New answers tagged

1

An simple argument goes via the Choi-Jamiolkowski isomorphism: The Choi state is obtained by applying the channel to the left half of a maximally entangled state, $$ \vert\Omega\rangle = \sum_{ij} \vert i,j\rangle_{AB}\vert i',j'\rangle_{A'B'}\ . $$ The Choi state is then $$ \sigma_{\mathcal E}= (\mathcal E_{AB}\otimes I_{A'B'})\,(\vert\Omega\rangle\langle\...


1

Pauli's Exclusion principle states 2 identical fermions can not occupy the same quantum state. In essence when we have identical particles, the probabilities will be invariant under exchange of particles. In other words, if we have 2 particles one in state $|{\phi}\rangle$ and another in $|{\psi}\rangle$ then we can’t distinguish between this and the state ...


0

The entries at time $t$ are $\psi(x,y,z,t)$. The arrangement of as a column vector, or as a three dimensional array, is essentially arbitrary. You may find it easier to think of it as a three dimensional array.


1

Why don't they commute? You can work out the maths. It does not have anything to do with the factor but the form. $$ \begin{align} [\mathbf L \cdot \mathbf S, \mathbf L_x] & = [L_xS_x+L_yS_y +L_zS_z, L_x] \\ & = S_x[L_x,L_x] + S_y[L_y,L_x] + S_z[L_z,L_x] \\ & = 0 - S_y (i \hbar L_z) + S_z (i \hbar) L_y \\ & = i\hbar (\mathbf L \times \...


0

This should be a comment, but it is too long. The amplitude, $Ψ$ as : $e^{i(\omega t-\vec k\cdot r)},$ The observable is the complex conjugate squared $Ψ$ , which gives the probability, the only measurable quantity. When $\omega t=k.r$ $Ψ$ becomes $e^i$, a complex number . It is $Ψ^*Ψ$ that becomes equal to 1, a real number. When the probability ...


0

This is a scattering problem as opposed to the eigenstates problem. In principle these must be treated in different chapters, since they point out to different settings and dufferent mathematical developments. So you are right to pose the question. What is hidden under the carpet is that for the bound states in a square well one cannot form a scattering ...


3

The Pauli principle states that the full many-body fermionic state must be antisymmetric (i.e. pick up a minus sign) under permutation of any two fermions. If you have 2 fermions occupying any two states $\psi$ and $\phi$, then the 2-fermion state will be (up to an overall phase and normalization) $$ \psi(1)\phi(2)-\psi(2)\phi(1)\, . $$ This generalizes to a ...


2

The space of spin states for a single electron is two-dimensional (spanned, say, by UP and DOWN in whatever direction you care to choose). Therefore (by simple algebra) the space of antisymmetric spin states for a pair of electrons is one-dimensional (spanned by the single vector $\hbox{UP}|\hbox{DOWN}-\hbox{DOWN}|\hbox{UP}$ ). [Edited to add in ...


1

If you are okay with approximations and some nasty follow-up calculations, you can do the following not-so-delightful recipe: 1- Expand the $V(r)$ in its analytical form using Maclaurin Seires. (Hint, use $r=\zeta + 1$) 2- Stop when you reach the order that you feel appropriate for you. (e.g. 3) 3- Solve the Schrödinger Equation for each term after ...


2

Just integrate by parts a few times to get it in the desired form. For the first term in the identity, $$\int d\mathbf{r} \, (\nabla^2 \psi^*) \psi = - \int d\mathbf{r} \, (\nabla \psi^*) \cdot (\nabla \psi) = \int d\mathbf{r} \, \psi^* \nabla^2 \psi.$$ For the third term in the identity, $$- 2 \int d \mathbf{r} (\nabla \psi^*) \cdot (\nabla \psi) = 2 \int ...


6

I do not understand why one can just swap inner products These inner products are just complex numbers and therefore commute with each other. The analogous thing with 3D vectors (where the inner product is a real number) is unsurprising: $$(\mathbf{a}\cdot\mathbf{b})(\mathbf{c}\cdot\mathbf{d})=(\mathbf{c}\cdot\mathbf{d})(\mathbf{a}\cdot\mathbf{b})$$


0

Assuming we are working with bosons, then: $$ \langle k_1,k_2|k_3,k_4\rangle=\delta^{(3)}(k_1-k_3)\delta^{(3)}(k_2-k_4)+\delta^{(3)}(k_1-k_4)\delta^{(3)}(k_2-k_3) $$ To see this, you can use the commutation relations to move all the creation operators to the left and annihilation operators to the right. Those terms with annihilation operators acting on the ...


0

The mistake is in the "little bit of algebra" at the end where it is transformed into an undesired and probably incorrect form. First note that: $$ \Delta \lambda = \lambda- \lambda_0 = \frac{h}{p} - \frac{h}{p_{0}} = h\big( \frac{{p_0}-p}{pp_{0}}\big), $$ so we will first try to get all terms that contain $p_0-p$ to the LHS. I will rewrite your last correct ...


0

Yes, your reasoning us correct. Two comments though: it is often more practical to act with all the operators on the right, rather than separate them in two groups. you need to account for the exchange effects - your answer is partial ($k2=k3, k1=k4$ and $k1=k3, k2=k4$ with appropriate signs).


1

There's no observable physical significance. Since you can only ever observe a probability density given by a wavefunction, the exact phase of the wave is not observable. Physically, it makes no difference whether at time $t$ and location $r$ the amplitude is $1$, $i$, or $\frac{1-i}{\sqrt{2}}$. What you can, however, measure, is the relative phase between ...


0

There is no physical significance of the complex amplitude being exactly 1 in the above example. It just means it will constructively interfere with an amplitude of 1 and destructively interfere with an amplitude of -1. Just as an amplitude of i will constructively interfere with an amplitude of i and negatively with an amplitude of -i. 1 is in no way ...


0

I think you're asking about how to find a probability of a particle, given its wavefunction? First of all, wavefunction MUST be normalized. That is, the first step you should do in any QM problems is to make sure that the integration (with respect to x, from minus infinity to infinity) equals to 1. In other words, the particle exists in this entire universe,...


0

There are many things in your question that are unclear. However, regardless of probability or not, what you have is $e^{i(kx-\omega t)}$ whose mod-square is always one. Your function $\psi$ isn’t a wavefunction because it isn’t normalised.


0

Let’s assume our quantum state is a superposition of $|0\rangle$ and $|1\rangle$: $$ |\psi\rangle = \alpha |0\rangle + \beta |1\rangle, $$ where $\alpha = r_1 e^{i \theta_1}$ and $\beta = r_2 e^{i \theta_2}$ are complex numbers. It is customary to say that the $\theta$ in the exponent is a phase, and the whole exponent $e^{i \theta}$ is a phase factor. ...


1

For low speed you could use Newton's relation between kinetic energy $E_k$ and momentum $p$: $$E_k=\frac{p^2}{2m}$$ But for high speed this is no longer valid. You need to use the relativistic energy-momentum relation instead: $$(mc^2+E_k)^2=(pc)^2+(mc^2)^2$$ Solving for momentum $p$ you get $$p=\sqrt{2mE_k+\left(\frac{E_k}{c}\right)^2}.$$ Applying energy ...


0

I believe the key fact is that the electronic transition dipole moment is a vector quantity, and the unit vector in its direction $\hat{p}$ is what is included in the angular integral. So the electronic contribution does not reappear in the second integral; this integral over a unit vector $\hat p$ in the direction of $\bf p$ can only be $\leq 1$. See ...


0

Nothing happens to it. The electron - or the nucleus or whatever other object that interacts with the photon to make pair production possible - just participates for momentum conservation, since a single photon producing two massive particles is kinematically forbidden. The usual pair production reaction is $$ \gamma + Z \to f^+ + f^- + Z,$$ where $Z$ is e....


-1

A photon does not need to collide with an electron to have pair production. During pair production, a high energy photon (near a nucleus) ceases to exist, and the photon's energy is transformed into a electron-positron pair. Energy needs to be conserved, thus, the energy of the incoming photon needs to be at least the energy (rest masses) of the electron-...


-1

An electron does not create an electron-positron pair. A photon creates an electron-positron pair. For a photon to do this it must be "off mass shell". It must be a high energy product of an earlier collision.


0

Fuchs directly addresses the issue of potential experimental distinctions in this interview: Quantum Physics is No More Mysterious Than Crossing the Street: A Conversation with Chris Fuchs https://www.discovermagazine.com/the-sciences/quantum-physics-is-no-more-mysterious-than-crossing-the-street


0

You only get an equation with a second-order time derivative for an antiferromagnet. If you have feromagnet there is only a first time derivative and you get a Landau-Lifshitz equation rather than a conventional wave equation. Deriving the antiferromagnetic equation is tricky. See F. D. M. Haldane, Phys. Lett. 93A, 464 (1983).


3

I think that the problem is that the square root of the Laplacian is a non-local opertor and non-locality is usully regarded as a bad thing in physics. The long range nature shows up in the general expression $$ (-\nabla^2)^s f(x)\equiv \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \int_{{\mathbb R}^n} d^ny \frac{\left\{f(x)-f(y)\right\}}{|x-y|^{...


1

I have to say I think the notation is horrible. I would not use it this way but it is clear that $$ \left|1_j \right> = \hat{a_j} ^\dagger \left|0\right> $$ and that this is not related to $\psi$. $\psi_j (x)$ are the coefficients of $\hat{\psi} (x)$ in the energy basis. One cannot think of $\psi_j (x)$ as referring to definite energy and position, ...


3

Every map CPTP $\mathcal E$ can be written as $\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}\newcommand{\calU}{{\mathcal{U}}}\newcommand{\calE}{{\mathcal{E}}}\newcommand{\calH}{{\mathcal{H}}}\newcommand{\tr}{\operatorname{tr}}\calE(\rho)=\tr_E[\calU(\rho\otimes\ketbra{0_E}) \calU^\dagger]$ for some (unique) unitary $\calU$. Explicitly, denoting with $K(\...


1

The Copenhagen interpretation has never been precisely defined and exists in many forms. One should mistrust any source which claims to state what it actually is. In particular, the dominant form, called the Dirac-von Neumann, or orthodox, interpretation by Jeffrey Bub, does not feature complementarity (wave-particle duality) and already treats quantum ...


2

A physical way to think about this is that, if you have a degeneracy then you have a symmetry. Namely, if $|1\rangle,|2\rangle$ are states with the same set of eigenvalues $\lambda_1,\ldots,\lambda_n$ under $A_1,\ldots,A_n$, then there exists an operator that rotates $|1\rangle$ and $|2\rangle$ into each other and leaves all other observables untouched $$ U(\...


1

As other answers have mentioned, the SE is invariant under Galilean transformation, and there are other PSE posts that cover this. However, I wanted to address some specific things in your question. Part of me says, no: due to the uncertainty principle, the probability of finding a free particle anywhere would seem to be uniform and negligible over all ...


3

The answer is negative if I correctly understand the question. There are observables which do not commute, whereas if each of them could be viewed in its own Hilbert space and the whole Hilbert space were the tensor product of all those Hilbert spaces, these observables would commute. What it is true is that there is a maximal set of pairwise commuting ...


3

Like all proofs about properties of "maximal" sets in any context, this one too proceeds by assuming that we have a set that lacks the property and then constructing something we can add to the set, showing it was not maximal: Assume we have a set of $n$ operators $A_i$ and that there is a degenerate common eigenvector, i.e. an (w.l.o.g.) two -dimensional ...


0

Even though you enlarge the unit cell in $x$ direction, you apply the magnetic periodic boundary condition in both $x$ and $y$. On the other hand, when you choose the magnetic unit cells to be $N$ times larger than the lattice ones, your Brillouin zone also shrinks $N$ times. Hence, the energy spectrum is usually depicted with respect to magnetic momenta. ...


0

The relation between the two cases is given by a unitary transformation. This means that $|\psi(x,t)|^2$ is necessarily preserved.


0

No $|\psi(x,t)|^2$ will not change. Reason Schrodinger equation is invariant under Galilean transformation. See this Galilean invariance of the Schrodinger equation


2

This is actually a bit easier to understand than for conventional particles. First let's define two things: A photon's momentum is determined by its wavelength: $p=h \lambda$ A photon's position is determined by the wavepacket The wave packet is an idea you don't normally think of in introductory physics. In order to have a position, the photon can't ...


0

Smooth binding potentials - Morse, Lennard-Jones, Poschl-Teller - that go to $0$ as $r\to \infty$ usually have a minimum. Expanding about this minimum gives a locally harmonic potential: hence the vibrational energy levels in molecules. For some molecules the anharmonic can be quite small so you can have 10s of vibrational level also exactly equidistant ...


0

The Hartree-Fock equation has the form of Eq.1 when to derived it, you use basis of the form $\Psi=P_{lm}(r)Y(\theta,\phi)$ where $P_{lm}(r)$ are solution of a spherical potential, that is Eq.2 is imposed.


2

In the Schrödinger picture, the Hilbert space $\mathcal{H}$ is physically the set of states at a given time. A function like $\psi(x,t)$ is not a state, but a time evolution of a state. Operators are also a priori not time dependent: they take functions of $x$ and return functions of $x$. A time dependent operator is really an operator valued function; you ...


3

A nice way to see what's happening is to construct an approximate "toy" state $\newcommand{\ket}[1]{\left|#1\right>} \ket r $. For instance you might construct a wavefunction that's "at" a location ten units to the right of your origin by $$ \ket{10}=\left\{ \begin{array}{cl} 1 & \text{where } 9<x<11 \\ 0 & \text{elsewhere} \end{array} \...


0

1) The equations of motion come before you have quantized the fields. You write down the Lagrangian(which is a classical object), find the equations of motion. Solve them, and THEN quantize your fields. Before this quantization procedure, your field is just a field-you haven't yet promoted it to an operator and therefore cannot talk about quantum mechanical ...


4

All of my work is based on the assumption that I can write the following:$$\Pi|\vec{r}\rangle = -|\vec{r}\rangle$$ This is wrong. $|{-\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $-\vec r$. $-|{\vec r}\rangle$ is an eigenvector of the position operator $\hat{\vec x}$ with eigenvalue $+\vec r$, since $$ \hat{\vec ...


0

In quantum mechanics one usually starts with the Hamiltonian and then finds the solutions (eigenfunctions and energies) from that. The Hamiltonian defines the problem, not the other way around. There are exceptions ofcourse but I suspect the Hamiltonian for this problem is somewhere in your notes. Anyway the usual Hamiltonian for the finite potential is ...


0

Suppose we have a square well of width $a$. If $s = x_1 - x_2$ is the separation, the density you want is given by $$P(s) = \int_0^{a-s} dx \,\left(|\psi_1(x)|^2|\psi_2(x+s)|^2+|\psi_1(x+s)|^2|\psi_2(x)|^2\right)$$ Where did this come from? You want the joint probability that you'll find particle 1 at position $x$ and particle 2 and a distance $s$ away from ...


0

The wave functions do not exist! They (alongside with the Hilbert space and its axioms) are mathematical tools for describing the reality. Moreover, this is not the only description possible: from the very beginning the matrix formulation of quantum mechanics of Heisenberg was competing with the wave mechanics of Schrödinger, until their equivalence was ...


-1

The answer to the question are there initially two singularities in a merged black hole or only one depends on which singularity we talk about. If Schwarzschild black holes are considered then the merged black hole has one singularity which is a point in time and which is not part of the manifold. It is well known that the Schwarzschild singularity is in ...


1

Your guess is correct: "einselection" refers to the fact the density matrix becomes diagonalized only with respect to a special basis (more generally, a special collection of mutually orthogonal subspaces), and these are the basis vectors that represent to the possible outcomes of the "measurement." We can then say that any observable which is diagonal in ...


0

I agree with aljg up to the third line from the bottom of his derivation. I would argue that from there, the derivation should go as follows. \begin{align} \langle \psi_{n \mathbf{k}} | \frac{\mathbf{p}}{\hbar} | \psi_{n' \mathbf{k}'} \rangle &= \frac{1}{V} \int d\mathbf{r} e^{i(\mathbf{k}'-\mathbf{k})\cdot \mathbf{r}} u_{n\mathbf{k}}^*(\mathbf{r}) \left[...


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