Tag Info

New answers tagged

2

List (to be completed with more references and/or items, details of the relation to physics) there is a notion of positive energy representation (cf. Haag-Kastler axiom, "Spectrum" or "stability" condition) in which generators of translations can be choosen in the von Neumann algebra associated to the representation of the observables, but not necessarily ...


3

In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.


0

You only need to use the anticommutations relations for dirac creation/annihilation operators and their linearity. Expand the definition of $P^\mu$ or $Q$ and use those relations on each term. Only one term per case will be non-zero, the one in the answers.


1

You are correct in assuming that there should be no more than $L$ Fourier modes. In book you specified considered two different sets of boundary conditions. For periodic boundary conditions it is the usual problem and you get multiples of $2\pi/L$ which they wrote as $q=j\pi/L$ with $j=-L,\dots,-2,0,2,\dots,L$ (note that only even values of $j$ appear). ...


1

Mathematically speaking they are the same operator. Usually we reserve the d'Alembertian for 3+1 dimensional spacetime (so in absence of curvature it takes the form $\partial_0^2 - \nabla^2$), while the Laplace-Beltrami operator is defined for an aribtrary dimensional manifold with arbitrary signature. The only possible difference is that sometimes (not ...


6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$ Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following: $$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$ Since you get tired of writing it over and over you just shorten it by an addition ...


2

Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.


10

You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients. A general matrix in the $SU(2)$ group is written as $$ M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )] $$ where $\alpha\in ...


7

The ladder operators do belong to the real Lie algebra $$\quad su(1,1)~\cong~ so(2,1)~\cong~sl(2,\mathbb{R}),$$ but they do not belong to the real Lie algebra $$su(2)~\cong~ so(3).$$ All the above real Lie algebras have complexifications isomorphic to $sl(2,\mathbb{C})$.


0

Yes, the missing dot in the dot product of the second term $$\tag{2} -2\frac{e}{c}\hat{\mathbf{p}} \cdot \mathbf{A}(\hat{\mathbf{x}}) $$ of eq. (2.34b) is a typo. The operators $\hat{\mathbf{p}}$ and $\mathbf{A}(\hat{\mathbf{x}})$ do not commute, due to the CCRs $$\tag{CCR} [\hat{x}^i~,~ \hat{p}_j]_{-}~=~i\hbar\delta^i_j~{\bf 1}.$$ The second and third ...


2

$\renewcommand{ket}[1]{|#1\rangle}$ As others have pointed out, you can go whole hog and solve the characteristic equation $\text{det}(\hat{Q} - \lambda \hat{I})=0$ and find repeated solutions for $\lambda$. However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry. When an operator $\hat{Q}$ has a symmetry there ...


1

As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side: ...


3

Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation $$(Q-\lambda I)\xi = 0$$ Let these solutions generate a vector space $V_\lambda$ and then compute the dimension of $V_\lambda$. If it is greater than 1 then the eigenvalue $\lambda$ is degenerate.


1

I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...


0

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\ad}[0]{\hat{a}^\dagger}$ $\newcommand{\ao}[0]{\hat{a}}$ Firs of all I'd like to tell you that the position operator $x$ is given in terms of the ladder operator by the following relation: $$\hat x = d (\ao+\ad )$$, where $d=\sqrt{\frac{\hbar}{2m \omega}}$. Computing $\hat{x}^3$ gives then, as you ...


3

There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to $$ P_j = \prod_{i\neq j} \frac{A-a_j}{a_i - a_j}, $$ where the product goes over all distinct eigenvalues $a_i$ of $A$. ...


1

Eigenvectors are undetermined up to a scalar multiple. So for instance if c=1 then the first equation is already 0=0 (no work needed) and the second requires that y=0 which tells us that x can be anything whatsoever. This is fine, and correct, as $x=re^{i\theta}, y=0$ is a fine solution when c=1. If $c\neq 1$ then none of the equations are already 0=0 so ...


3

The answer is Yes. Define function $g(q):= \frac{1}{f(q)}$ for later convenience. Then the classical Hamiltonian reads $$2h~=~g(q)p^2.$$ One may show that the Weyl-ordered Hamiltonian reads $$2H_W~=~ (g(q)p^2)_W ~=~ \frac{1}{4}P^2 g(Q)+\frac{1}{2} Pg(Q)P+\frac{1}{4} g(Q)P^2$$ $$~=~ Pg(Q)P - \frac{1}{4}\hbar^2g^{\prime\prime}(Q),$$ see e.g. Ref. 1 and this ...


2

Assuming $\Delta$ is the Laplacian operator, i.e. $-\Delta=D_x^2$, where $D_x=-i\partial_x$, this goes as follows (but the result is different from the one you give). Choose a suitable dense domain of $L^2$ where $x$ and $-\Delta$ are well defined, e.g. the functions that are $C_0^\infty$. Let $\psi\in C_0^\infty$, then $$e^{itD_x^2}xe^{-itD_x^2}\psi=x\psi ...


1

A rigorous version of the time-ordered product is given in the Epstein-Glaser causal approach to quantum field theory. The key to getting rid of difficulties is to make sure that distributions are always multiplied only with sufficiently regular expressions to make sure that the product is well-defined. See, e.g., Pinter's article Finite renormalizations in ...


1

Unitarity of the time-evolution operator is exactly the point: Stone's theorem (see e.g. Reed, Simon: Theorems VIII.7, VIII.8) tells us If $A$ is self-adjoint, the spectral theorem holds. This gives us a functional calculus which makes it possible to define $U(t) = e^{itA}$ in the first place. A such defined $U(t)$ is a strongly continuous unitary group. ...


7

In QM, a real valued observable $A$ is mathematically represented by a projector valued measure over $\mathbb{R}$, $P^{(A)}$, i.e., if $E$ is a Borel subset of the real line, then $P^{(A)}(E)$ is a projector representing the proposition "the outcome of measuring $A$ falls in $E$". In principle, that's all you need for representing, mathematically, ...


-1

$$\begin{align} \left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}\epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = ...


0

Yes, eq. (2.193) is a classical formula, and the symmetry of the (Hilbert) stress-energy-momentum tensor $T^{\mu\nu}$ is only valid classically. Quantum mechanically, the symmetry of $T^{\mu\nu}(x)$ is broken by the presence of other fields in positions $x_1,x_2,\ldots$ in the (time-ordered) correlator $$\langle T\left\{ (\hat{T}^{\mu \nu}(x) - \hat{T}^{\nu ...


6

I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$ ...


6

It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.


2

The time-ordered product is not an operation on operators. It is an operation on time-dependent operators. Let $\cal{O}$ denote the algebra of linear operators on a Hilbert space $\cal{H}$. If $A, B: [0, T] \to \cal{O}$ are two time-dependent operators, we can define an operator $$ AB: [0,T]^2 \to \mathcal{O}, \ \ (AB)(t_1, t_2) = A(t_1) B(t_2) $$ which ...


0

At least a partial answer to your question is that commuting hamiltonians help you to solve the physical system described by one of them: in particular, if your system has $N$ degrees of freedom and you have $N$ commuting hamiltonians, there is good hope that you can trivialize the problem and solve it exactly. In classical mechanics, this is known as ...


2

Adding to Lubos' answer, let me address this part more specificially: I am very confused about this - actually - I'm guessing that transformations that can be written $A⊗B$ are a subset of all possible transformations with all 16 coefficients free. This is correct. First some notation: Let $H_1$ and $H_2$ be two (finite-dimensional) Hilbert spaces for ...


6

First of all, the equation $$ \begin{equation}A\otimes B=A\otimes \mathbb{1}+\mathbb{1}\otimes B,\end{equation} $$ is a claim about an identity, and this claim is incorrect. Note that for $1\times 1$ matrices, the matrices are numbers and the equation above reduces to $$ a\cdot b = a\cdot 1 + 1 \cdot b$$ which is clearly wrong because the addition (the right ...



Top 50 recent answers are included