Questions tagged [commutator]

A mathematical construct quantifying the difference in effect of applying two operators in two alternate successions. It is the defining product of a Lie algebra, the efficient underlying description of Lie groups, of use in several areas of physics, most notably quantum field theory.

Filter by
Sorted by
Tagged with
2
votes
1answer
42 views

Why $\int dx \partial_\mu\neq \partial_\mu \int dx$ but $\int dp \partial_\mu=\partial_\mu\int dp$?

It's well known that $\int dx \partial_\mu\neq \partial_\mu \int dx$. But I have a hard time understanding $\int dp \partial_\mu=\partial_\mu\int dp$, because $[p,x]\neq0$ do not commute. However, ...
1
vote
0answers
37 views

Proving that $$ [\phi(\vec{x}, 0), \phi(\vec{x}, t)] \sim e^{-i m t}-e^{+i m t} $$ in QFT

So far, I get the following (for the left term in the integral, $d$=3): \begin{equation} \begin{aligned} \Delta_{+}(x) &= \int \frac{\mathrm{d} \vec{p}^{d}}{(2 \pi)^{d} 2 e(\vec{p})} \exp (-i t e(...
2
votes
1answer
39 views

Simple question on computing commutation relation

In bosonization, one faces with the following commutator: $$[\phi(x_1), \theta(x_2)]=\sum_{q\neq 0} \frac{\pi}{Lq} e^{iq(x_2-x_1)-\alpha |q|}\tag{1}$$ where $q$ is an non-zero integer multiple of $2\...
10
votes
5answers
249 views

Does an Operator that neither commutes with $\hat{X}$ or $\hat{P}$, nor can be expressed as a “function” of $\hat{X}$ and $\hat{P}$ make sense?

When you come from classical hamiltonian mechanics (which is based on the phase space), observables are introduced as functions $f$ on the phase space $(q, p)$. There can't be a classical observable ...
0
votes
1answer
230 views

If the mean of the commutator is zero, then what we can say about the commutator itself?

Suppose we have \begin{equation} \langle[H,N]\rangle=0 \tag{1} \end{equation} where both $H$ and $N$ are hermitian. Under which assumption I can claim that then $$[H,N]=0~?\tag{2}$$
6
votes
2answers
141 views

If $E$ and $P$ don't commute, why could we have an $E$-$K$ diagram?

If $E$ (energy) and $P$ (momentum) only commute in constant potential, how could we have an $E$-$K$ diagram in a solid material? $[E,p] \neq 0$. Then we cannot prepare electrons whose $E$ and $P$ are ...
2
votes
0answers
32 views

Can you argue without explicitly calculate the eigenenergies that one Hamiltonian is gapped and another is not?

Consider a pair of one dimensional four band model $H_1$ and $H_2$, which read as: $$ H_1 = \begin{pmatrix}k\sigma_x-E_0&0\\0&k\sigma_x+E_0\end{pmatrix} + \alpha \begin{pmatrix}0&\sigma_x\...
0
votes
1answer
28 views

Do Equal Time Commutation / Anticommutation relations automatically also apply for spatial derivatives?

The question is basically in the title. My naive thought was that when a commutation relation holds for all field operators $\Psi(\vec{x})$ (by "all" I mean "at all positions $\vec{x}$") on a fixed ...
3
votes
1answer
41 views

Generalized commutator/anticommutator via phase factor

We know that the commutator between two operators $A$ and $B$ reads $[A,B]_{-}=AB - BA$, while the anticommutator reads $[A,B]_+=AB + BA$. I am wondering if someone has ever used a generalized ...
0
votes
0answers
43 views

Commutators of dot and cross products

I apologize if this question is too basic, but I am wondering if identities for commutators such as $[AB,C]=A[B,C]+[A,C]B$ also hold for dot and cross products within the commutator (i.e., $[A\times ...
0
votes
1answer
88 views

Commutation relation for Dirac field

In "Quantum Field Theory" by Peskin and Schroeder, I couldn't understand the commutation relation calculation for Dirac field (pg. 53): $$ \begin{align} \psi(x) &= \int \frac{d^3p}{(2\pi)^3} ...
1
vote
2answers
219 views

Why is it “disconcerting” if the components of an operator do not commute?

A symmetrized operator for the operators $\hat{H}$ and $\hat{N}$ is given by $$\hat{R}=\frac{1}{2\hat{H}}\hat{N}+\hat{N}\frac{1}{2\hat{H}}.$$ When $\hat{H}$ is the Hamiltonian and $\hat{N}$ is the ...
0
votes
1answer
173 views

Commutation of currents in QED

In an outline of a proof of the Ward identities in QED, the authors Green, Schwarz, and Witten in their book "Superstring theory", vol. I, Section 1.5.1, claim that in the QED the electromagnetic ...
1
vote
1answer
137 views

Domain space of compatible and incompatible operators (observables)

Sakurai (Modern Quantum Mechanics, by J.J. Sakurai) states in the section on compatible operators: Let us first consider the case of compatible observables A and B. As usual, we assume that the ket ...
3
votes
3answers
1k views

Example of two linearly independent, nowhere vanishing vector fields in $\mathbb{R}^{2}$

I knew that two linearly independent and nowhere-vanishing vector fields provide a basis for the tangent space at each point in $\mathbb{R}^{2}$. Is it necessary that these two vector fields commute? ...
0
votes
1answer
73 views

What is the value of the commutator $[\vec{S}, H]$?

What is the commutation relation between $[S, H]$ where Hamiltonian $H= - \vec{S} \cdot \vec{B}$ , $\vec{S}$ is the spin and $\vec{B}$ is the magnetic field. I am getting $0$ but it seems wrong.
1
vote
1answer
285 views

Mutually Commutative

What is the definition of a Mutually Commutative set of operators? I've found articles describing a complete set of mutually commutative operators, but I can't actually find what mutually commutative ...
1
vote
2answers
10k views

Commutator of $X$ and $P$ in quantum mechanics [closed]

So, this is my first contact with Quantum Mechanics and I'm having trouble with this exercise. One of the steps involves calculating $[X,P]$, and I stuck there. Can anyone give me some help?
0
votes
2answers
563 views

Commutator relationships and the exponential

I am currently trying to prove that the two following commutator relationships are equivalent (for an operator $\hat{A}(s)$ that depends on a continuous parameter $s$), so if one holds the other one ...
1
vote
2answers
64 views

Commutation of position four-vector with spacetime derivatives

I am trying to understand a simple demonstration in Ashok Das' Lectures in QFT. He does the following on p. 134 $$[P_\mu,M_{\nu\lambda}]=[\partial_\mu ,x_\nu\partial_\lambda-x_\lambda\partial_\nu]=\...
1
vote
2answers
958 views

Projection operators and their subspaces (of Hilbert space)

I've been watching Susskind's lectures on Quantum Entanglement, and something he said regarding (non-)commuting projection operators confused me. Consider two subspaces {$|a>$} and {$|b>$} of ...
0
votes
0answers
41 views

Holonomic basis

Is the following definition correct? Given a differentiable manifold $M$ and an ordered basis $\{e_j^m\}$ of the tangent space $T_m M$ with $m\in M$ (they are vectors and not vector fields). An ...
1
vote
1answer
60 views

How creation operator pops out while expanding field operator?

While doing QFT when we try to canonically quantize the Klein Gordon equation $\Box \phi =0$ we promote the $\phi $ to an operator field and impose the commutation rule $[\phi(x,t),\pi (y,t)]=i\hbar\...
1
vote
1answer
45 views

The conmutator of the square of Pauli-Lubanski vector and the generators of Poincare group

I'm working on trying to solve the following problem: Using the following expressión for the square of Pauli-Lubanski vector:$$W^2=-\frac{1}{2}M_{\mu\nu}M^{\mu\nu}P_{\alpha}P^{\alpha}+M^{\mu\nu}M_{\...
2
votes
2answers
68 views

One question about BRST symmetry in reading Srednicki’s book: Why should the BRST charge $Q_B$ be nilpotent?

In p.453, Srednicki claims that since the BRST transformation of a BRST transformation is zero, $Q_B$, the BRST charge, must be nilpotent: $$Q_{B}^{2}=0.\tag{74.32}$$ I don't know why.
-1
votes
3answers
252 views

The use of the commutators in quantum mechanics: explanations [duplicate]

Considering that I've never studied quantum mechanics before I have need to understand the operator commutator. My start is: $[A,B]=AB-BA \tag{a}$ Now, why must be $$\left[\frac{\partial }{\partial ...
5
votes
3answers
1k views

Hamiltonian with position-spin coupling

I am solving a Hamiltonian including a term $(x\cdot S)^2$. The Hamiltonian is like this form: \begin{equation} H=L\cdot S+(x\cdot S)^2 \end{equation} where $x$ is the position operator, $L$ is ...
0
votes
2answers
246 views

Prove conservation law in quantum mechanics

I major in Math, and I am studying Quantum Mechanics (QM). I see the conservation law in QM as a mathematical theorem. Please check if my understanding is right, and help me to prove the theorem? ...
1
vote
1answer
160 views

Commutation relation in quantized electromagnetic field theory

I have a question regarding a proposed problem (Problem 4.8) in Rodney Loudon's book "The Quantum Theory of Light". Let $U(t)$ be an operator defined by $$ U(t)=\exp\left\lbrace\frac{i}{\hbar}\int\...
2
votes
1answer
96 views

Common observables and associated operators: operator momentum [duplicate]

Starting from my previous question Commutators in quantum mechanics and considering that the commutator $$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar, \tag{1}$$ the associated linear ...
0
votes
2answers
59 views

Commutation with unspecified potential function

Instead of a potential given like $V(r) = k r^2$ or $V(r) = y^2$ , if the potential is given like in the form a function but not clearly specified, can we tell that if that commutes with the ...
6
votes
1answer
138 views

If Poisson Bracket of Momentum and Position is non-zero, why no Uncertainty Principle?

In Hamiltonian classical mechanics, we have that the Poisson bracket of position and momentum satisfies $$\{q_i, p_j\} = \delta_{ij}$$ But this implies that momentum and position 'generate' changes ...
1
vote
2answers
102 views

Quantum Harmonic Oscillator Raising and Lowering operators

The commutator of the operators, $[a,a^\dagger] = 1$ is useful in rewriting the Hamiltonian in a neat way in terms of the creation and annihilation operators. So my question is, Is there a physical ...
8
votes
4answers
2k views

Fermions, different species and (anti-)commutation rules

My question is straightforward: Do fermionic operators associated to different species commute or anticommute? Even if these operators have different quantum numbers? How can one prove this fact in a ...
0
votes
1answer
67 views

Number operator - annihilation operator commutation

Is there a rigorous way to prove that $$ (N+1)^{-1/2} a = a N^{-1/2} $$ where $a$ is a bosonic annihilation operator and $N=a^\dagger a$ is the corresponding number operator?
1
vote
1answer
179 views

Propagator Causality with commutators all the way

We know that two fields commute - by locality and causality - iff there is spacelike separation $\left[\phi_l^k(x) , \phi_m^{k'}(y)\right] = 0$ for $(x-y)^2<0$ In the canonical quantization of ...
1
vote
1answer
58 views

What is the implication of overlap between eigenstates of two operators in Quantum Mechanics?

For instance, what does it mean that a certain position eigenstate is also an energy eigenstate? I understand that measurable (Observables) in Quantum mechanics are the operators. Their eigenvalues ...
0
votes
0answers
44 views

Covariant derivative with respect to commutator

I have some confusion with the notion of $\nabla_{[A, B]}\bf{v}$, that expression, with a commutator of vector fields as the subindex of the connection appears for instance in the definition of the ...
32
votes
4answers
9k views

How does non-commutativity lead to uncertainty?

I read that the non-commutativity of the quantum operators leads to the uncertainty principle. What I don't understand is how both things hang together. Is it that when you measure one thing first ...
0
votes
1answer
264 views

Commutator of canonical fields in Quantum Field Theory

Let $\phi(\vec{x},t)$ denote the canonical fields and $\pi(\vec{x},t)$ denote the canonical impulses where they're given by: \begin{equation} \phi(x)=\int\frac{d^3\vec{p}}{(2\pi)^3\sqrt{2\omega_{\...
1
vote
0answers
78 views

Why is the Schrödinger field an annihilation operator?

The relativistic scalar field operator is not a ladder operator. Its commutation relations are $$\begin{align} \left[\hat{\phi}\left(\vec{x}\right), \hat{\phi}\left(\vec{y}\right)\right] = \...
1
vote
1answer
119 views

Quantum mechanics, Fourier transformation

Why do we use $p=-i\hbar\frac{\partial}{\partial x}$ in quantum physics? (I know the reason for $i\hbar$, quantization). Is this right to say we can't measure velocity and position of electrons at the ...
1
vote
0answers
24 views

Explicit form of Klein factors in Giamarchi

In Giamarchi, Quantum Physics in One Dimension, Appendix B, I don't understand how he did his last step in equation B.8, as shown below. If anyone has gone over the derivation, I would really ...
1
vote
2answers
120 views

How does one obtain $\hbar$ as $\frac{h}{2\pi}$?

I'm reading Dirac's Principles of Quantum Mechanics. He defines $\hbar$ to be the real number satisfying the following relation $$ uv - vu = i\hbar[u,v]$$ where $u$ and $v$ are dynamical variables, ...
2
votes
1answer
62 views

Why does equal commutator relation imply equal operator?

In 1d bosonization, Giamarchi (Quantum Physics in One Dimension) Chap 2, shows that fermionic Hamiltonian $$H_f=\sum_k k(R_k^* R_k -L_k L_k)$$ is equal to the bosonic representation $$H_b = \sum_k |...
1
vote
1answer
79 views

Baker-Hausdorff for normal ordering exponential

Let $A=A^+ +A^-$ where $A^+,A^-$ denote the creation and annihilation portion of the field. Then in Eduardo Fradkin, Field Theories of Condensed Matter Physics, equation (5.284), it states that $$ :e^...
2
votes
1answer
41 views

Commutation relations in Gupta-Bleuler quantization

Quantization of the free electro-magnetic field has essential differences in comparison to quantization of say scalar or massive vector fields. In fact there are different approches to it. One of ...
0
votes
0answers
101 views

Virasoro algebra commutation (part 2)

This was a sub-question in my previous post that I ask separately now. In Introduction to Conformal Field Theory by Blumenhagen and Plauschinn (springer link) the Virasoro algebra is introduced the ...
4
votes
1answer
59 views

Ehrenfest theorem and correlation among observables at the quantum scale

I am studying quantum mechanics and I encountered the famous Ehrenfest Theorem, which states that given an observable $A$, its expectation value time evolution is governed by $\partial_t\langle A\...
0
votes
1answer
83 views

Show that when angular momentum $L_x$ and $L_y$ commute with operator $G$, then $L_z$ also commutes with $G$

I want to prove that if Angular momentum $L_x$ and $L_y$ commute with an operator $G$, angular momentum $L_z$ also commutes with $G$. if $[L_x , G] = [L_y, G] = 0$ then $[L_z , G] = 0$ I know that $...