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.

learn more… | top users | synonyms (1)

0
votes
1answer
88 views

Calculations with operators - Proof: Equation of Operators [duplicate]

I have a problem in Quantum mechanics 1 with Operators. I have to prove the following equation. I tried it for about 4 hours without any result: Condition: $[[\hat A,\hat B],\hat A]=[[\hat A,\hat B],\...
3
votes
1answer
431 views

Commutation Relations for Creation & Annihilation Opertors of Two Different Scalar Fields

Let us consider two different scalar fields $\phi$ and $\chi$. The commutation relations for the creation and annihilation operators of the scalar field $\phi$ are given by $$ [a(\textbf{k}), a(\...
8
votes
2answers
439 views

A question about causality and Quantum Field Theory from improper Lorentz transformation

Related post Causality and Quantum Field Theory In Peskin and Schroeder's QFT p28, the authors tried to show causality is preserved in scalar field theory. Consider commutator $$ [ \phi(x), \phi(y) ]...
-1
votes
1answer
138 views

How does $p_x$ commute with $p_y$, i.e. $[p_x,p_y]=0$? [closed]

I know it's a simple and basic question but would someone show me how to evaluate $[\hat{p}_x,\hat{p}_y]$?
11
votes
2answers
488 views

Why uncertainty principle is not like this?

In Griffiths' QM, he uses two inequalities (here numbered as $(1)$ and $(2)$) to prove the following general uncertainty principle: $$\sigma_A^2 \sigma_B^2\geq\left(\frac{1}{2i}\langle [\hat A ,\hat B]...
5
votes
1answer
1k views

Finding the creation/annihilation operators

Using Minkowski signature $(+,-,-,-)$, for the Lagrangian density $${\cal L}=\partial_{\mu}\phi\partial^{\mu}\phi^{\dagger}-m^2\phi \phi^{\dagger}$$ of the complex scalar field, we have the field $...
3
votes
1answer
190 views

Is obtaining the coordinate representation of momentum operator from commutator more fundamental than generator of translation

Related post: What is the most general expression for the coordinate representation of momentum operator? There are two methods of obtaining the coordinate representation of momentum in quantum ...
0
votes
2answers
360 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
1answer
156 views

How to derive the commutation relationship between $\hat{L}^2$ and $\hat{\textbf{p}}$ [closed]

How to prove that $$[\hat{L}^2,\hat{\textbf{p}}] = i\hbar(\hat{\textbf{p}}\times\hat{\textbf{L}} - \hat{\textbf{L}} \times \hat{\textbf{p}})$$ I tried to expand $\hat{L}^2$: $$[\hat{L}^2,\hat{\...
1
vote
1answer
87 views

Position and potential Energy

Why are the position and potential energy of a particle able to be measured precisely in Quantum Mechanics? I mean why do they commute with each other?
1
vote
1answer
79 views

Is $\langle k \vert k_1k_2\rangle=0$

Using that $$ \vert k_1k_2\rangle = a^\dagger({\bf k_1})a^\dagger({\bf k_2})\vert 0 \rangle$$ and the commutation relations $$[a({\bf k}),a^\dagger({\bf k'})]=(2\pi)^32\omega\delta^3(\bf {k}- \bf {...
3
votes
1answer
337 views

The Physical Meaning behind a Commutator [duplicate]

I've just been introduced to the idea of commutators and I'm aware that it's not a trivial thing if two operators $A$ and $B$ commute, i.e. if two Hermitian operators commute then the eigenvalues of ...
-1
votes
1answer
140 views

Apply the Heisenberg Equation to the Hamiltonian [closed]

$\frac{d}{dt}$$\hat{H}$ = $\frac{i}{\hbar}$$[\hat{H},\hat{H}]$ +$\frac{\partial{\hat{H}}}{\partial{t}}$ That's as far as I've got. I do not know much about the Heisenberg equation or even what it ...
0
votes
3answers
133 views

Commutator summation notation

I have the relation $ e^L M e^{-L}=\sum_{n=0}^\infty \frac 1{n!} [L,M]_{(n)}$ where $L$ and $M$ are operators. What does the subscript $n$ after the commutator bracket denote?
5
votes
4answers
405 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? ...
3
votes
4answers
173 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 ...
-1
votes
1answer
648 views

Commutator with Pauli spin matrices and the momentum operator

How is $\left[\vec\sigma \cdot \vec p, \vec \sigma \right]$ proportional to $\vec \sigma\times \vec p$, where $\sigma$ are the Pauli spin matrices and $p$ is the momentum operator?
5
votes
1answer
189 views

Why don't we use Hamilton-Jacobi method in QM?

In classical mechanics, we usually try to find a set of coordinates by Hamilton-Jacobi method to transform the Hamiltonian to zero such that the coordinates are conservations. However, we never try ...
6
votes
3answers
2k views

What is the commutator of an operator and its derivative?

Is it possible to calculate in a general way the commutator of an operator $O$ which depends on some variable $x$ and the derivative of this $O$ with respect to $x$? $${O}={O}(x)\\ \left[\partial_x{O}(...
1
vote
1answer
133 views

Deriving commutation relations in second quantisation

I am trying to start from: \begin{align*} [\phi(x),\pi(x')] = i\hbar\delta(x-x') \\ [\phi(x),\phi(x')] = [\pi(x),\pi(x')]=0 \end{align*} to derive: \begin{align*} [a(k),a(k')^\dagger]=\delta_{kk'}\\ [...
3
votes
1answer
174 views

prove: $[p^2,f] = 2 \frac{\hbar}{i}\frac{df}{dx}p - \hbar^2 \frac{d^2f}{dx^2}$

I need to prove the commutation relation, $$[p^2,f] = 2 \frac{\hbar}{i}\frac{\partial f}{\partial x} p - \hbar^2 \frac{\partial^2 f}{\partial x^2}$$ where $f \equiv f(\vec{r})$ and $\vec{p} = p_x \...
4
votes
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 ...
4
votes
1answer
271 views

Causality in QFT from vanishing commutator and the EPR paradox

The question relates to this post. As shown in Peskin and Schroeder's introduction to quantum field theory p. 28., $$[\phi(x),\phi(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$, which implies the ...
6
votes
2answers
8k views

Fundamental Commutation Relations in Quantum Mechanics

I am trying to compile a list of fundamental commutation relations involving position, linear momentum, total angular momentum, orbital angular momentum, and spin angular momentum. Here is what I have ...
5
votes
1answer
230 views

Does This Really “Prove” Spin-statistics Theorem?

In quantization of scalar field theory we impose commutation relation between the field operators by hand and similarly we impose anti-commutation relation between Dirac field operators by hand. As a ...
2
votes
1answer
271 views

Help Simplifying a Commutator Equation

For the SHO, our teacher told us to scale $$p\rightarrow \sqrt{m\omega\hbar} ~p$$ $$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$ And then define the following $$K_1=\frac 14 (p^2-q^2)$$ $$K_2=\frac ...
2
votes
3answers
2k views

Commutators involving functions

I am looking for the commutator: $$[e^{aq},p]$$ My approach is to Taylor expand the function: $$[\sum_n \frac{1}{n!}(aq)^n,p]$$ I know that $[q^n,p]=ni\hbar q^{n-1}$ So how do I account for $n$ ...
4
votes
1answer
150 views

Simple Commutator question

For some reason this is really tripping me up: $$[q_rq_sp_r,q_sp_rq_s]$$ Where $r$ and $s$ are different. Is this just zero because $p_r$ on $q_s =0$. I am trying to simplify this and I feel like 0 ...
3
votes
1answer
240 views

2D Harmonic Oscillator Commutators

So I am given a 2-dimensional harmonic oscillator with $H=H_1+H_2$ where $$H_i=\frac{p_i^2}{2m}+\frac{1}{2}m\omega^2x_i^2$$ Additionally, $$L=x_1p_2-x_2p_1$$ If we define $$A=\frac{1}{2\omega}[H_1-...
4
votes
3answers
372 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. $\left[Q,P\right]=i$ ...
3
votes
3answers
1k views

EQUAL TIME commutation relations

Why is equal time commutation relation used in canonical quantization of free fields?
4
votes
2answers
433 views

Commutator not transitive

I noticed the following: $$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad [L^2,L_3]=0.$$ This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ ...
5
votes
2answers
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, $AB+BA$. ...
0
votes
1answer
421 views

Commutators in bra-ket notation

2-d Hilbert space, with 2 (orthogonal) kets $|a\rangle$ and $|b\rangle$ Operator $A=|a\rangle\langle b| + |b\rangle\langle a| $ Operator $B=-i|a\rangle\langle b| +i|b\rangle\langle a| $ Commutator $...
0
votes
0answers
143 views

Change of QM Momentum operator under coordinate transformation

Can any one please let me know what is the general procedure to construct the momentum operator under some coordinate transformation? For example, I understand that if $${\bf{r}}\rightarrow{\bf{r'}}=...
3
votes
2answers
993 views

Commutator of $L^2$ and $X^2$, $P^2$

In our quantum mechanics script, it states that $[L^2, X^2] = 0$ and $[L^2, P^2] = 0$, therefore for the following Hamiltonan $$H = \frac{P^2}{2m} + V(X^2)$$ it is that $[H, L^2] = 0$ therefore $H$ ...
2
votes
1answer
447 views

Can the quantum angular momentum operator be derived from its commutation relations with position and momentum?

Exercise 12.2.2 in Shankar's Principles of Quantum Mechanics asks to derive the expression for the angular momentum operator $L_z$ \begin{equation} L_z = XP_y-YP_x \end{equation} using its ...
3
votes
1answer
106 views

Does this commutation relation hold?

I was wondering whether it is true that $[L_x^2,x^2+y^2+z^2]=0$. I could not find it in the internet and therefore I wanted to ask here whether anybody here knows that this is true or false.
5
votes
0answers
133 views

Commutator as a time-ordered product

I'm reading through Seiberg and Witten's paper "String Theory and Noncommutative Geometry," and one part in $\S$2.1 isn't quite clear to me. (Sorry, in advance, for the length.) My question is about ...
5
votes
3answers
388 views

Evaluating commutator of $[\operatorname{sign}(X),\, \operatorname{sign}(P)]$

I wish to evaluate the following commutator: $[\operatorname{sign}(X),\, \operatorname{sign}(P)]$. Is there a general method for evaluating $[\operatorname{f}(X), \operatorname{f}(P)]$? I thought of a ...
1
vote
2answers
699 views

Canonical equal time commutation relations in QED

I understand that to quantize the classical electromagnetic field one needs to impose commutation relations and express the field in terms of creation and annihilation operators. I notice that the ...
4
votes
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 ...
-2
votes
1answer
89 views

What is the commutator? [closed]

$e$ and $f$ are unit vectors, $L_e$ is defined by $L_e=eL$, where $L$ is of course the angular momentum operator. A similar definition for $L_f=fL$ The commutator that I can't solve: \begin{...
2
votes
0answers
214 views

Commutators with function

I have following exercise: If $[C,D]$ is a c-number and $f(x)$ is a well-behaved function (i.e. all derivatives exist and are finite), show that: $$[C, f(D)]=[C,D]f'(D)$$ where $f'(D) = \frac{d}{dx}f(...
1
vote
1answer
241 views

Observables still commute even if fields only anti-commute

In Peskin & Schroeder page 56, after introducing anti commutation relations for the fields instead of commutation relations (in order to fix the negative energy problem as well as to have proper ...
4
votes
1answer
366 views

Quantizing the Dirac Field: which commutation relations are more fundamental?

When quantizing a system, what is the more (physically) fundamental commutation relation, $[q,p]$ or $[a,a^\dagger]$? (or are they completely equivalent?) For instance, in Peskin & Schroeder's ...
1
vote
1answer
107 views

Commutator evolution operator and position operator

Let $H= \frac{p^2}{2m}$, then I am supposed to calculate $[x,e^{-iHt}]$. My idea was to use $[x,p^n]=i \hbar n p^{n-1}$ and so I ended up by using the series for the exponential function with $-\frac{...
6
votes
4answers
860 views

Is uncertainty principle a technical difficulty in measurement? [duplicate]

Is the uncertainty principle a technical difficulty in measurement or is it an intrinsic concept in quantum mechanics irrelevant of any measurement? Everyone knows the thought experiment of measuring ...
7
votes
2answers
645 views

How does the proof of operator commutativity work with non-continuous operators?

In some books, a proof that if two self-adjoint operators $A$ and $B$ share a common eigenbasis $\{\phi_n\}$, then they commute is given as follows : For any $\phi_n$, $$AB\ \phi_n = a_n\ b_n\...
1
vote
1answer
76 views

How does the following commutator for measured observables and this operator relation imply the following relation?

$$ \hat{\Omega}_j{(\tilde{q}_j)}=\Omega_j(\tilde{q}_j-\hat{q}_j) $$ $$ [\hat{q}_j,\hat{q}_l]=ik_{jl} $$ Implies $$ [\hat{q}_j,\hat{\Omega}_l]= \frac{\partial\Omega_l(\tilde{q}_l-\hat{q}_l)}{\...