A mathematical construct used to study the effect of applying two operators in succession.

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7
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2answers
531 views

What does the Canonical Commutation Relation (CCR) tell me about the overlap between Position and Momentum bases?

I'm curious whether I can find the overlap $\langle q | p \rangle$ knowing only the following: $|q\rangle$ is an eigenvector of an operator $Q$ with eigenvalue $q$. $|p\rangle$ is an eigenvector of ...
3
votes
1answer
400 views

What physical significance has the Heisenberg Group?

I read that the canonical commutation relation between momentum and position can be seen as the Lie Algebra of the Heisenberg group. While I get why the commutation relations of momentum and momentum, ...
14
votes
4answers
3k views

What is the connection between Poisson brackets and commutators?

The Poisson bracket is defined as: $$\{f,g\}_{PB} ~:=~ \sum_{i=1}^{N} \left[ \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial ...
4
votes
2answers
645 views

Why are anticommutators needed in quantization of Dirac fields?

Why is the anticommutator actually needed in the canonical quantization of free Dirac field?
3
votes
2answers
359 views

How to prove that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space?

How to prove that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space?
30
votes
4answers
2k 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 ...
4
votes
1answer
1k views

Momentum as Generator of Translations

I understand from some studies in mathematics, that the generator of translations is given by the operator $\frac{d}{dx}$. Similarly, I know from quantum mechanics that the momentum operator is ...
1
vote
2answers
528 views

Prove $[A,B^n] = nB^{n-1}[A,B]$

I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of ...
8
votes
5answers
9k views

What is the Physical Meaning of Commutation of Two Operators?

I understand the mathematics of commutation relations and anti-commutation relations, but what does it physically mean for an observable (self-adjoint operator) to commute with another observable ...
3
votes
1answer
279 views

What conservation law corresponds to this local $U(1)$ symmetry of the CCR?

It is known that canonical commutation relations do not fix the form of momentum operator. That means that if canonical commutation relations (CCR) are given by ...
1
vote
3answers
372 views

Simple Quantum Mechanics Question about The Commutator of Translation Operators

Say there is $\hat{J} = \exp[-i \hat{p} l/ \hbar]$ and $\hat{U}= \exp[-i\hat{H}t/ \hbar]$, where $\hat{H}$ is time-independent. Can we say anything about $[\hat{J},\hat{U}]$? Is it zero? How do we ...
10
votes
2answers
698 views

In QFT, why does a vanishing commutator ensure causality?

In relativistic quantum field theories (QFT), $$[\phi(x),\phi^\dagger(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$ On the other hand, even for space-like separation $$\phi(x)\phi^\dagger(y)\ne0.$$ ...
7
votes
1answer
271 views

Canonical quantization in supersymmetric quantum mechanics

Suppose you have a theory of maps $\phi: {\cal T} \to M$ with $M$ some Riemannian manifold, Lagrangian $$L~=~ \frac12 g_{ij}\dot\phi^i\dot\phi^j + \frac{i}{2}g_{ij}(\overline{\psi}^i ...
3
votes
1answer
130 views

State space of QFT, CCR and quantization, and the spectrum of a field operator?

In the canonical quantization of fields, CCR is postulated as (for scalar boson field ): $$[\phi(x),\pi(y)]=i\delta(x-y)\qquad\qquad(1)$$ in analogy with the ordinary QM commutation relation: ...
4
votes
2answers
237 views

What is the most general expression for the coordinate representation of momentum operator?

I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$. One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th ...
3
votes
2answers
237 views

Unitary spacetime translation operator

Srednicki writes: We can make this a little fancier by defining the unitary spacetime translation operator $$ T(a) \equiv \exp(-iP^\mu a_\mu/ \hbar) $$ Then we have $$ T(a)^{-1} \phi(x) T(a) = ...
4
votes
4answers
278 views

Is uncertainty principle a technical difficulty in measurement?

I have searched for an answer to this question on physics SE but I have not seen a question in which it is addressed properly. Please let me know if there is an answer already. My question briefly ...
4
votes
3answers
317 views

Does the canonical commutation relation fix the form of the momentum operator?

For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar $$ Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on ...
4
votes
2answers
327 views

Classical Limit of Commutator

In Dirac's book Principles of quantum mechanics (4th ed., pgs 87-88), he seems to give a very elementary argument as to how the commutator $[X,P]$ reduces to the Poisson brackets ${x,p}$ in the limit ...
3
votes
1answer
289 views

Commutator of Lorentz boost generators : visual interpretation

I have always struggled to visualize the correctness of the commutation relation for the generators of the boost in the Lorentz group. We have $$[K_i,K_j] = i \epsilon_{ijk} L_k$$ I fail to picture ...
2
votes
1answer
146 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 ...
5
votes
1answer
280 views

Do mutual eigenkets imply commutation of two operators?

I have been working on this question. I have solved it, and I would like to check whether my line of reasoning is right or wrong Question: Prove that if there exists a mutual complete set of ...
4
votes
2answers
313 views

Heisenberg picture of QM as a result of Hamilton formalism

Let's have formula of full time-derivative of physical value in Poisson's formalism: $$\tag{1} \frac{df}{dt} = -[H, f]_{P. br.} + \frac{\partial f}{\partial t}, $$ where $[A, B]_{P. br.}$ is Poisson's ...
2
votes
0answers
73 views

commutator to entropy in an uncertainty relationship?

Question: Does there exist a commutator to entropy in an uncertainty relationship? Similar Energy and time for instance.
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vote
3answers
122 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 $n$ ...
1
vote
2answers
203 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 ...
1
vote
1answer
570 views

Commutator of Momentum with a Position dependent function

I heard from my GSI that the commutator of momentum with a position dependent quantity is always $-i\hbar$ times the derivative of the position dependent quantity. Can someone point me towards a ...
1
vote
0answers
54 views

commutators in an uncertainty relationship derived from a partition function?

The maximum information principle for the discrete case gives rise to a partition function (>>> see details here) $$Z(\lambda_1,\ldots, \lambda_m) = \sum_{i=1}^n \exp\left[\lambda_1 f_1(x_i) + \cdots ...
1
vote
2answers
298 views

Full time-derivative of a function and Schrodinger equation

From Hamiltonian formalism there is well known equation, $$ \frac{d F}{dt} = \frac{\partial F}{\partial t} + \{F, H\}_{PB}, $$ where $ \{H, F\}_{PB}$ is the Poisson bracket. After using Hamiltonian ...