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0

A closed extension $A_c$ of an operator $A$ is an operator whose action is the same as $A$, the domains satisfy $D(A_c)\supset D(A)$ and $A_c$ is closed. Given that, the smallest closed extension of a symmetric (densely defined) operator is its double adjoint $A^{**}$. We call it the closure of $A$, and denote it by $\overline{A}$. An operator $A$ is ...


1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$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 $$ ...


1

Well, this is a problem of linear algebra substantially. The basic idea is that a square matrix $A$ is invertible if and only if $detA\neq 0$. Then, if the matrix is diagonalizable, since the determinant is invariant under coordinate transformations $A'=C^{-1}AC^1$, when you compute the determinant you get that, you get $detA=\lambda_1\dots\lambda_n$. (Note ...


3

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


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.


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

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 ...


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 ...


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 ...


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 ...


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 ...


0

This doesn't directly answer your question of orthogonality, but may still address your concern. I need to point out that you seem to be working on a scattering problem and resonant states. Resonant wave functions DO NOT belong to the Hilbert space. They are not even eigenfunctions of the Hamiltonian in an usual sense. You already know that they do not ...


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First, you took the second order Taylor's expansions of $e^{-itH}$ and $e^{-itH_0}$, so, if your following calculations were right, $H$ and $H_0$ would commute with each other just to second order of time $t$, not for all order of $t$ (or just approximatively commute). Second, for 4 operators $A,B,H,K$: $$ AHB=AKB \rightarrow A(H-K)B=0. $$ From this ...


1

The fact that $$ U_I(t_1,t_2)U_I(t_2,t_3) = U_I(t_1,t_3)\tag{1}$$ in the interaction picture does not rely on $H_0$ and $H_\text{int}$ commuting, but can be derived without that assumption from the Tomonaga-Schwinger equation $$ \mathrm{i}\partial_t U_I(t,t_0) = H_I(t)U_I(t,t_0)$$ with $H_I(t):=\mathrm{e}^{\mathrm{i}H_0 ...


1

Yes. Nonweak measurements correspond to Hermitian (or self adjoint) operators. The results are 1) an eigenvalue and 2) you project the state vector onto the corresponding eigenspace. The projections onto different eigenspaces produce eigenvectors with different eigenvalues, and eigenvectors of a symmetric operator with different eigenvalues are orthogonal. ...


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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. ...


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 ...


0

The volume normalization gives a simple form for the Hamiltonian. Try substituting the field operator into the Hamiltonian. The field must approach zero as the volume tends to infinity. To do otherwise would imply an infinite energy. This is not a quantum field phenomenon. That is the obvious boundary condition. You either have a specific volume over which ...


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 ...


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

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 ...


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 ...


0

This just means that any hermitian Operator $K$ will fullfill the requirement of unitarity for $U(s)$: $$s\left[\frac{dU}{ds}+\frac{dU^\dagger}{ds}\right]_{s=0}=0$$ This is because for any hermitian operator $K$ you will get $$s\left[\frac{dU}{ds}+\frac{dU^\dagger}{ds}\right]_{s=0}=s\left[iK+(iK)^\dagger \right]=0$$.


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.


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

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

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 ...


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.


1

I am confused. Why would $\langle x \rangle = \langle x \rangle + \delta x$? Because you acted with the translation operator on the state. This is by definition what we want the translation operator to do. If it doesn't do this then we are in trouble. Shouldn't it equal $\langle x \rangle?$ Nope. Since, $\langle x\rangle = \langle ...


2

Yes, that's what a commutator is. Remember that the momentum operator is $\vec p = -i\hbar\vec\nabla$, so $\nabla^2 = -p^2/\hbar^2$; also that $\nabla^2$ is a scalar, not a vector. Usually when you find a commutator you need a "test function" for the operators to operate on; that's how you find for instance that $[\hat x, \hat p_x]=+i\hbar$.


1

You made a mistake somewhere. My approach was to write $xp=i\hbar +px$ and $px=xp-i\hbar$ on the side of my paper and just substitute away. I first wrote $$\{x^2,p^2\}-\frac{(\{x,p\})^2}{2}=x^2p^2+p^2x^2-\frac{xpxp+xp^2x+px^2p+pxpx}{2}$$ The terms in the fraction are $$xpxp=x^2p^2-i\hbar xp$$ $$xp^2x=x^2p^2-2i\hbar xp$$ $$px^2p=x^2p^2-2i\hbar xp$$ ...


2

Prove $$(e^{i\lambda A})^\dagger=e^{-i\lambda A^\dagger}$$ where $A$ is an operator. Can anyone explain how to go about this question? Writing it as a power series gets confusing. So basically I get: $(e^{i\lambda A})^\dagger=\sum_{n=0}^\infty({(i\lambda)}^n\frac{A ...


4

I believe the difficulty stems from the archaic notation. What he is trying to do is show that $-\mathrm{i}\mathrm{D}$ is Hermitian, where $\mathrm{D}=\mathrm{d}/\mathrm{d}q$. Note the first equation you wrote. It is just saying that $\mathrm{D}$ acting "backwards" on the bra is equivalent to $\mathrm{D}$ acting "forwards" on the ket. He is trying to derive ...


1

The mean value of the position is given by $\langle \psi|\mathbf{X}|\psi\rangle$. Now insert the completeness relation for position states to get $$\langle \psi|\mathbf{X}|\psi\rangle=\int_{\mathbb{R}^3} \mathbf{x} |\psi(\mathbf{x})|^2\,\mathrm{d}V$$ From the OP, we see that the modulus squared of the wave function is ...


0

Preliminaries Recall that a representation of an algebra on a Hilbert space is a map from the algebra to the bounded operators on a certain Hilbert space. Also recall the Heisenberg canonical commutation relations $$[q_i,p_k]=i\delta_{ik}I$$ A representation of such relations is a set of operators on some Hilbert space that satisfy to the same commutation ...


7

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$): $$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$ Which ...


3

The operator $K$ on the Hilbert space $H$ is antilinear and it satisfies to $$(\eta,K^*\psi)=\overline{(\eta,K\psi)},\qquad\forall \eta,\psi\in H,$$ which is not what you would expect for self-adjointness. You can consider the conjugate Hilbert space $\overline H$ and turn $T$ into a linear map by regarding it as $T:H\to\overline H$. If you then take ...


0

$\hat{K} \Psi = \Psi^*$ is not a well defined operator at all. On the one hand it is clear that $\hat{K^2} = 1$ and $\hat{K}$ is therefore real. On the other hand by your definition $$ <\Psi| K |\Psi> = \int \Psi^{*2} d^3r \not\in \mathbb{R} $$ It is ill defined because it does not map from the space of the Kets into the space of the kets but instead ...


1

$\newcommand{ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 \big| #2 \right> }$ Though I am not sure 100% if what I am going to do is legitimate I would suggest the following (I am about 90% sure that it is legitimate): The confusion arises because the author has used $x'$ for two distinct ...


7

Don't be fooled by the zero inside the ket. That is just a label. For example in Scalar QFT, the vacuum state of interacting theories is usually denoted as $|\Omega\rangle$ rather than $|0\rangle$. So no, the vacuum state does not represent the null vector of the Hilbert space. Rather, the vacuum state is defined to be the state with the lowest possible ...


1

The vacuum state is (or mathematically looks like) the ground state of a ficticious harmonic oscillator. So you have a state you might label $| 0 \rangle$, but that is not a zero (0), but a basis state, the one for zero quanta in the harmonic oscillator.


1

Your concern is justified; a linear operator will always have an output of zero when the input is a zero vector. However, when you operate on the vacuum state with the annihilation operator, you do get a zero vector. What the vacuum state look like will then depend on your representation of creation and annihilation operators, but the rule that ...


2

No. I might be misremembering my linear algebra here, but the zero vector should be additive identity, correct? But the state $| 0 \rangle + | 1 \rangle$ is clearly distinct from the state $| 1 \rangle$, as can be shown by taking expectation values for any observable that differs between vacuum and first excited state.


0

I would like to determine the general expansion of $(A+B)^n$, where [A,B]≠0 The expansion of $(A+B)^n$ for non-commuting A and B is the sum of $2^n$ different terms. Each term has the form $$ X_1X_2...X_n\;, $$ where $X_i=A$ or $X_i=B$, for all the different possible cases (there are 2^n possible cases). For example: $$ ...


0

if $[A,B]=0$ then as you know you get the usual $$ (A+B)^n = \sum_{p=0}^n C^n_p A^{n-p}B^p $$ Now if $[A,B]\neq 0$ each term in the sum (for each $p$) splits into a sum of $C^n_p$ terms of all possible permutations of $(n-p)$ $A$s and $p$ $B$s, without regard to the order of $A$s and $B$s. Equivalently to the sum of all possible permutations of $(n-p)$ $A$s ...


1

I think you have the answer for your second question. For your first question let me clarify: $$\langle x\ |\ \hat{x}\ |\ p\rangle \overset{(1)}{=} \int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(2)}{=} \int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(3)}{=} x\langle x\ |\ p\rangle ...


2

There are many good answers already. This answer is basically an expanded version of Emilio Pisanty's answer. Let us start by recalling the standard convention to write the position wavefunction $$ \tag{1} \psi(x)~=~\langle x | \psi \rangle$$ as an overlap with a position bra state $\langle x |$. The CCR $$\tag{2} [\hat{x},\hat{p}]~=~i\hbar{\bf 1}$$ is ...


0

Second, if I try to do the same thing by using the $p$-basis representation of $\hat{x}$, I get into more trouble: $$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'i\hbar\frac{\partial}{\partial p'}\langle x\ |\ p'\rangle\langle p'\ |\ p\rangle$$ $$=\int dp' ...


1

This question (v6) [concerning the overall minus sign in OP's calculation] is essentially a Fourier transformed version of e.g. this Phys.SE post, see Emilio Pisanty's answer and my answer. The main point is again that the derivative in the momentum Schrödinger representation $$\hat{x}~=~i\hbar\frac{\partial}{\partial p}, \qquad \hat{p}~=~p,$$ acts on the ...


1

How do the eigenfunctions of the total angular momentum operator analytically look like? I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can we explicitely say what they are? The eigenfunctions of $J$ are going to be made up of linear combinations of tensor-product states of the ...



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