New answers tagged

1

For context, any general quantum operation $\Phi$ on a bipartite system $AB$ with finite dimensional Hilbert space $H_{A} \otimes H_{B}$ has a Kraus representation $\Phi(\rho) = \sum_j K_j \rho K_j^{\dagger}$ where the "Kraus operators" $K_j$ are linear operators on $H_{A} \otimes H_{B}$ such that $\sum_j K_j^{\dagger} K_j = I_{AB}$. The Kraus ...


0

Teleportation and entanglement do not involve FTL communication of any kind. There is a local description of the evolution of any given quantum system in terms of its Heisenberg picture observables. The observables change only when the system changes by itself or through a local interaction with another system. Entanglement and teleportation can be ...


5

When you have a bunch of interrelated phenomena in physics, trying to figure out which one is the "reason" for the other ones is often just a recipe for confusion. Different people will start from different postulates, so they will disagree on which results are trivial and which aren't, but hopefully everyone agrees on what's true. In a first course on ...


1

Yes, this is not the "true" reason. The direct reason is that they are not commuting operators. See the Robertson-Schrodinger Relation. You end up getting that the product of the uncertainties is bounded by 1/2 of the absolute value of the failure to commute. In the case of position and momentum this is $\hbar/2$.


0

It is quantum tunneling if the step potential has finite length. In this case, the continuity conditions on both sides of the step potential will show you that there is a nonvanishing amplitude behind the potential. If the potential has infinite length, there is no space to tunnel into.


1

Direction of deflection of electrons in magnetic field It is not the full picture you are describing. A moving electron in a magnetic field gets deflected according the rule $ \vec F = q \vec v \times \vec B $. This vector product has a direction and and this is what we observe in natur: all electrons get deflected in the same direction. Otherwise no ...


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I would like to add to @M.Enns that the magnetic lines of force is a brilliant discovery of the great mind like Faraday. They are just the visualization tool. You can visualize the direction and strength of the magnetic field from the direction and density (which is also virtual) of magnetic lines. Sometimes $1\times10^8$ field lines are associated with 1 ...


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The Schrodinger equation provides the definite energy space an electron can occupy inside a shell in the atom. The wave function, one of its variables, actually gives the probable location of an electron in space.


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No, magnetic lines of force don't flow. They have a direction, which shows the direction of the magnetic field but there is nothing flowing. If you were to place a small magnetic dipole at the location of the magnetic field line its north pole would feel a force in the direction of the line of force. The phrase "line of force" was introduced by Michael ...


0

Erasure doesn't work the way you think it does. Basically it goes like this... The most obvious way to 'erase' entangled information is to just undo the operation that created it: Notice how the top qubit has been restored to a pure state by the end, despite that nasty entanglement in the middle. But you can get a bit fancier than just repeating the ...


1

Yes. We can do this for any number of qubits using N dimensional spherical coordinates. For two qubits if we can write a general density matrix as a linear combination of direct products of Pauli matrices and identity , $$ \rho = \sum_{ij=0}^3 a_{ij}~ \sigma_{i} \otimes \sigma_{j}$$. Here $\sigma_{0} = I$, and the rest are the usual Pauli matrices. For ...


4

Considering only the spin, ignoring translational DOF, the Hamiltonian is $$H = -\mu ~\mathbf{B} \cdot \mathbf{S}$$ If $\mathbf{B}$ is directed along $z$, it's easy to see that $S_z$ states are energy eigentstates and thus are stationary. Applying time evolution and taking the expectation value shows that if the spin is not oriented along $z$ initially, ...


1

A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix. Your qubit representation could be rewritten, more suggestively as: \begin{align} \rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \...


-1

A one qubit state may be written in general as \begin{equation} |\psi_1\rangle=\alpha|0\rangle+\beta|1\rangle \end{equation} where $\alpha,\beta \in \mathbb C$ and there is the further restriction that $\langle \psi_1|\psi_1\rangle=|\alpha|^2+|\beta|^2=1$. However, only the relative phase between $|0\rangle$ and $|1\rangle$ is physically meaningful (quantum ...


1

Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are not used because measurements must use real numbers, but rather because we almost always decide to use real numbers. As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able ...


1

Quantum field theory is based on quantum mechanics. The ground state which describes the fields is the free particle solution of the corresponding Dirac/KleinGordo/Maxwell equation. QFT is a theory developed to be able to calculate the many body interactions, seen even in the simplest feynman diagram. It posits fields of each type of particle described by ...


-1

As far as I know, QFT does modify QM. Single particle description is problematic when you want to incooperate with Special Relativity, e.g. negative probability. Hence a quantum field theory is needed. I recommend the book Quantum Field Theory for the Gifted Amateur, which is very useful for learning QFT.


3

When in doubt go back to the masters. From Dirac's Principles of QM When we make an Observation we measure some dynamical variable. It is obvious physically that the result of such a measurement must always be a real number, so we should expect that any dynamical. variable that we can measure must be a real dynamical variable. One might think ...


1

The word "coherent" is used in Physics in a rather sloppy way. Your first state is a linear combination of harmonic oscillator eigenvectors that turns into a gaussian in momentum/position representations. In a more general background, a coherent state is just a state where coherences (off-diagonal terms in the density matrix) are non-zero, which means the ...


0

You're right, the usual language of band theory doesn't apply in the out-of-plane direction. The system really becomes a quantum well in that direction, so you will have a discrete spectrum of energy levels and there won't be any dispersion with $k_z$ ($dE/dk_z = 0$) if $z$ is out of the plane. The relevant tunneling processes in that paper are basically ...


0

I'm assuming your input photon has a known polarization (say horizontal). You won't see interference, because the polarizers act as a "which-path" measuring device. If you erase the polarization information, the interference pattern will appear.


0

You can model the wave function as collapsing into either $|slit1\rangle$ or $slit2\rangle$ when the electron passes through the slits, or when the electron hits the detector, or when a human experimenter comes along and examines the results. The corresponding theories make all the same predictions (this is a theorem of von Neumann), so it's entirely up to ...


1

Let's imagine the double slit experiment you proposed as something simpler, yet equivalent. In the electron beam experiment you have a free particle's wavefunction that suddenly faces a decision: go left or right, and then collapses at a screen, giving you a result. I propose finding an analogous example with spin: suppose we have a hydrogen atom trapped ...


1

Two points: A radio or tv signal is certainly not a thermal state. A thermal EM state looks like blackbody radiation, and is indeed incoherent. An EM field that has a well-defined phase and frequency, like an FM radio wave or a monochromatic laser, is certainly coherent. Whether it should be said to have 'quantum coherence' is debatable. It could be ...


3

In the Heisenberg picture, one simply has $$ A(t) = \exp(-Ht/i\hbar) A(0) \exp(+Ht/i\hbar) $$ The Hamiltonian $$ H = \frac{E_1+E_2}{2}\cdot {\bf 1} + \frac{E_1-E_2}{2} \cdot \sigma_z $$ while $$ A(0) = a\sigma_x $$ The term in $H$ proportinal to ${\bf 1}$ cancels in $A(t)$ so we have $$ A(t) = a\cdot \exp(-(E_1-E_2)t\sigma_z/2i\hbar) \sigma_x \exp(+(E_1-E_2)...


2

Let us make things clear. Protons and electrons are quantum mechanical entities and there is little meaning to project classical electrical attractive behavior to the micro framework of quantum mechanics, nor classical electric field calculations . Classically, a negative charge attracted to a positive charge will experience acceleration, and accelerating ...


3

Both $$\sum_i |i\rangle \langle i | $$ and $$\sum_j |j\rangle \langle j | $$ are summations over basis vectors. The indices $i,j$ run over the same values – values of indices that identify the basis vectors in the same basis (set of vectors) – but the particular values of the indices $i,j$ are independent. Can you calculate how much is the expression below?...


2

When don't you condition on the result, measurement of a qubit can only decrease its purity (you end up with less information than you started with). When you do condition on the result, measurement of a qubit will make it 100% pure but there are two possible results. One possible result is along the measurement axis you measured. The other is against the ...


1

$f(E)$ is the probability that a quantum state of energy $E$ is occupied. There are two quantum states (for two spin states) at each energy. The probability cannot be doubled, since that could then exceed 1. All that happens for a spin $1/2$ particle is that the number of available quantum states is doubled.


0

Maybe it is not relevant to you now, but it will be to someone. $[x,p]$ is Heisenberg algebra and you can easily see that this algebra is solvable. You have Lie theorem that says that every finite-dimensional irreducible representation of solvable algebra has to be one dimensional. So if you have one dimensional representation everything would commute and ...


0

Following Garyp's comment; "r , being the radial coordinate, has no knowledge of direction. Another way to see this is to note that the dipole matrix element has to be a vector. Calculating the rr matrix element gives you just one number. Perhaps more to the point: you want to find the matrix element of r⃗ r→, not rr. – garyp" The dipole element should be; ...


0

The annihilation of an electron positron pair most commonly creates either two or three photons in the final state (depending on the original angular momentum of the pair). In each case both energy and momentum must be conserved so there are kinematic limits on the energies of the photons as measured in the center of momentum frame of the original pair. ...


0

One can indeed see the reason behind the Heisenberg uncertainty principle as a mathematical propeties. The link between the physics and the mathematics is provided by the foundational work of Planck and de Broglie. They established the link between energy/momentum and frequency/k-vector. The general textbook on quantum mechanics therefore always starts by ...


1

Basically yes, up to your inaccurate articles. When you say "gives us the eigenstate $|\alpha+2\vartheta\rangle$", the word "the" indicates that you believe that the eigenstate with this eigenvalue has to be unique. But you haven't proven so and it doesn't have to be the case in general. So you wanted to say "an eigenstate with the eigenvalue that would be ...


1

Indeed, the quantum mechanical evolution of a strict 2-level system won't be exponential: it will be sinusoidal i.e. periodic. The probability of being in one of the states is $a+b\cos\omega t$ for some constants $a,b$. The coefficient $b$ is nonzero whenever the two states deviate from the energy eigenstates. The exponential decay only appears when the ...


5

Short answer: of the order of a nanosecond for hydrogen for "allowed" transitions, and the emission rate scales roughly as $Z^2$, where $Z$ is the atomic number. For an oxymoronically named "forbidden" transition, these times increase to tens of milliseconds or fractions of a second. So let's elaborate: what sets these times? A point not made enough is ...


1

The state is still $ \psi (t=T) $ right after the rapid change of B field to y-direction because the system doesn't have enough time to response the change.


1

Remember that $\eta^{q}$ are Fermionic operators, therefore $\eta^{2}_{q}=0=\eta^{\dagger 2}_{q}$.


2

It is better to see an explicit example. Consider a vector $\vec v\in\mathbb R^3$. There are three linearly independent subspaces of $\mathbb R^3$ generated by the unit vectors $\vec e_i$, $i=1,2,3$. If you act with the most general block diagonal $3\times 3$ matrix on $\vec v=\sum v_i\vec e_i$ you get $$ \begin{bmatrix} a&0&0\\ 0&b&c\\ 0&...


1

You said "Fermi level is constant throughout the junction" - that's correct. But fermi level is "A" (see top right in the table). So A is constant (you can set it to zero if you like). B is not constant.


2

Let us label the state spaces clearly as $\mathcal{H}_1$ and $\mathcal{H}_2$ for the first and second particle respectively and denote the canonical isomorphism sending a state in $\mathcal{H}_1$ of the first particle to the exact same state of the second particle by $\phi : \mathcal{H}_1\to\mathcal{H}_2$. Let us further denote the canonical "flip ...


2

You did all right, but you forgot about delta function's property. $$ \delta(x)f(x) = \delta(x)f(0) \, . $$ That's correct for every smooth function (it is hard to tell what $\theta(x)\delta(x)$ is, but there is a way to define it too). Physicists usually explain it superficially. Therefore $$ \delta(r'-r) \exp \left[-\frac{\gamma}{2}(r'-r)^2 \right] = \...


0

The exponential factor is a phase factor, where a change in the exponent represents a rotation in the complex plane. Because $\xi^2$ multiplies a very large number (both $\epsilon$ and $\hbar$ being very small), when $\xi$ is not small any $d\xi$ makes the rotation "fast" compared to the corresponding change in the other factor (the phase is proportional to ...


0

The evolution of the system is govern by the Hamiltonian. Let the system is evolved from the initial state $\rho_{i}=\frac{1}{Z}\exp^{-\alpha H(t)}$. Here $Z=\mbox{Tr}\exp^{-\alpha H(t)}$ and $\alpha$ is the temperature of the initial state. Suppose the Hamiltonian $H(t)$ is varying in time. If $\rho_{f}=\frac{\exp^{-\beta H(t\rightarrow\infty)}}{\mbox{Tr} ...


1

Consider the probability of state $n$ divided by the probability of state 0: $$\frac{\text{prob of }\left \lvert E_n \right \rangle}{\text{prob of }\left \lvert E_0 \right \rangle} = \frac{\exp \left( -E_n / k T \right)}{\exp \left( - E_0 / k T \right)} = \exp \left( - (E_n - E_0) / kT \right) \, .$$ If $E_n > E_0$, then as $T \rightarrow 0$ the ...


1

To normalize this function you have $$ 1~=~\int_0^{2\pi}\Phi^*\Phi d\phi ~=~|C|^2\int_0^{2\pi} e^{-im\phi}e^{im\phi}d\phi~=~|C|^2\int_0^{2\pi}d\phi~=~2\pi|C|^2. $$ The result is then obvious.


2

Particle interpretation indeed can be understood from the point of quantization of electromagnetic field. Basically, as the Maxwell equations are linear, each Fourier mode of the electromagnetic 4-potential $A^\mu$ is independent from others. This greatly simplifies the description, as the system is effectively diagonalized. Each Fourier mode satisfies the ...


0

I figure that given the rather extensive question that it does deserve some answer. I don't know if there is stack exchange on the history of science or physics, but that might in fact be a more appropriate place for this. The grandfather of quantum mechanics was Max Planck. His assumption that distributions of energy occurred in discrete units is what lead ...


2

The expression you give, $$\rho_G=\frac 1Z e^{-H/kT},$$ is already explicit. If you want an explicitly diagonalized expression, then you can use the fact that a function $f:\mathbb C\to\mathbb C$ is defined as acting on Hilbert space operators $A$, using the eigenvector route, as giving $f(A)|a⟩=f(a)|a⟩$ whenever $A|a⟩=a|a⟩$. Thus if $$ H=\sum_n E_n|E_n⟩⟨...


1

Planck's work was not in resolving the ultraviolet catastrophe, it was in alignment with Wien's law (a thermodynamic approach). Douglas Stone goes into this issue quite in depth in his book on the history of quantum mechanics. Planck's role is seminal but also overstated (and he disagreed vehemently with Einstein's arguments in his 1905 and 1906 papers on ...



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