# Tag Info

1

I can see some mistakes : "I realized that we must have $|F\rangle=\sum_\alpha\int\;\mathrm{d}\textbf{k}\,\Theta(k_F-k)\,\hat{a}^\dagger_{\textbf{k},\alpha}|0\rangle$" This is wrong : the ground state of an non-interacting fermion gas (Fermi sea) is not some sort of superpositions on the individual $\textbf{k}$ states as you seem to suggest. It has to ...

1

Write the terms $U |i \rangle$ and $V | i \rangle$ explicitly in the matrix form, e.g., $U | i \rangle = \sum_{k} | k \rangle U_{ki}$ and analogously. Then play with exchanging the three summations of indices, you should get it proved.

1

This is one of the mysteries of quantum mechanics. If you could measure the velocity of an electron, you would get a non-zero value. But what you can't do is use that velocity to predict where to find it next. The act of measurement disturbs the electron in an essential way. One popular interpretation of quantum mechanics is a statistical one. This says ...

0

I don't know where you have trouble, but it's straightforward if you just assume $R(x,t)$ and $S(x,t)$ in the form given for $\psi$ and then expand out the operations, where: \begin{align} \psi &\equiv R(x,t)e^{i S(x,t)\hbar} \\ \partial_t \psi &\rightarrow \Big(\frac{\partial_t R}{R} + \frac{i \partial_t S}{\hbar} \Big)\psi \\ \nabla \psi &\... 0 As a addition to the reply by @Jose Javier Garcia and in the spirit of the reply by @anna v, in molecular spectroscopy, anharmonic potentials are often approximately described by the Morse Potential, but this is insufficient in most cases and then the Rydberg-Klein-Rees (RKR) numerical method is very widely used to obtain the potential energy profile. This ... 2 Your question is posed in classical language, yet it involves quantum systems (many-body systems in fact) that cannot be treated with classical physics. The best that can be said in this framework is that the probability of such a swap is greater than zero. In fact, quantum many-body systems are theoretically formulated in such a way to allow such swapping ... 4 ... an electron is point sized Here you find what John Rennie says about this: Although it's commonly said that fundamental particles are point particles you need to be clear what this means. To measure the size of the particle to within some experimental error d requires the use of a probe with a wavelength of λ=d or less i.e. with an energy of ... 1 To answer this question, you'd have to agree on what model of the electron you're talking about. Quantum mechanical? Classical? Electrons can have force exerted on them by electric fields. If this causes the electron to move, then work is done to it. Thus, energy is transfered "to" the electron. 4 Use the Baker-Campbell-Haussdorf formula to commute the displacement operators, e^{\alpha a^\dagger} e^{-\alpha^* a} = e^{|\alpha|^2} e^{-\alpha^* a} e^{\alpha a^\dagger} $$and rearrange the matrix element as$$ \langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |m \rangle = \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \langle 0 | a^n e^{-\alpha^* a} e^{\alpha a^\...

0

People say the "electron" releases the energy for shorthand, but once again the energy exchanges in the photoelectric effect have to do with photon energy and the electron's orbital energies.

2

Assuming you have it set up that $\langle n\mid m\rangle=\delta_{nm}$ Then taking the sum: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\langle n-l|m-k\rangle$$ This simplifies to: $$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{... 1 It is at least possible to simplify your second expression. Note that (if I've not miscalculated)$$\langle m|e^{{\alpha}a^{\dagger}}e^{-\alpha^*{a}}|n\rangle = \left( \langle n|e^{{-\alpha}a^{\dagger}}e^{\alpha^*{a}}|m\rangle \right)^{*}$$so we can without loss of generality assume n > m. Now, in your second sum, we can use that$$\langle n-l|m-k\...

1

Interaction between a physical object in a superposition, and another object, usually causes the second object to also go into a superposed state, that will be correlated in some way with the original superposed state. For example, this is true of the quantum-mechanical description of a measurement interaction - some part of the measurement device has to ...

0

Being a layman, probably better answers are also possible. Here is what I know/think. The problem is, that there is no evidence. The LHC is actively on the search to find extra dimensions, without any result (until now). The branes are beyond SM things, i.e. there are a lot of papers about them, but no experimental data proving or disproving them. A ...

3

when I open the box and see the cat dead, there should be a timeline split, creating a universe where the cat is alive instead of dead. Is this possible? I know we can't really know for sure, but is it possible? One has to keep in mind clearly that physics is about mathematical models that fit data and predict new observations. One can have an infinity of ...

1

Suppose $V$ is a real or complex vector space with inner product, $\{\left|\phi_i\right\rangle\}_i$ is a basis, not necessarily orthogonal, $A$ is an operator on this space, so $A:V\to V$. Since $\{\left|\phi_i\right\rangle\}_i$ is a basis we can certainly expand as follows $A\left|\phi_i\right\rangle =\sum_j a_{ji}\big|\phi_{j}\rangle$ The numbers $\{a_{... 1 I'm going to adjust notation so that the whole thing becomes much clearer for you. Let's work with linear operators$A$,$B$and$C$, which act on our Hilbert space, and use no notation for matrices. So instead of "$\mathcal{A} \mathcal{B} = \mathcal{C}$", I write$AB=C$. Instead of "$\mathbf{A} \mathbf{B}= \mathbf{C}$", I will write that for all$i$,$k$: ... 0 I think it may be helpful for the answer to take a look at an intermediate theory. Namely, consider a scalar field, and, instead of this field being defined on$\mathbb{R}$or$\mathbb{R}^3$, using a lattice on$\mathbb{Z}$or$\mathbb{Z}^3$, or, even better, a finite lattice on on$\mathbb{Z}_N$or$\mathbb{Z}_N^3$with periodic boundary conditions. In ... 16 Rigorously speaking, the probability to find the electron at a distance exactly equal to$r$from the nucleus is$0$. On the other hand, we can define the probability to find the electron in a volume$d \mathbf{r}$as $$P(d \mathbf r) =| \Psi(\mathbf r)|^2 d \mathbf r = |\Psi(r,\theta,\phi)|^2 r^2 \sin \theta \ dr \ d\theta \ d \phi$$ where I have ... 2 Your initial state is entangled, which means it exhibits a correlation between the spin of the two particles. Each particle is spin up or spin down with equal probability, but the spins are (anti)correlated so that they cannot have the same spin. The states$\left|\uparrow \downarrow\right>$and$\left|\downarrow \uparrow\right>$are in agreement ... 0 First of all, notice that you can either Taylor-expand first, and then integrate, or first integrate and then Taylor-expand. You wrote down the latter, so I'll write down only the former. You can write the operatorial identity (assuming$H$is analytic on the states) $$H(t')=H(t+(t'-t))=H(t)+\frac{dH}{dt'}(t)\ (t'-t)+\frac{1}{2}\ \frac{d^{2}H}{dt'^{2}}\ (t'-... 4 "The Program of QFT" might be specified as follows. Specify a Lagrangian L for the particle content of the model in question. For instance, a scalar field of mass m would be$$L=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi+\frac{1}{2}m\phi^2$$or a Fermion field might be$$L=i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi-m\bar{\Psi}\Psi$$(those are four-... 2 "To be at rest" in classical mechanics means "to have definite position and zero momentum", the two properties being (again, in CM) equivalent: if something has definite position, then it must have zero velocity thus zero momentum, and if it has zero momentum, then it must have zero velocity thus definite position. Depending on which of the two properties ... 3 The time-independent Schrodinger equation \hat{H} \psi = E \psi only holds when the Hamiltonian does not depend explicitly on time. If you start with a time-independent Hamiltonian and make a time-dependent gauge transformation, then the new Hamiltonian will depend explicitly on time, and there is no reason to expect that the (time-dependent) eigenvalues ... 0 The maximum entanglement entropy of a system whose Hilbert space is N-dimensional is S = \ln N: https://en.wikipedia.org/wiki/Von_Neumann_entropy#Properties. 2 The answer can be found in section 6 of http://www.sciencedirect.com/science/article/pii/0550321381902376. Above the Hagedorn temperature, the strong interaction becomes long-ranged and falls of as 1/r, just like the Coulomb interaction. As you braid a charge-e/3 quark around a Dirac string, it picks up an Aharonov-Bohm phase from the quark's electric ... 5 Measurement is a process in which by help of interactions one quantum entangles the quantum system under study and the environment and by rapidly utilizing decoherence cleans up the "off diagonals" of the entanglement into an apparent collapsed state. After this process, it effectively appears as the wave function of the system has been set to a distinct ... 0 Ok I think I miss a really important thing. in discrete express. I use primed for discrete and umprimed for continuous$$|P'\rangle = \sum^n |p'_n\rangle$$|p'\rangle is a different from continuous one. in case of position, the discrete eigenket is probability but for continuous, its probability density. for continuous |p\rangle dp is analogy to |p'\... 1 Thanks to Ján Lalinský for getting me to double check my faulty memory. My assumption of \langle\chi_n|[H_0,z]|\chi_m\rangle = 0 was wrong. Here is the solution after double checking my work.$$\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\chi_n|H_0\,\,z|\chi_m\rangle-\langle\chi_n|z\,\,H_0|\chi_m\rangle\langle\chi_n|[H_0,z]|\chi_m\rangle=\langle\... 1 I had the same doubt as I was figuring out if a certain coupling of angular momenta in the 3$\gamma$could be found such that parity would be conserved. Turns out (if I am correct) you cannot have three photons with coupled total angular momentum$J=0$in the first place. For the coupled photon angular momentum$L(3\gamma)$and spin$S(3\gamma)$to be able ... 6 OK you can do it. GP Thomson have done it long before. You need a low pressure vacuum tube to generate the cathode rays (electrons). An essential requirement and rather costly one. I am afraid that you have to make one (size ~500 mm). To do this you need electrocathode (thermionic), resistive heating element, acrylic or glass tube, anode, a rotary vacuum ... 2 Your confusion comes from unfortunate terminology Quantum numbers are supposed to denote every individual orbital quantum numbers denote the state of a quantum system as solution of the Schrödinger equation; in particular they often refer to the eigenvalues of a maximal set of operators used to describe the physics at hand. As an example the hydrogen ... 4 The orbitals that you are taught about at school are energy eigenstates i.e. eigenfunctions of the Hamiltonian. For a confined system like an electron in an atom the eigenstates of the Hamiltonian have discrete and precisely defined energies so the transition energies are all precisely defined. In the real world electron states are only approximately ... 1 It is understood how the process works. In a nutshell, the energy of a photon is converted to the mass of an electron and positron as is given by Einstein's$E = mc^2$. This is spelled out as follows:$γ → e^− + e^+$There are two musts for this: The photon must have a higher energy than the sum of the rest mass energies of a positron and electron for ... 10 The theory behind the trick is based on the Hellmann-Feynman (HF) theorem $$\frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$ which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value$\langle\frac{1}{r^2}...

1

Someone might give a more precise answer in terms of group theory but I'll give it a go anyway ; feel free to edit my post. Instead of considering the case of an odd number of fermions, one can consider just a single spin $1/2$ - fermion to discuss $2 \pi$ rotations. Spin $1/2$ is a representation of dimension 2 of the rotation group, which is called the ...

-1

Is quantum mechanics on a measurement level a deterministic theory or a probability theory? If we know the quantum state of a (ideal) closed system, then we have the probability distribution of measurement outcomes for any of its observables. The state evolution is deterministic, measurement predictions are probabilistic. The probabilities here do not ...

1

From what I understand, you want to characterize the solution space of those $a,b\in\mathbb C$ such that \begin{align} a|00⟩+b|11⟩ & = a|{++}⟩+b|{--}⟩ \\ & = \frac{a}{\sqrt 2}\left(|00⟩+|01⟩+|10⟩+|11⟩\right) \\ & \quad + \frac{b}{\sqrt 2}\left(|00⟩-|01⟩-|10⟩+|11⟩\right) \\ & = \frac{a+b}{\sqrt{2}}\left(|00⟩+|11⟩\right) +\frac{a-b}{\sqrt{2}}\...

1

Write $$U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right)$$ where $U'_{ij} \in G_n$. In terms of the latter, $$U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \... 0 The statement is almost true, although one has to redefine both H_S and H_{SB} to make it work. You need to also make use of the standard assumption of factorising initial conditions,$$ \rho(0) = \rho_S(0)\otimes \rho_B,$$where \rho_B is the bath reference state, which must commute with the bath Hamiltonian, i.e. [\rho_B,H_B]=0. Often (but not ... 2 The radius of the wavepacket is unrelated to the radius of the particle. For instance you could have a particle of zero size (as the electron is suspected) but whose probability of being found somewhere has a certain size. For instance a wave packet: The wave packet only gives you the probability of finding a particle at that place (that is, more likely ... 2 Your second equation isn't quite right. If you have a continuous complete set of states \{|p⟩\}, then the correct expansion of a given arbitrary state |P⟩ in that basis is of the form$$ |P⟩ = \int\mathrm dp \: f(p)|p⟩, \tag 1 $$with a single arbitrary function f(p) over the indexing variable p as a (continuous) coefficient. Here the dp denotes ... 2 For question 1, it comes down to probability. I have two distinguishable particles, a and b. The probability to find particle a at x_1 is$$P_a (x_1)=\int dx_1 \Psi_a(x_1) \Psi_a^*(x_1),$$and we have a similar expression for particle b at x_2. The probability to find particle a at x_1 and particle b at x_2 is just the product of ... 1 If the state of two particles is the tensor product of the two single particle states, then the wave function of the two particles is the product of the two single particle wave functions. For indistinguishable particles it is an experimental fact that the final state must be either symmetric or antisymmetric with respect to the exchange of the two ... 1 About the limit: \frac{\sin[\frac{\pi}{d}(1/N+1)x]}{\sin[\frac{\pi x}{dN}]}= \sin[\frac{\pi}{d}(1/N+1)x]\times \frac{\frac{\pi x}{dN}}{\sin[\frac{\pi x}{dN}]}\times \frac{dN}{\pi x}\approx \sin[\frac{\pi}{d}x] \times 1 \times \frac{Nd}{\pi x}  in the last step I used \lim_{x\to 0}\frac{\sin x}{x}=1 and 1/N+1\approx 1. after rearranging the terms it ... -1 Interesting question. Electrons have electric charge and are therefore a source of electric field. An atom can be thought as a series of energy levels in which electrons can be in. When the electron transit between these levels, the energy will be emitted as electromagnetic radiation, since the fundamental interaction involved here is the electromagnetic ... 1 The atomic model developed starting from the light spectrum emitted by the hydrogen atom. It was known that hydrogen was one proton and one electron. The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using ... 0 Your first expression should have a additional factor in the integrand called a density of states (DOS) and is usually denoted \rho(E), i.e.,$$U(t)=\sum\int\rho(E)dE|E,\alpha\rangle\langle E,\alpha|e^{-iEt/\hbar}. The density of states is the number of states in the energy range $(E,E+dE)$. Look at the parabolic dispersion section of the density of ...

4

There's no analytic proof, but numerical evidence suggests that if you know that the Hamiltonian is local, and it satisfies the Eigenstate Thermalization Hypothesis (which most local Hamiltonians do), then you can extract the entire Hamiltonian from a single excited eigenstate, though not from the ground state: https://arxiv.org/abs/1503.00729.

4

Yes, light diffraction may be viewed both as a classical phenomenon and as a quantum mechanical consequence of Heisenberg's uncertainty principle. However, since both explanations work equally well, it doesn't provide any direct evidence for quantum mechanics. Let me explain why the two explanations are equivalent. I'll do the classical uncertainty bound ...

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