# Tag Info

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Note carefully Nick's comment. Suppose I send two plane EM waves on some collision course so they interfere. The waves will pass through the region where they meet, generating some interference pattern in that region, then they will exit that region and continue on their separate ways unchanged. In other words neither the energy nor the momentum of the waves ...

6

This is an estimation tool not uncommon in theoretical physics. Namely, one knows the value of some quantity for a given problem and therefore assumes that the scale of the problem with regards to that quantity is of the same order of magnitude as the known value. In other words, we assume that the error in our known value must not be too much greater than ...

6

This argument just replaces one axiom by another. It assumes that if a quantum system consists of identical particles, then the state of the system should not change (it get's multiplied by a phase) under exchange of quantum numbers. Although this is (perhaps) a more intuitive way of thinking about states of identical particles, it's still a strong ...

4

Neuneck's answer is the pithiest description of how you get normalisable states as superpositions of non-normalisable states, but the following is more of a "why" these states happen. Hopefully, you should see that this discussion is independent of the number of dimensions. Practically speaking, the reason why there are always such states it is because ...

3

Photons have some conditions to have an evanescent wave, e.g. total internal reflection. Suppose we have some material with index of refraction $n_1$ and a layer of another material, with smaller $n_2<n_1$. At some angle we'll see total internal reflection, i.e. when the light totally reflects, but leaves some exponentially decaying trails in layer with ...

3

The best way to answer the question "How are anyons possible" is to use the "dynamical" path integral formalism, rather than the "static" wave function formalism. The permutation group action on the wave function is "static" in the sense that only initial and final states are specified. It will be ambiguous if there are more than one non-equivalent ways to ...

3

The modern version of Pauli's principle requires completely antisymmetry of a state of $n$ fermions. Instead, the argument discussed in the body of the question only implies that a state of $n$ fermions has to be either symmetric or antisymmetric under interchange of a pair of particle. It is impossible, following this way, to prove that the full state is ...

3

There is no need for high order mechanism. It is simply because a single photon can interfere with itself. If you remember the double slit experiment, they are indeed looking for a single photon passing through a slit and interfere with itself. Now if, instead we have billions of billions photons, the same single photon interference still happen ...

3

I'm going to explain roughly what the Born Rule, following Stan Liou's comment. One of the Postulates of Quantum Mechanics relates a mathematical quantity, the wave function (or state $\psi$ of a Hilbert space, $\mathcal{H}$) to a measurable entity, the probability of a given event to happen. The idea goes like this: if you want to measure a quantity ...

3

I think the answer is it depends on distance (relative to the size of your system). Another well known example of a boson which is comprised of fermionic components is the helium-4 atom, which has integer spin (both the nucleus and the neutral atom itself). Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared ...

3

I supposed you are in a context of bound states, with normalized eigenfunctions $\psi_n(x,t) = \phi_n(x) e ^{iE_nt}$. Of course, if you calculate $\langle x(t)\rangle_{\psi_n} = \int dx \bar \psi_n x \psi_n$, you will find a position expectation value which does not depend on time. Now, this is not the general case, if you take a linear combination of the ...

3

You've forgotten one crucial thing when you've written your superposition: the separate $\psi_{k,\ell,m}(x,y,z)$ are eigenfunctions of the Hamiltonian with different eigenvalues. The superposition will no longer be an eigenstate because of this. In fact, by taking an appropriate superposition, you would be able to get any function you like (in your case, ...

2

For first question, you must simly take $l = 0$(in your notations, $l$ is curved). It's because angular momentum is conserved in radial-symmetry field problem, and you can simply take $L^2 = \hbar^2 l(l+1)$. So, for $l = 0$ states you simply ommit $\frac{L^2}{2mr^2}$ term. Potential is given in problem. It behaves like constant for some $r$ values, and ...

2

There is a thread in Physicsforums.com which states due to Quantum Mechanics, if you wait long enough diamonds will appear in your pocket, it also states its possible for all your atoms to spontaneously re-arrange themselves so you turn into a Boeing airplane. Surely this is fiction? No, it could be possible. That's why we say that in Quantum Mechanics ...

2

You can work it out from the Taylor series $$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots$$ where $x = \lambda a_n^{(1)} + \lambda^2 a_n^{(2)} + \cdots$. Each term can then be expanded in a power series in $\lambda$: \begin{align} -x &= -\lambda a_n^{(1)} - \lambda^2 a_n^{(2)} - \lambda^3 a_n^{(3)} - \cdots \\ x^2 &= \lambda^2 ... 2 Inside vs outside, there is a sign change inside the square root, so that changes the nature of the "phase" \phi(r). Normally, when you match wave functions you require that \psi_\mathrm{left}(x) = \psi_\mathrm{right}(x) (continuity) and that the derivative changes according to what you get when you integrate the Schrodinger equation: ... 2 Let's put it clear first: for Raman scattering there is no excited state at all, the light just bounces of a molecule. If the photo has the right energy, it can bring the molecule to an excited state. Different things can happen to a molecule in this state - in most of the cases the energy will be dissipated through collisions, but in a rare case the ... 2 The basic idea for this is to use the momentum space version of the Schroedinger equation: \hat{p}\to p,\quad\hat{x}\to i\hbar\frac{\partial}{\partial p} $$and then solve the system1,$$ \left[\frac{p^2}{2m}+img\hbar\frac{d}{dp}\right]\phi=E\phi $$which should be solvable (e.g., complex exponentials). You can then Fourier transform to physical space to ... 2 I'd like to perhaps a slightly different viewpoint to your question and maybe turn it around a little. Probability is hard. Very hard. Defining the foundations of probability and statistics so that they are altogether sound and rigorous is actually a work in progress. It definitely is not complete. On the other hand Quantum mechanics is easy. Very easy! I'm ... 2 Let me try to give you a kitchen-table explanation. I can't help you with statistics vis-a-vis quantum mechanics, but probability is very basic. The underlying "real stuff" in quantum mechanics are numbers that, when squared, produce probabilities of seeing things. Typically, these numbers are complex, but they don't always have to be. These numbers are ... 2 Does this mean, that the probability of detecting the particle it the SAME everywhere? No, it does not. This is quite a common mistake, stemming from the idea that the Green function \mathcal{M} can be used in the role of the \psi function of free particle with the Born interpretation of |\psi|^2 as probability density. But that is not possible, ... 2 The number of neutrons is even, so it indeed means that they contribute spin zero and positive parity. The spin and parity comes from the "last proton" because the number of protons is odd. The dependence of the energy on the angular momentum is such that the pairs at a high value of J are preferred (lower in energy) due to the special, spin-dependent ... 2 Don't think of it as leaping. An object's location is not well-defined unless and until a measurement of its position has been made. In Classical Mechanics, the motion of a particle is given by its position and its momentum. In Quantum Mechanics, the fundamental "thing" that controls a particle's motion is its wave function, and wave functions are mucho ... 1 Yes, the summation is taking over all possible integer value of m=0,1,2,... except m=n. It can be easily seen by following the derivation of the first order perturbation theory. In your example \psi_1^{(1)}, it is sum over m=0,2,3,4,.... Note that sometimes the index start from 1 instead of 0 such as infinite square well, then you should skip 0. 1 The potential barrier problem and solution in quantum mechanics is discussed within the solutions of Schrodineger's equation in which there exist potentials, and the solutions of the equations with the boundary conditions give the wave function of a particle, i.e an entity with a mass. In addition it is a non relativistic equation.Thus in this framework: ... 1 I should say that the comments and answers so far have nothing to do with the effect that is shown on the plot. Note, in particular, how the intensity of broadband background depends on the wavelength of the laser used for Raman scattering. This indicates that the background is not related to the properties of the molecule itself, but rather to its ... 1 As I had written in the comments, it is the second term that you have gotten incorrect. Focusing exclusively on this term (leaving aside the 1/8 factor), we have$$ |\psi_2\rangle \langle \psi_2| = \frac{1}{3}(|00\rangle \langle 00|+ |10\rangle \langle 10|+ |11\rangle \langle 11| - |00\rangle \langle 10|-|10\rangle \langle 00| + \ ...), $$where I have ... 1 Let's say you've prepared an electron n the state:$$|\psi\rangle= \alpha|0\rangle + \beta|1\rangle$$And you want to measure it in the state:$$|\phi\rangle= \alpha'|0\rangle + \beta'|1\rangle$$Then the probability of the electron collapsing into the state \phi upon measurement is:$$|\langle\phi|\psi\rangle|^2$$This means that it is possible for an ... 1 Your solution seems ok. You should compute the transmission coefficient T and reflection coefficient R. It is the sum of transmission coefficient and reflection coefficient that give one.$$T + R = 1 I obtained $k_{left} = 6.2746×10^{9} \text{ m}^{-1}$ and $k_{right} = 3.62264×10^{9} \text { m}^{-1}$ from your solution. When I used them to compute sum ...

1

There is some restrictions on previous answer. Hamiltonian must be time-independent to use $U = e^{-iHt}$ rule. For time-dependent hamiltonian, time-evolution in form $\psi(t) = U(t,t_0) \psi(t_0)$ takes U in more general form $U(t,t_0) = \mathcal{T}\,\exp(-i \int _{t_0} ^ t H(\tau) d\tau)$. Of course, if your potential is constant over $t$ to $t_0$ period, ...

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