$, then correlation functions satisfy
$$
\langle \phi(...
0
votes
How to show that orthogonal states can always be distinguished?
Your answer works for $dim H = n$ and $dim H > n$ as well. The only difference is that in first case, your set of measurements forms a complete set of projectors, and in the second case it should ...
- 3,335
2
votes
Does Haag's theorem lack dynamics?
Haag's theorem indeed only talks about the n-point functions (Wightman functions) of the fields, and yet it is about dynamics.
The short answer to the question is: Of course Haag's theorem knows about ...
- 129k
5
votes
How do I see that the different definitions for the SYK Hamiltonian are equivalent?
In the operator formalism there is an implicitly written normal ordering of the 4-fermion term. The fermions anticommute under the normal order symbol.
In the path integral formulation, the fermions ...
- 213k
0
votes
Computing descendants of primary operator in 2D CFT
For any field $V$, the identity $\partial V = L_{-1}V$ allows you to compute correlation functions of $L_{-1}$-descendants. For correlation functions of $L_{-2}$-descendants or higher, use the local ...
- 2,792
0
votes
Diagonalisation of a quadratic Hamiltonian for bosons
You need positive definiteness. The essential result is Williamson's theorem..
- 56.5k
2
votes
Proving that a Hamiltonian transformation is canonical - Condensed Matter, Altland and Simons
A quantum canonical transformation is an adjoint group action $\hat{F}\mapsto {\rm Ad}(\hat{U})\hat{F}=\hat{U}\hat{F}\hat{U}^{-1}$ with a unitary operator $\hat{U}$, cf. e.g. this Phys.SE post.
...
- 213k
3
votes
Possible errata Landau and Lifshitz in $\S29$ Matrix elements of vectors in Quantum Mechanics Third Edition
I don't know what is wrong with your reasoning, since it seems to return the right result in the special case where $A_{yy}=A_{zz}$. That said, I never saw this approach, and I searched in Landau'book ...
- 3,335
3
votes
Accepted
Possible errata Landau and Lifshitz in $\S29$ Matrix elements of vectors in Quantum Mechanics Third Edition
The identity to use is:
$$[p_i,r_j]=-i\hbar\delta_{ij}$$
You are doing something wrong:
$$[L_z,r_x]=[r_xp_y-r_yp_x,r_x]=r_x[p_y,r_x]-r_y[p_x,r_x]$$
$$=0-r_y (-i\hbar \delta_{x,x})=i\hbar r_y$$
It ...
- 5,831
0
votes
Accepted
How to prove the Hermiticity of the angular momentum operator's X-component?
How to Prove the Hermiticity of the Angular Momentum Operator's X-Component?
I understand that the momentum operator is Hermitian...
I assume you also understand that the position operator is ...
- 23.3k
1
vote
Is there an Ehrenfest-like result for the expectation value of orbital angular momentum?
Yes, indeed, it is possible to derive such a result... although it might look somewhat trivial, if compared to position and momentum in an arbitrary potential.
Hamiltonian of angular momentum in ...
- 65k
3
votes
Clarification of Ehrenfest theorem
The Newton's equation of motion in potential $V(x)$ are:
$$
\dot{x}=\frac{p}{m},\dot{p}=-\frac{dV(x)}{dx}
$$
or simply
$$m\ddot{x}=-\frac{dV(x)}{dx}=F(x),$$
where $F(x)= -\frac{dV(x)}{dx}$ is called (...
- 65k
2
votes
Accepted
Is Kubo formula the integrated version of Ehrenfest?
TL;DR Wikipedia article on Kubo formula is wrong. See article on Green-Kubo relations.
The Heisenberg equation of motion for an arbitrary operator is
$$
\frac{d}{dt}A = \frac{1}{i\hbar}[A,H] + \frac{\...
- 65k
0
votes
Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
But I cannot see how I can ignore the eigenvectors as if they do not exist and use chain rule
Do you want to prove it without using the spectrum or Taylor expansion?
Let $U(t) = e^{-iHt}$. By ...
- 3,335
0
votes
Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
If $H$ is a hermitian matrix with eigenvectors $$H|n\rangle=\epsilon_n|n\rangle,$$
(note that the eigenvectors and eigenvalues do not depend on time)
then we could write
$$
H= \sum_n|n\rangle\langle n|...
- 65k
0
votes
Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
Since the Hamiltonian is a self-adjoint operator, you can use the spectral theorem to write
$$H = \int_\mathbb{R} \lambda \, dP_\lambda.$$
Then, as the exponential is a measurable function,
$$
\exp(-...
- 1,314
4
votes
The problem with commutation relation
Classic error. The inequality holds only if the states are normalized, and the eigenstates of $Q$ are not - at least not normalized in the usual sense as they are outside the usual Hilbert space. ...
- 47.8k
1
vote
Accepted
Confusion about integral form of interaction potential in Dyson series
In short, it is important (for us all) to distinguish between operators defined in various pictures, and how these definitions relate to each other. P&S define the interaction in equation 4.12, ...
- 3,749
1
vote
Accepted
Inversion on fermionic operators
The matrix $𝑈_{𝑙𝑚}$ is not translationally invariant, i.e. it is not simply a function of $Δ=𝑙−𝑚$. This is because you pick an origin when you think of inversions.
This is also clear from writing ...
- 1,058
1
vote
How to show that $\int d^3 x\; e^{ikx}(x^1 \partial^2-x^2\partial^1)\varphi (x)=0$?
The integral doesn't always identically vanish. Strictly speaking it only vanishes if you approach spatial infinity using a cylindrical geometry.
I will do the problem in 2D because the third ...
- 438
2
votes
Accepted
Contradiction in the value of $[\phi(x), v^{\mu} \partial _{\mu} \phi(x)]$ for space-like $v^{\mu}$
There is no contraddiction.
First of all $[\phi(x), \partial_\mu \phi(x)]$ is not defined as the two arguments coincide. What we can do is to define the normal order of it
$$:[\phi(x), \partial_\mu \...
- 78.1k
4
votes
Accepted
Omitting regular terms in OPE
"We can ignore regular terms in an OPE" is often made as a blanket statement but we need to be more careful. It's better to say that we can do this if we're trying to compute correlation ...
- 8,900
0
votes
How are matrices used to represent quantities, and what is the meaning of a matrix?
It never hurts to cite chapter and verse in a question. You are apparently asking about problem 7.1 in Jordan's Quantum Mechanics in Simple Matrix Form. In this problem he asks about two 3 x 3 ...
- 642
0
votes
OPE of two quasi-primaries involves only other quasi-primaries and their derivatives
Here we consider the case of unitary highest weight representations: there is a primary state $\left|\phi\right>$ and the representation $V$ is spanned by $L_{-k_1}\dots L_{-k_n}\left|\phi\right>...
0
votes
Difference of OPE in "Conformal Field Theory" and "Introduction to Conformal Field Theory With Applications to String Theory"
I've basically figure out why they are different: In "Intro" they are considering quasi-primary field, which is a larger space, and those together with partial derivatives, i.e. $L_{-1}$ ...
- 287
2
votes
Accepted
Commutator between $P^\mu$ and $\phi$ (complex scalar field)
The commutator of the momentum operator with the field gives you the result an infintesimal translation. For a scalar field $\Phi$ (let's assume it is complex although it doesn't matter), under a ...
- 55.3k
1
vote
Commutator between $P^\mu$ and $\phi$ (complex scalar field)
A complex field has essentially two independent types of components (e.g. "holomorphic" and "antiholomorphic" modes, or alternatively real and imaginary modes), and these ...
- 3,749
0
votes
What does the absence of a ground state physically mean?
This is likely not what you are looking for, but it is an important point nonetheless. Consider a free, non-relativistic particle in 1 dimension (or more). The Hamiltonian $H = - \nabla^2$ has ...
- 218
7
votes
What does the absence of a ground state physically mean?
The free particle Hamiltonian has spectrum $\sigma(H_0) = [0,\infty)$, but $0$ is not an eigenvalue. More generally, any system that possesses only scattering states (For example $H = H_0+ V$ where $V$...
- 1,120
3
votes
What does the absence of a ground state physically mean?
I am not aware of any example where a Hamiltonian would be bounded from below but do not have a ground state.
On the other hand, one can easily come up with Hamiltonians not bounded from below and ...
- 65k
2
votes
What does the absence of a ground state physically mean?
Consider a Hamiltonian bounded from below. If there is no ground state, any state will not be the lowest energy state, which means that it could suffer a transition to a lower energy state. We can ...
- 3,335
11
votes
What does the absence of a ground state physically mean?
Very interesting question. I don't believe there is any principle prohibiting a scenario where there is no ground state, but the Hamiltonian is bounded from below. It's just that for most systems ...
- 3,477
2
votes
How does Sean Carroll swap the operators and sign on the index when calculating the observed number of particles in the Unruh Effect?
As you wrote, we have the operators
$$
{\hat{b}_k}^{(1)}~=~\frac{1}{\sqrt{2\sinh\left(\frac{\pi\omega}{a}\right)}}\left(\mathrm{e}^{\pi\omega/(2a)}{\hat{c}_k}^{(1)}+\mathrm{e}^{-\pi\omega/(2a)}{{\hat{...
- 66
1
vote
How can a QFT field act on particle states in Fock space?
Think of a piano:
The piano keys represent the field. They are fundamental and always present, even if silent.
Pressing a key generates a note (an excitation of the piano's field). This is analogous ...
- 11
2
votes
Contradiction in the value of $[\phi(x), v^{\mu} \partial _{\mu} \phi(x)]$ for space-like $v^{\mu}$
TL;DR: OP's paradox seems to be caused by ill-defined mathematical manipulations on a distributional object.
To give an idea of what may go wrong consider the following calculation for a scalar field $...
- 213k
5
votes
Why does $J^2$ only commute with one of $Jx, Jy, Jz$ and not all?
It's certainly true that $J^2$ commutes with each of $J_x$, $J_y$ and $J_z$. Additionally, it is true that for each of $J_x$, $J_y$ and $J_z$, a set of simultaneous eigenstates between that operator ...
- 416
2
votes
Why does $J^2$ only commute with one of $Jx, Jy, Jz$ and not all?
$\vec{J}^2$ actually commutes with all three $J_x$, $J_y$ and $J_z$. So $\vec{J}^2$ have common eigenstates with $J_x$ and have common eigenstates with $J_z$. How is it possible? Those are different ...
- 8,739
1
vote
Accepted
Number of pairs of indistinguisable particles
Number of pairs of indistinguishable particles
I don't think the particles are indistinguishable. I think the particle pair is created by choosing one of $n$ distinguishable particles, then choosing ...
- 60.4k
Top 50 recent answers are included