New answers tagged operators
1
Yes, if it's in a pure momentum eigenstate, which is the only way you would obtain a single eigenvalue of the momentum operator. There are complications due to the fact that you can't really have a system in a pure momentum state (though you can get arbitrarily close). But conceptually, that's what it means to have a single eigenvalue.
Note: applying the ...
4
Under a unitary $U$, for which $U^\dagger U = UU^\dagger = \mathbb I$, any arbitrary operator $A$ transforms as
$$
A \mapsto A' = U^\dagger AU.
$$
You're being asked to show that if $C=[A,B]$, then $C'=[A',B']$, or in other words, that
$$
U^\dagger[A,B]U = [U^\dagger AU, U^\dagger BU].
$$
The proof is simple but it's for you to work out.
4
No. Mathematically multiplying your state vector by an operator is not how you determine the outcome of a measurement. First, that would assume you can get a deterministic value, which we know isn't true of measurements of quantum systems. Second, if your system is (most likely) in a superposition of momentum states, then this operation will not even give ...
1
The number operator does not commute with the position operator. We have $$\hat{H}=\frac{\hat{P}^2}{2m}+\frac{1}{2}m\omega^2\hat{X}^2=\left(a^\dagger a+\frac{1}{2}\right)\hbar\omega=\left(\hat{N}+\frac{1}{2}\right)\hbar\omega$$
Then, $$\hat{N}=\frac{\hat{H}}{\hbar\omega}-\frac{1}{2}$$
where $\hat{H}$ is the Hamiltonian for the harmonic oscillator.
Using $[\...
2
Each pure State $|\psi_n\rangle$ of the quantum system is a multi-particle state describing all particles of the system. So your pure states are all possible states of the system. Since it is not prepared in a state $|\psi\rangle$ there are plenty of possibilites of states in which your systems could be.
Let us say you prepared your system in $|\psi\rangle$...
0
You're almost there. Note that
$$T(\delta x) \delta x = \delta x + \mathcal O(\delta x^2)$$
so if you stick to first order (i.e. consider infinitesimal $\delta x$), your result and the more general result from wikipedia agree.
3
Question (1)
Yes, $A$ is in a pure state. By definition, a state $\rho$ is pure if and only if it is a projector, that is $\rho^2 = \rho$. This is equivalent to $\mathrm{Tr}(\rho^2) = 1$. From this, it is clear that $\rho_A$ is a pure state.
Question (2)
I suppose you want an explanation of the definitions behind the notion of entanglement. If $\rho$ is ...
2
The requested lower limit is zero already for $X$ and $P$ as I am going to prove.
Let us consider the Fourier-Plancherel transform $F: L^2(\mathbb{R},dx)\to L^2(\mathbb{R},dx)$, formally for integrable functions (otherwise a further extension is necessary)
$$(F\psi)(x) = \frac{1}{(2\pi)^{1/2}} \int_{\mathbb R} e^{ixy} \psi(y) dx$$
It is clear that if $\psi$ ...
0
The Hamiltonian operator in quantum field theory is constructed by using the fields in question. For example, the Hamiltonian for a free scalar field reads
\begin{equation} \notag
H = \frac{1}{2} \int_V d^3x \left( \pi^2 + ( \partial_i \phi )^2 + m^2 \phi^2 \right) \, .
\end{equation}
The key observation is now that a given field evaluated at different ...
2
As mentioned in the comments,
are there any operators where I cannot simplify $\langle \mathbf{r}|\hat O|\mathbf{r}'\rangle$ to $O(\mathbf{r})\delta(\mathbf{r}-\mathbf{r}')$ and obtain
$$
\hat{O}=\int d^3r\,\hat\psi^\dagger(\mathbf{r})O(\mathbf{r})\hat\psi(\mathbf{r})?
$$
Yes, there are ─ the momentum operator is the trivial example, and it cannot be ...
1
Basically there is nothing that prevents you from writing such a term. The question is whether such non-local one particle operator will occur in a physical system. On the face of it, this operator 'teleports' a particle from ${\bf r}$ to ${\bf r}'$ instantly. Therefore, normally we would suspect any such single particle operator as unphysical.
However, in ...
1
Yes. You are correct. We can write operator $A$ as
$$ A = \left[\lambda_1 |\phi_1 \rangle \langle \phi_1| + \lambda_2\sum |\phi_n\rangle \langle \phi_n| \right] \otimes I_b$$
as it acts as the identity operator on subsystem $b$. Consequently, it doesn't matter what is the basis in which we write subsystem $b$, as the identity operator is the same in every ...
3
It may not be an eigenstate of $\hat L_z$ but, if the system is in a pure state, it will be an eigenstate of $\hat L_{\hat n}=\hat n\cdot \vec L$, i.e. it will be an eigenvector of the projection of angular momentum in some direction $\hat n$.
It may be tricky to find this direction but one way might be to make a beam that travels through weak magnetic ...
6
There seems to be some sort of misunderstanding here. Making a measurement of observable $A$ of a system in the state $|\psi\rangle$ does not mean we need to get a number from the calculation $A|\psi\rangle$. The issue here is that $A|\psi\rangle$ is still a vector. If you are expecting the measurement to give a value of $a$ for $A|\psi\rangle=a|\psi\rangle$ ...
5
We can always move to the eigenbasis given a hermitian operator like $L_z, H, L^2, \cdots.$ There is a theorem called Spectral theorem, which states that there exists an eigenbasis given hermitian operator, and this applies to finite dimensional, infinite dimensional, including Hilbert spaces.
Thus, since we have an eigenbasis, we can expand $|\psi\rangle=...
0
Briefly, the action (Lagrangian$^1$ density) in the path integral are functionals (functions), respectively, as opposed to operators. This is a consequence of how the path-integral formalism is derived from the operator formulation (by inserting infinitely many completeness relations).
The action & Lagrangian density usually don't depend on Planck's ...
0
The main point is that the 2-pt function $\langle {\cal O}_1{\cal O}_2\rangle$ acts a non-degenerate bilinear form on the infinite-dimensional vector space of linear operators (which consists of primary operators and descendants thereof), i.e. $$ [\forall {\cal O}_2:~~ \langle {\cal O}_1{\cal O}_2\rangle~=~0]\quad\Rightarrow\quad {\cal O}_1~=~0. $$
It ...
1
I think you have misunderstood the notation.
When you write
$$
\left( Q^{\dagger}\phi_1\right)^*
$$
the order of operations matters. First, one has to get image of $\phi_1$, through the operator $Q^{\dagger}$, which is a function in a Hilbert space of complex functions, and then the complex conjugate is obtained by complex conjugation of the function $Q^{\...
1
Recall that a Hermitian operator is its own conjugate, that is, $Q^{\dagger} = Q$. Then, we have: $\displaystyle \int \phi^{*}_{1}(x) Q \phi_2 (x) dx = \int Q^{\dagger} \phi^*_1(x)\phi_2(x)= \int Q \phi^*_1(x)\phi_2(x)$.
1
It might help to apply these relations to a familiar example such as the cartesian components of the 3D position operator, which forms a complete set of commuting observables.
In the 1st relation, I think he is using the Dirac delta as opposed to the Kronecker because he is assuming the eigenvalues are continuous as opposed to discrete. If we are in 3D ...
1
Stick to one dimension to avoid superfluous complication.
The operator $\hat p$ has eigenvalues, but on specific eigenvectors, labelled as such,
$$ \hat p | p\rangle = p|p\rangle,
$$
But unless your state $|\psi\rangle$ were one such, your first equation will not hold. You may always compute $\langle \psi | \hat p | \psi\rangle$, some sort of momentum of ...
4
In the number basis of the harmonic oscillator (where the number operator energy is diagonal), your text must (and certainly Wikipedia does) have
$$a^\dagger =\begin{pmatrix}
0 & 0 & 0 & \dots & 0 &\dots \\
\sqrt{1} & 0 & 0 & \dots & 0 & \dots\\
0 & \sqrt{2} & 0 & \dots & 0 & \dots\\
0 &...
4
Short answer: $x$ and $p$ in quantum mechanics are operators, not numbers. It means that, for example, $\hat{x}|\psi\rangle = |\phi\rangle$ where $|\psi\rangle$ and $|\phi\rangle$ vectors in our Hilbert space (up to normalization). As they are operators that correspond to physical observables, they must be Hermitian operators. The rules of how to take the ...
3
Since the Operator $H_0$ is Hermitian, $\langle\psi_p|H_0|\chi\rangle$ is equal to its adjoint $\langle\chi|H_0|\psi_p\rangle^*$. Now, we act to the right, and since $E^{0}_p$ is real, we obtain $E_p^0\langle\chi|\psi_p\rangle^*$ which is equal to $E_p^0\langle \psi_p|\chi\rangle$ since the unit operator is also hermitian.
2
The energy eigenvalue is a number - so many eV’s or whatnot - and not a matrix so it does not change when one goes from the $x$- to the $p$-representation, or when one makes a change of basis.
It is possible that, under a basis transformation, the Hamiltonian matrix no longer takes a diagonal form but the eigenvalues of the transformed Hamiltonian are ...
1
I think this is what you are looking for:Schwinger-Keldysh formalism. Part I: BRST symmetries and superspace
As mentioned the Keldysh formalism is the way to go. In this work they aim to provide a more mathematically elegant formulation of such formalism by means of a set of underlying BRST symmetries. It also has a chapter on computation of Out-of-time-...
1
Not to leave this nice question unanswered, a truly thorough analysis of the subject of self-adjoint extensions of symmetric operations is offered in the whole 3rd chapter of the book ”Self-adjoint Extensions in Quantum Mechanics. General Theory and Applications
to Schrödinger and Dirac Equations with Singular Potentials” by D.M. Gitman, I.V. Tyutin, B.L. ...
2
Consider
$$
\langle p|\hat x |x\rangle = x \langle p |x\rangle =x \frac{1}{\sqrt{2\pi \hbar }} e^{-ipx/\hbar}=i\hbar \partial_p e^{-ipx/\hbar} \frac{1}{\sqrt{2\pi \hbar }}= i\hbar \partial_p \langle p |x\rangle ,
$$
and note it holds for all x.
Can you now sandwich the above between $ | p \rangle $ and $\langle x| $ and integrate over x and p to get ...
1
Indeed, the eigenvalues of a tensor product of operators are just the product of the individual operators' eigenvalues, for the reason that you state.
Do you have Healey's exact quote? I think when he says "the product of the measurements" he means each of the four products of the three single-particle measurements, not the product of the four three-...
2
I'm answering the specific question posed in the final paragraph of the post. If this doesn't clear up your overall confusion, please feel free to comment or edit your question.
I will denote eigenvectors of $\widehat{\diagup}$ with eigenvalues of $+1$ and $-1$ as $\left|+\right>$ and $\left|-\right>$ respectively. I'll also use the notation $\left|+++...
0
Here a method based on generating function is given for calculating matrix elements of the sort (scheme is analogous to this)
$$\langle\{\alpha_{i}^{}\}|\prod_{i=1}^{N}\left[\lambda_{i}^{}a_{i}^{\dagger}+\mu_{i}^{}a_{i}^{}\right]_{}^{n_{i}^{}}|\{\alpha_{i}^{'}\} \rangle,$$
where $a_{i}^{\dagger}$ and $a_{i}^{}$ are bosonic creation and annhilation operators ...
5
There is no unique canonical notion of complex conjugation $C:H\to H$ of vectors in an abstract complex Hilbert space $H$. However, given a notion of complex conjugation $C:H\to H$, it is naturally to demand that it is an antiunitary map
$$\forall v,w\in H:~~\langle C(v) | C(w)\rangle~=~\overline{\langle v | w\rangle}.\tag{1}$$
(This is e.g. the case for the ...
1
First unitarily diagonalize $\sigma_{y}$:
$$
\sigma_{y} = U^{\dagger} D U
$$
where $U$ is a unitary matrix satisfying $UU^{\dagger} = U^{\dagger} U = \mathbb{I}$. It's always true that $D = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right]$, and I pick $U = \frac{1}{\sqrt{2}}\left[ \begin{matrix} 1 & 1 \\ i & - i \end{matrix}\right]$ ...
4
The relation is shown using a taylor series of the exponential:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ so that $e^{-i\pi/2\sigma_y}$ can be expanded.
$e^{-i\pi/2\sigma_y}=1+(-i\pi/2\sigma_y)+\frac{(-i\pi/2\sigma_y)^2}{2!}+\frac{(-i\pi/2\sigma_y)^3}{3!}+\frac{(-i\pi/2\sigma_y)^4}{4!}+\frac{(-i\pi/2\sigma_y)^5}{5!}+...$
Noting that $\sigma_y^2=\begin{...
1
Hint: Recall that $|\alpha\rangle$ is an eigenstate of $a$:
$$ a|\alpha\rangle = \alpha |\alpha\rangle \quad \text{and} \quad \langle \alpha | a^\dagger = \alpha^\ast \langle \alpha | . $$
Therefore, an expression like $a^{n-2k-s} |\alpha\rangle$ is substantially easier to evaluate if you do not expand $|\alpha\rangle$ in the number basis ($|n\rangle$).
In ...
1
Note that later in the text on p. 34 the formula (4.54) explains exactly how Zwiebach's coherent states $|\tilde{x}_0\rangle$ are related to the standard coherent states.
The intuition is that actual position-eigenstates does not belong to the Hilbert/Fock space, so instead we roughly speaking consider Gaussian wavepackets, centered around $x_0$.
1
Let's recall a simple but very powerful trick: $e^{k\hat{a}}|0\rangle=\big(1+k\hat{a}+...\big)|0\rangle=|0\rangle$ where $k$ is a scalar, $\hat{a}$ is the annihilation operator, and $|0\rangle$ is the ground state. You see, when we open the bracket, only the first term survives because the rest of them just act on the ground state and annihilate it. So, we ...
0
Adding the identity to your observable does not change the eigenvectors at all; it merely shifts both associated eigenvalues by a constant. So the operators
$$P_{A}=\frac{1}{2}(\mathbb{I}\pm\hat{a}\cdot\vec{\sigma})$$ still project out the eigenvectors of $A$.
11
Operators and derivatives
Let $X$ and $P$ be individual operators satisfying
$$
[X,P]=i.
\tag{1}
$$
The operator $X$ is just one operator, so there is nothing to differentiate here.
Now consider a one-parameter family of operators, with one operator $X(t)$ for each time $t$. This is often called an operator-valued function. That's just a fancy name for ...
0
Well, how we would find the probability of being in the same initial state (say $|\psi_n\rangle$) is as follows:
$$P(|\psi_n\rangle, t) = |\langle\psi_n| \Psi \rangle|^2$$
where $|\Psi\rangle = e^{-i H t}|\psi_n \rangle$. Physically, how you should think about this expression is that I am finding the projection of my quantum state at time $t$ on my initial ...
1
Hints:
Whenever you have a summation that goes over a Kronecker delta, you can just cancel out one of the indices, so that e.g.
$$
\sum_{ij}a_i b_j \delta_{ij} = \sum_{i}a_i b_i.
$$
For your specific case, you can do this term-wise, so that e.g.
\begin{align}
\sum_{ablm} \epsilon_{iab} \epsilon_{jlm} \delta_{bl} |{a}\rangle\langle{m}|
=
\sum_{alm} \...
1
The dynamics of every quantum system can be expressed in both the Schrödinger and Heisenberg pictures, and the two descriptions are formally equivalent.
In other words, for whatever calculation you would like to perform, the outcome of the calculation does not depend on the chosen picture.
(This situation is quite analogous to mechanics, where you are ...
1
The problem is a rather fundamental one in fact, and it cannot be overcome so easily, and it is not per se about having different Hilbert spaces, but rather inequivalent representations.
All separable, infinite dimensional, Hilbert spaces are isomorphic (they are thus the same mathematical structure essentially). Nonetheless, they accommodate infinitely ...
3
In linear algebra an operator is something which acts on a vector and returns another vector. A linear operator can be represented by a matrix. So in this sense you should think of differentiating an operator like differentiating a matrix. Take for example this time dependent rotation matrix:
$$\partial_tR=\partial_t\begin{pmatrix}
\cos\omega t & \sin\...
3
To define an operator it is sufficient to define its action on an arbitrary state $\Psi$. The definition of $d \hat{x} / d t$ is as follows:
$$ \left( \frac{d \hat{x}}{dt} \right) \left| \Psi \right> = \frac{d}{dt} \left( \hat{x} \left| \Psi \right> \right). $$
Equivalently, the matrix elements of the derivative of an operator are derivatives of the ...
3
Consider the Hilbert space $\mathcal H = L^2(\mathbb R)$ - roughly, the set of square integrable functions on $\mathbb R$. The momentum operator $P$ is a linear map
$$ P : \mathcal D_P \rightarrow \mathcal H$$
$$ \psi \mapsto -i \psi'$$
where $\mathcal D_P \subseteq \mathcal H$ is some densely-defined domain. First question - what is the appropriate ...
2
I think that you should get familiar with Quantum Mechanics's groundings from functional analysis. I can suggest you lectures from Frederic Schuller, it will be easy for you to find it in internet. The key point is there a lot of ways how to determine definitions in Quauntum Mechanics and in your course you were introduced to nonrigorous way. Most rigorous ...
0
Yes, the non-unitary Lindbladian time-evolution of an open quantum system can be formulated in the Heisenberg picture: https://en.wikipedia.org/wiki/Lindbladian#Heisenberg_picture.
0
$\newcommand{\bra}[1] {\left< #1 \right|}
\newcommand{\ket}[1] {\left| #1 \right>}
\newcommand{\bracket}[2] {\left< #1 \vert #2 \right>}
$
Your calculation of the Hamiltonian
$$\hat{H} = \hbar^2 \ket{-z}\bra{-z}$$
is correct so far.
One eigenvector is $\ket{-z}$, with eigenvalue $\hbar^2$.
But you missed (as Dani already wrote in his comment):
...
-1
My quantum mechanics is a little rusty but I will attempt an answer. The problem is that the Hamiltonian is not Hermitean. A Hermitean operator in this case would have two eigenvectors. You need to add the hermitean conjugate of the S_{-}S_{+} term to the hamiltonian. From what I remember this is common practice in condensed matter physics.
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