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10

Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O ...


6

If $|\phi⟩$ and $|\psi⟩$ are linearly independent, then it is always possible to assign them to the column vectors $$ |\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix}, $$ but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis. ...


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 ...


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 $r \times p$.


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

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, ...


4

Suppose the space-time group includes dilatations which expand or contract space. Points in space $x^{i}\in V_{3}$ transform under a small dilatation $\epsilon$ near the identity as, \begin{equation} x'^{i}=x^{i}+\epsilon x^{i} \ . \end{equation} The change in the coords is, \begin{equation} \frac{d x^{i}}{d\epsilon}=x^{i} \end{equation} In the Hamiltonian ...


4

In this case $p_0$ is just a number. It's the amount of phase that you add to the wave function. $\text{e}^{ip_0 x/\hbar}$ is not a translation operator. It's just multiplying the wave function by a complex number of norm $1$. In you previous question you had $\hat{p}$ which is an operator. It acts on the wave function by differentiating it.


3

It is invertible iff its determinate doesn't vanish $$ \det([H]_B) \ne 0 $$ Note that this property of the determinate is invariant under a change of basis since: \begin{align} \det(S^{-1} \cdot [H]_B \cdot S) & = \det(S^{-1}) \cdot \det([H]_B) \cdot \det(S) = \frac{1}{\det(S)} \cdot \det([H]_B) \cdot \det(S) \\ & = \det([H]_B) \end{align} with ...


3

I know of no such publication. However, this issue may simply be on one hand too trivial and on the other hand too far removed from practical relevance. Let's derive what you are after to see: A creation ladder operator $\hat{a}^\dagger$ for arbitrary states would have to be of the form $$\sum_{n=0}^\infty c_n \left| n+1 \right\rangle \left\langle n ...


3

your result is correct $$ [a_k, a_q] = -2 a_k a_q $$ which is consistent with $$ [a_k, a_k ]= - 2 a_k a_k = 0 $$ because $$ a_k a_k = \frac{1}{2}\{a_k, a_k \} = 0 $$ And in general you can use $$ [A,B] = 2AB - \{A,B\}$$ which would also give $$[a_k^\dagger, a_q] = 2a_k^\dagger a_q - \delta_{kq}$$


3

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 | #2 \right>}$ $\newcommand{bok}[3]{\left< #1| #2 |#3\right>}$ It means basically that all of the energy eigenstates has zero energy eigenvalue. Ups... Let $\left| \psi \right>$ be a normalized energy eigenstate with energy ...


3

Let's take the canonical commutation relations (CCR), in their exponentiated form (Weyl's relations): $$V(\eta)T(q)=e^{-i\eta\cdot q}T(q)V(\eta)\; ,$$ where $\{V(\eta)\}_{\eta\in \mathbb{R}^d}$ and $\{T(q)\}_{q\in \mathbb{R}^d}$ are objects of a given normed algebra with involution. This is a very general notion, that is nowadays taken as the definition of ...


3

An antilinear map (scalars are factorized "out of the map" as their complex conjugates) $C:\mathscr{H}\to \mathscr{H}$ is called a conjugation if it preserves the norm of $\mathscr{H}$ and $C^2=\mathrm{id}$ (the identity operator). Theorem [von Neumann]. Let $A$ be a symmetric operator. If there exists a conjugation $C$ such that $C: D(A)\to D(A)$ and ...


3

Repeatedly applying this relation to the ground state is exactly what you need to do. There's nothing more to it.


2

$\langle a|i\hat{C}|a\rangle=\langle a|[\hat{A},\hat{B}]|a\rangle = \langle a| \hat{A}\hat{B}-\hat{B}\hat{A}|a \rangle = \langle a|\hat{A}\hat{B}|a\rangle - \langle a|\hat{B}\hat{A}|a \rangle = a^{*}\langle a|\hat{B}|a \rangle - a \langle a|\hat{B}|a\rangle = 0 $ (Since $a^{*}=a$ because $\hat{A}$ is Hermitian and Eigen Values of Hermitian operators are ...


2

$\newcommand{\ket}[1]{\lvert #1 \rangle}$You seem to be confused about what measuring an operator means. Let $A,B$ be two commuting self-adjoint operators as in your question, and let $\{u_n\}$ be a basis of simultaneous eigenvectors, that is $$ A\ket{u_i} = a_i \ket{u_i} \ \vee \ B\ket{u_i} = b_i \ket{u_i}$$ Now, a generic state $\psi$ can be written 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 ...


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 ...


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 ...


2

Ref. 1 writes the correct formula $$ U(t,t^{\prime})~=~e^{iH_0(t-t_0)} e^{-iH(t-t^{\prime})}e^{-iH_0(t^{\prime}-t_0)} , \qquad t~\geq~ t^{\prime},\tag{4.25}$$ which satisfies $$ U(t_1,t_2)U(t_2,t_3)~=~U(t_1,t_3) , \qquad t_1~\geq~ t_2~\geq~ t_3.\tag{4.26}$$ Here $t_0$ is an arbitrary but fixed fiducial initial instant where operators and states in the ...


2

You're right, there is no eigenfunction. The eigenfunctions of a self-adjoint operator form a complete basis for the Hilbert space, but this is simply not true for symmetric operators. Therefore if an operator is not self-adjoint, it may not have any eigenfunctions.


2

A hint on this could be the fact that a superposition of stationary states of different energies is NOT a stationary state, because you can not express the wave function as the product of a single time-dependent exponential tiames a spatial function.


1

$\newcommand{ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$ $\newcommand{\bk}[3]{\left< #1| #2 |#3\right>}$ In a one dimensional problem $\langle \hat p \rangle$ is always zero. $$\langle \hat p \rangle = \bk{\psi}{\hat p}{\psi}=\int \mathrm{d}x \,\psi^* \hat p \, \psi \propto \int \mathrm{d}x \, \psi^* \psi' \overset{(1)}{=}-\int ...


1

Given a densely defined pre-closed operator $T$ on a Hilbert space $H$, you can define its transpose (more properly called the adjoint) $T^*$ by requiring it to be the operator with the property that $$(\eta,T\psi)=(T^*\eta,\psi)$$ for any $\eta$ in the domain of $T^*$ (which is dense) and $\psi$ in the domain of $T$. Using Dirac's notation we can rewrite ...


1

It was an incorrect statement, as it is explained here by Griffiths himself. Anyways, the mathematical explanation is straightforward: given a self-adjoint operator $A$ with domain $D(A)$, any sufficiently regular real function $f(A)$ of it (and the square is perfectly ok for $-\Delta=p^2$) is self-adjoint on some domain $D(f(A))$ by the spectral theorem ...


1

Without specifying the domains of the involved operators all the discussion has not much sense. Let me say that, if $A :D(A) \to H$ with $D(A)\subset H$ a linear dense subspace of the Hilbert space $H$, $A$ is self-adjoint if $D(A)=D(A^\dagger)$ and $A=A^\dagger$. Notice that consequently (I stress that the converse is false) self-adjointness implies ...


1

Let me assume $\hbar = 1$ We have the following expression for the commutator $$ [\nabla^2, L_z] = -[p^2, L_z] = -[p_x^2 + p_y^2 + p_z^2, L_z]$$ Now using the fact that the action of the commutator is linear we can write $$ [p_x^2 + p_y^2 + p_z^2, L_z] = [p_x^2, L_z] + [p_y^2, L_z] + [p_z^2, L_z]$$ Now I'm going to use the property $$ [AC, B] = A[C, B] + ...


1

I think I'm essentially supposed to show... Why do you think this? Is this a homework question? If so then it should be tagged as such. If this is not a homework question and you are interested to read an explanation involving no explicit equations then consider the following: The operator $\vec L$ is the generator of rotations. Therefore, any ...


1

$\frac{dU}{ds}+\frac{dU^\dagger}{ds}=0$ implies that $\frac{dU}{ds}$ is anti-selfadjoint: set $A = \frac{dU}{ds}$, then what you have is $A + A^\dagger = 0$. Therefore by setting $A = iK$ you have that $K$ is a self-adjoint operator.



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