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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Key Assumption to derive the Uncertainty Principle — Canonical Conjugate Operators [on hold]

From this related question, Rigorous Mathematical Proof of the Uncertainty Principle from First Principles The key assumption to derive the uncertainty principle seems to be the relationship ...
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1answer
53 views

Relationship between position and momentum matrix elements

I am going through Faist's Quantum Cascade Lasers book. In the first section of the fourth chapter, he asserts that (in the context of a perturbative analysis of electronic transitions) the ...
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34 views

Why can the y and z-components of spin be measured simultaneously? [duplicate]

I have a gut feeling that this is wrong. By the uncertainty principle where $x,y,z$ are the $x,y,z$ components of spin $$ \sigma_{y}\sigma_{z}\geq \frac{\hbar}{2}\langle x \rangle $$ and it can be ...
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1answer
129 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 ...
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1answer
326 views

In QFT, why do fermions have to anticommute in order to insure causality?

I have seen this question and I believe I understand the answer to it. However, AFAIK, only for bosons the causality condition is a vanishing commutator. For fermions we expect the anticommutator $[\...
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64 views

Does Microcausality follow from Lorentz Invariance?

In a Lorentz Invariant theory, does microcausality automatically hold? In a free theory this is obvious. In an interacting theory I found some 'proof's in this paper: http://arxiv.org/abs/0709.1483 ...
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75 views

What happens to Pauli's argument (that says that there is no time operator) when applied to $X$ operator for some simple systems?

An argument by Pauli is usually referred to in the literature when it is stated that there cannot be a time operator in quantum mechanics. This argument can be found as a footnote to P63 of W. Pauli, ...
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Commutation Relation between Annihilation & Creation Operators and Ascending & Descending Operators

I am currently working on a QD-Cavity system. After the point Heisenberg Equation of motion is obtained from corresponding Hamiltonian of the system, in order to find the expression for bosonic ...
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17 views

Properties of non-commuting (and particularly anti-commuting) operators

The commutator of 2 operators describes in some sense "how much" do they commute or actually its more about "how far" are they from commuting. The case where the commutator of 2 operators (say ...
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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 ...
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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 $$ ...
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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), \...
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1answer
72 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^...
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37 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 ...
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1answer
47 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 ...
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1answer
73 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 ...
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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 ...
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1answer
77 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 $$ a_{sk}^\dagger $$...
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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, ...
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2answers
104 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 $C=[A,B]$. ...
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212 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 ...
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1answer
40 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 ...
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62 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\...
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1answer
44 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 ...
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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 $ ?
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1answer
40 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 ...
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406 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? ...
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2answers
71 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(...
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286 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, $AB+BA$. ...
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1answer
32 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 ...
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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 ...
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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{...
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2answers
485 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-...
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3answers
399 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 ...
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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 ...
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1answer
50 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 ...
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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 ...
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4answers
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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
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1answer
297 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 ...
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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 ...
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1answer
63 views

Lorentz force derivation in quantum mechanics [closed]

In Sakurai and Napolitano, chapter 2, there's a derivation of the QM Lorentz force. Given $$H=\frac{1}{2m}\left(\mathbf{p}-\frac{e\mathbf{A}}{c}\right)^2+e\phi = \frac{\mathbf{\Pi}^2}{2m}+e\phi$$ ...
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1answer
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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 $[H,k]=0$...
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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. $\left[Q,P\right]=i$ ...
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1answer
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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
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3answers
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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 ...
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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 ...
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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^{...
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2answers
79 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
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3answers
109 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 ...
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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 ...