# Tag Info

18

IF you know that your Hamiltonian is of the form $$\hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1$$ for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state $\Psi_0(\mathbf r)$ obeys $$\hat H\Psi_0(\mathbf r) =... 16 Rigorously speaking, the probability to find the electron at a distance exactly equal to r from the nucleus is 0. On the other hand, we can define the probability to find the electron in a volume d \mathbf{r} as$$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$$where I have ... 10 The theory behind the trick is based on the Hellmann-Feynman (HF) theorem$$ \frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value \langle\frac{1}{r^2}... 6 OK you can do it. GP Thomson have done it long before. You need a low pressure vacuum tube to generate the cathode rays (electrons). An essential requirement and rather costly one. I am afraid that you have to make one (size ~500 mm). To do this you need electrocathode (thermionic), resistive heating element, acrylic or glass tube, anode, a rotary vacuum ... 6 Assume for simplicity that all the operators are bounded. If you know the wave function \psi associated with the ground state of the unknown Hamiltonian H, then H has the form$$H = E_0|\psi\rangle\langle\psi| \oplus K$$where K is another Hamiltonian defined on a subspace of the original Hilbert space of co-dimension 1, and E_0 is the energy of ... 5 I don't have enough reputation to comment but please note that Weinberg uses the (-,+,+,+) metric, which means you need a change of sign in$$e^{iP.x}$$. The field transforms as, with x'=x+a,$$\phi'(x')=U^{-1}(a)\phi(x')U(a)=\phi(x)$$. For (+,-,-,-) signature this is implemented by U(a)=e^{iP.x}, see Peskin & Schroeder, page 26 for example. In ... 5 Measurement is a process in which by help of interactions one quantum entangles the quantum system under study and the environment and by rapidly utilizing decoherence cleans up the "off diagonals" of the entanglement into an apparent collapsed state. After this process, it effectively appears as the wave function of the system has been set to a distinct ... 4 \hat{X} = i\partial_{k_x} in momentum space is wrong. The left hand side is an operator, the right hand side is the representation of an operator when acting on functions, taken with respect to the scalar product with the basis |\ p\rangle. In particular we have$$ \langle x\,|\,\hat{x}\,|\,\phi\rangle = x\phi(x) $$and$$ \langle p\,|\,\hat{x}\,|\,\...

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 ...

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 ...

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

"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-...

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^\... 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 ... 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 ... 3 In a nutshell, no. Part of the problem seems to be that you misunderstand the fundamentals of string theory. The strings do vibrate. The frequency of these vibrations determines the type of particle and the energy of the string determines the energy of the particle. Second, your understanding of the uncertainty principle isn't quite right. Yes, we cannot ... 3 If unknown part of Hamiltonian is potential V({\bf{r}}), then you can write a stationary Schrodinger equation and figure out what the potential should be. 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 ... 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 ... 2 Yes - these experiments have been conducted, most famously by Aspect et al., but also by others, see Wikipedia. They all observed violations of Bell's inequalities - Our world is therefore not local-realistic in the sense of Einstein. An extension of Bell's inequalities by Legett (Legett inequalities) holds for non-local realistic theories. Their violation ... 2 I'm not sure what you mean by "the rest of the blue light." Basically three things can happen when low energy light passes though matter. The light excites a dispersion free mode, like your electronic energy state example, in that case the light will be reemitted at the same wavelength most often, you can work out the exact distribution with the Bohr-Dirac ... 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 $$with a single arbitrary function f(p) over the indexing variable p as a (continuous) coefficient. Here the dp denotes ... 2 The radius of the wavepacket is unrelated to the radius of the particle. For instance you could have a particle of zero size (as the electron is suspected) but whose probability of being found somewhere has a certain size. For instance a wave packet: The wave packet only gives you the probability of finding a particle at that place (that is, more likely ... 2 For question 1, it comes down to probability. I have two distinguishable particles, a and b. The probability to find particle a at x_1 is$$P_a (x_1)=\int dx_1 \Psi_a(x_1) \Psi_a^*(x_1),$$and we have a similar expression for particle b at x_2. The probability to find particle a at x_1 and particle b at x_2 is just the product of ... 2 Your confusion comes from unfortunate terminology Quantum numbers are supposed to denote every individual orbital quantum numbers denote the state of a quantum system as solution of the Schrödinger equation; in particular they often refer to the eigenvalues of a maximal set of operators used to describe the physics at hand. As an example the hydrogen ... 2 "To be at rest" in classical mechanics means "to have definite position and zero momentum", the two properties being (again, in CM) equivalent: if something has definite position, then it must have zero velocity thus zero momentum, and if it has zero momentum, then it must have zero velocity thus definite position. Depending on which of the two properties ... 2 Your initial state is entangled, which means it exhibits a correlation between the spin of the two particles. Each particle is spin up or spin down with equal probability, but the spins are (anti)correlated so that they cannot have the same spin. The states \left|\uparrow \downarrow\right> and \left|\downarrow \uparrow\right> are in agreement ... 2 Assuming you have it set up that \langle n\mid m\rangle=\delta_{nm} Then taking the sum:$$\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!}}{...

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 ...

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\...

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