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.

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22
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6answers
2k views

Is there something behind non-commuting observables?

Consider a quantum system described by the Hilbert space $\mathcal{H}$ and consider $A,B\in \mathcal{L}(\mathcal{H},\mathcal{H})$ to be observables. If those observables do not commute there's no ...
0
votes
1answer
105 views

Commutation relations in second quantization

I know that for operators $a(\chi_1), a(\chi_2)$ of the same type (fermionic or bosonic) $$ [a(\chi_1), a(\chi_2)]_{-\xi} = [a^\dagger (\chi_1), a^\dagger (\chi_2)]_{-\xi} = 0 \tag{1}$$ where $$\xi ...
1
vote
2answers
37 views

Second Quantization: Do fermion operators on different sites HAVE to anticommute?

In second quantization, we assume we have fermion operators $a_i$ which satisfy $\{a_i,a_j\}=0$, $\{a_i,a_j^\dagger\}=\delta_{ij}$, $\{a_i^\dagger,a_j^\dagger\}=0$. Another way to say this is that $$ ...
0
votes
0answers
37 views

How can $\hat p = - i \hbar \partial_q$ be derived starting from the definitions of $\hat q$ and $\hat p$ in terms of creation/destruction operators? [duplicate]

Consider the position and momentum operators $\hat q$ and $\hat p$, defined respectively in terms of creation and destruction operators in the usual way: $$ \hat q = c (\hat a + \hat a^\dagger), \...
4
votes
1answer
71 views

Non-abelian current commutators

There many articles, in which non-abelian current commutators are computed. The general result is that quantum corrections lead to additional term in commutator $$[J^a_\mu (x), J^b_\nu (y)] \delta (x^...
0
votes
1answer
34 views

Cross-product within commutator

I've been trying to prove some commutator identities of angular momentum, and I don't want to go brute force and prove for each coordinate seperately. So I tried using the Levi-Civita formalism for ...
0
votes
1answer
46 views

Verification of proof of complete set of commuting operators

Hi I am interested in the validity of the following proof. I am interested in the validity of this particular proof as I am aware of how to prove this result in a different way. Theorem: If two ...
0
votes
1answer
69 views

Yang-Mills field strength tensor

In basically every QFT book the Yang-Mills strength tensor $F_{\mu\nu}$ is defined as $$F_{\mu\nu}=[D_\mu,D_\nu]$$ where $D_\mu$ is the covariant derivative $$D_\mu=\partial_\mu-A_\mu$$ and $A_\mu$ is ...
0
votes
0answers
35 views

Glauber formula, Baker

somewhere I read here: $[A,F(B)]=[A,B]F'(B)$ is used to prove Glauber's formula $\exp(A+B).\exp([A,B]/2)$ I have tried and looked everywhere to try and understand this to no avail. The first is in ...
4
votes
1answer
73 views

Computing the density operator commutation relations (Atland & Simons)

I'm trying to work through Altland and Simons' example of interacting fermions in one dimension. It's in chapter 2, page 70 (you can find it here). They define fermionic operators $$ a_{sk}^\dagger $$...
1
vote
1answer
68 views

Under what condition is angular momentum conserved in both classical and quantum physics?

Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it? If this is the case, ...
-2
votes
2answers
103 views

Does $[A^2,B]=0$ imply $[A,B]$=0? [closed]

The commutator $[A^2,B]$ can be written as $A[A,B]+[A,B]A$. So if $[A,B]=0$, $[A^2,B]$ is also zero. But is the converse also true? If $[A^2,B]$ is given to be zero, then is [A,B]=0? Let $C=[A,B]$. ...
3
votes
2answers
209 views

Why do the ladder operators in harmonic oscillators work?

The Hamiltonian can be diagonalized by transforming $x$ and $p$ to $a$ and $a^\dagger$. I understand how one proceeds from there to find the spectrum of $a^\dagger a$, the ground state $|0\rangle$ and ...
1
vote
1answer
39 views

Why the constancy of an observable w.r.t time depends on whether it commutes with $H$ or not?

I have been reading Modern Quantum Mechanics by J.J.Sakurai. Under the chapter Quantum Dynamics, the author says if an observable $A$ initially commutes with the Hamiltonian operator $H$, then it ...
0
votes
1answer
61 views

Question about a formula in the book by Green, Schwarz, Witten

In chapter 7 of superstring theory, it is written $$ g\langle0;k_1|\zeta\cdot\alpha_1V_0(k_2)\zeta_3\cdot\alpha_{-1}|0;k_3\rangle=g\langle0;k_1|\zeta\cdot\alpha_1e^{k_2\cdot\alpha_{-1}} e^{-k_2\cdot\...
0
votes
1answer
42 views

Does Heisenberg's uncertainty hold for any two quantum measurements?

Heisenberg's uncertainty principle is most commonly expressed in terms of the uncertainty in measurement of position and momentum of a particle, $$\Delta x\Delta p \geq \hbar$$and uncertainty in ...
1
vote
0answers
55 views

Uncertainty Principle with the corresponding operators

Why does the corresponding operator do not commute if there is uncertainty related to two observables A and B that states $\Delta A\,\Delta B > 0 $ ?
0
votes
1answer
36 views

Placement of indices in canonical commutation relations of coordinates and conjugate momenta as well as fields and conjugate momenta

The canonical commutation relations between generalised coordinates $q_a$ and their conjugate momenta $p^a$ are given by $[q_a,q_b]=[p^a,p^b]=0$ $[q_a,p^b]=i\delta^b_a$. Furthermore, the canonical ...
5
votes
4answers
399 views

Is commutation relation an equivalence relation?

I'm now learning quantum mechanics with Liboff. In the book it deals with "a compete set of mutually compatible observables" in order to make a state maximally informative. How can one find such set? ...
0
votes
2answers
67 views

How to recognize a Complete Set of Commuting Operators (CSCO)

A question about 'completeness'. These two operators are commuting, but I want to know more about their completeness. How do you know if {H}, {B}, {H,B} and/or {$H^2$,B} are forming (a) Complete Set(...
5
votes
2answers
285 views

Tricky operator identity: $[L^2,[L^2,\vec{r}]]=2 \hbar ^2 \{ L^2, \vec{r}\}$?

This operator identity showed up in a course I was taking, and it was given without proof. $$[L^2,[L^2,\vec{r}]]=2 \hbar ^2 \{ L^2, \vec{r}\}$$ The curly brackets denote the anticommutator, $AB+BA$. ...
0
votes
1answer
31 views

Tensor products of Hilberts spaces: definition of outer products and commutators

Suppose one has two single-particle Hilbert spaces $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ and consider the tensor product of these such that $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is a two-particle ...
4
votes
0answers
56 views

Elegant method to show $[L^2,[L^2,\vec{r}\,]\,] = 2\hbar^2\{L^2, \vec{r}\}.$ [duplicate]

Show that $[L^2,[L^2,\vec{r}\,]\,] = 2\hbar^2\{L^2, \vec{r}\},$ where $\vec{r} = x\, {\hat x} + y\, {\hat y} + z\, {\hat z}.$ "Edit: $\{A,B\} = AB + BA$ is the anti-commutator." I am able to solve ...
-1
votes
1answer
47 views

Deriving eigen values of $\hat{N}$

So let's say we have an operator $\hat{a}$ (ladder operator), where $\left[\hat{a},\hat{a}^\dagger\right] = 1$, and $\hat{a}^2 |\phi\rangle = 0$. How do I show that the eigenvalues of $\hat{N}=\hat{...
0
votes
2answers
470 views

Expectation values of commutator and anti-commutator (momentum and position)

What are the expectation values of commutator and anti-commutator for momentum and position operators? In the case of commutator: $$\langle[x,p]\rangle=\langle i\hbar\rangle=~?$$ In the case of anti-...
4
votes
3answers
391 views

Schroedinger field operators and their commutation relations

I've got several questions regarding the so called second quantization of the Schroedinger equation. My professor introduced the field operators for the Schroedinger field by simply stating them as ...
1
vote
2answers
61 views

Commutation Relations in Second Quantization

I understand that if I have the field operators $\psi(r)$ and $\psi^\dagger(r)$, then I have the canonical commutation relation (in the boson case) $$[ \psi(r) , \psi^\dagger(r')]=\delta(r-r').$$ My ...
1
vote
1answer
45 views

The fifth gamma matrix and fermion fields

I am aware of the various relations with Dirac spinors and chirality but how does the fifth gamma matrix $\gamma^5$ behave with fermion fields, $\psi$? Does the fifth gamma matrix have any particular ...
0
votes
0answers
28 views

Construct any Hamiltonian that is the linear combination of existing constructable Hamiltonians

In the paper Quantum Computation over Continuous Variables, it states that since $$e^{iAt}e^{iBt}e^{-iAt}e^{-iBt} = e^{-[A,B] t^2} + O(t^3)$$ when $t\rightarrow 0$, if one can apply a set of ...
3
votes
4answers
170 views

Help understanding proof in simultaneous diagonalization

The proof is from Principles of Quantum Mechanics by Shankar. The theorem is: If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors ...
4
votes
1answer
291 views

Virasoro operators commutation relations [closed]

For the commutation relation in quantising the bosonic string $$\left[L_n,L_{m}\right]=(n-m)L_{n+m}+\frac{D}{12}n(n^2-1)\delta_{n+m,0}$$ we can then calculate this for $m=-n$ in between the vacuum ...
1
vote
0answers
91 views

Physical significance of non-commutativity of ladder operators in Quantum Harmonic Oscillator

If we apply the raising (creation) operator to $Ψ_n(x)$ and the apply to it the lowering (annihilation) operator, we get $Ψ_n(x)$ times a constant. Does it physically say something? Can we get any ...
1
vote
1answer
62 views

Lorentz force derivation in quantum mechanics [closed]

In Sakurai and Napolitano, chapter 2, there's a derivation of the QM Lorentz force. Given $$H=\frac{1}{2m}\left(\mathbf{p}-\frac{e\mathbf{A}}{c}\right)^2+e\phi = \frac{\mathbf{\Pi}^2}{2m}+e\phi$$ ...
2
votes
1answer
53 views

Why do we use the anticommutation relation for particle-hole and chiral symmetries?

In physics we say that a quantity is conserved if its operator commutes with Hamiltonian. For example, in condensed matter systems, when the momentum $k$ commutes with the Hamiltonian $H$ as $[H,k]=0$...
4
votes
3answers
370 views

Relation between representations of boson operators?

I have a simple (I think !) question about the representations of boson operators and how they are related. First of all let's define two conjugate observables $Q$ and $P$ (i.e. $\left[Q,P\right]=i$ ...
1
vote
1answer
36 views

Structure constant of the commutators of generators in broken symmetry

When I read a paper related to spontaneously global symmetry breaking, I cannot understand a statement: If we use the notation $T^i$ for the unbroken group generators in $H$ and $X^a$ the broken ...
3
votes
3answers
130 views

Does the canonical commutation relation relate to the fact that momentum is the generator of spatial translations?

In classical mechanics momentum is the generator of spatial translations. This remains true in quantum mechanics. The way we define the momentum operator in one-dimension, for example, already shows ...
2
votes
0answers
110 views

Commutation relations in quantum mechanics

As we know, simple harmonic oscillator can be solved only by commutation relations between creation and annihilation operators, and the Hamiltonian expression. The spin energy is either solved only ...
-1
votes
1answer
79 views

Commutation relations in Quantum Field Theory [closed]

\begin{align} [a, a^\dagger] =& \left[\int d^3 x e^{-ikx} (\omega \phi(x) + i \Pi^\dagger(x)), \int d^3 x' e^{ikx'} (\omega \phi^\dagger(x') - i \Pi(x')) \right] \\ =& \int d^3x \, d^3x' \, e^{...
0
votes
2answers
76 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 ...
4
votes
3answers
108 views

Eigenstates of Ladder Operators

According ot Griffith's Intro to Quantum Mechanics (page 147), if some function $f$ is an eigenfunction of $L^{2}$, then $L_{-}f$ is also an eigenfunction of $L^{2}$. Is $f$ also an eigenfunction ...
0
votes
1answer
49 views

Sum and Different and angular momentum operators

Why is $\overrightarrow{L_{1}}+\overrightarrow{L_{2}}$ an angular momentum operator, but not $\overrightarrow{L_{1}}-\overrightarrow{L_{2}}$? What does this show about the applicability of the vector ...
2
votes
2answers
121 views

Mathematical Proof the angular momentum and Hamiltonian commute?

I'm in a quantum mechanics class, and it is given in the book that the operators $\hat{L^{2}}$ and $\hat{H}$ commute for the 3D Harmonic Oscillator, but no definite mathematical proof is given, and I'...
0
votes
1answer
45 views

Spherical Polar Proof for Non-Commutativity of Indivdual Quantum Angular Momentum Operators

How can the following commutation relation be solved through spherical polar coordinates $[\hat{L}_{z}$,$\hat{L}_{x}]$ = $\imath\hbar\hat{L}_{y}$ I understand the derivation through partial ...
1
vote
1answer
121 views

Ladder operators - commutation relations and their properties

At the beginning of Fetter, Walecka "Many body quantum mechanics" there is a statement, that every property of creation and annihilation operators comes from their commutation relation (I'm ...
1
vote
1answer
59 views

How does this quantisation relation come about?

I'm currently doing a course in string theory and in the lecture notes it is stated: $$ [x^-, p^+]~=~-i \tag{1}$$ I am fine with this. However, after trying (and failing) a question, I looked at ...
3
votes
1answer
232 views

What is the physical importance of the commutation relations of angular momentum?

What is the physical meaning of these commutation relations: $$[L_{z},L_{\pm}]=\pm\hbar L_{\pm}\tag{1}$$ and $$[L_{+},L_{-}]=2\hbar L_{z} ~?\tag{2}$$
4
votes
2answers
96 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 ...
51
votes
5answers
5k views

Trace of a commutator is zero - but what about the commutator of $x$ and $p$?

Operators can be cyclically interchanged inside a trace: $${\rm Tr} (AB)~=~{\rm Tr} (BA).$$ This means the trace of a commutator of any two operators is zero: $${\rm Tr} ([A,B])~=~0.$$ But what about ...