# Tag Info

## New answers tagged quantum-mechanics

0

First of all, normalization is not a condition to fix the constant. Schodinger equation is a linear equation because the solutions form a linear space - Hilbert space. A physical state is an equivalent class of the vectors in Hilbert space, where two vectors belong to the same class if they differ by only a nonzero constant complex number. And it's clear ...

3

I am not entirely sure what you are asking, but since you seem to be sincerely interested in understanding some of the fundamentals of Quantum Mechanics, I'll do my best to answer what I think you are asking. The answer to why we don't consider a wave function to be a "real, deterministically evolving matter wave" is simply that such an interpretation ...

2

The 3D case is different than the 1D case, so don't feel too committed to the analogy. However, in the 1D case, you get only sine functions as solutions if you place a "hard wall" (an infinite potential barrier) at $x=0$. That is basically what is happening here: the origin is a sort of "hard" boundary for the radial equation since "negative radius" makes no ...

1

Suppose $A$ is at the space-time origin $0$, and $B$ is at space-time event $x$. You suppose that a real photon could go from $A$ to $B$, so this means that $A$ and $B$ are separated by a light-like interval, that is $x^2 = (x^0)^2- \vec x^2=0$. This means that $x^0>0$, too. Now, the propagator $D_{\mu\nu}(x)$ represents the amplitude for a photonic ...

-3

Below this energy, a particle can't go. That's what zero point energy means.

0

In addition to the standard estimation mentioned in other answers, it could be exact estimate when the mean value $\langle p \rangle=0$ with the definition $\Delta p=\sqrt{\langle p^2 \rangle - \langle p \rangle^2}$, which may happen in some case in the ground state and free particle.

1

Suppose that your decay does not violate parity, and let us take the example of strong interaction process involving mesons. Looking at this wiki paragraph and array, you will understand that each meson has a total angular moment, and a parity, noted $J^P$, and there are several possibilities for $S$ and $L$, for a given $J^P$ ($J^P$ is a characteristic ...

1

Look at $\Delta p$ as some momentum borrowed from the vacuum over the small distance $\Delta x$. The author has equated the minumum required amount of momentum to be borrowed to $\sqrt{2mE}$ which is related to the energy required to perform this task.

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 ...

0

The authors have this idea that the quantities of interest (size of the smallest possible binary switch $x_{min}$ and its smallest possible switching time $t_{min}$ ) are somehow connected to the uncertainty relations. They do not explain why this should be so. They do no explain what they mean by equation 1c, which has no standardized meaning in physics ...

7

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 ...

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 ...

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, ...

1

As you have already stated, the polyeder in the upper part of the figure is plotted in 3D k-space and visualizes the first Brillouin zone. The dotted line nicely visualizes a closed (one-dimensional) path in k-space that -- by convention -- runs through the special points you already mentioned. Now, this closed path is precisely the horizontal axis in the ...

1

There is absolutely physical meaning in it -- the sum is what gives you the final probability of the particle being at a particular location. But, often knowing just the single number is not illustrative enough to get much information from, which is why we look at the eigenfunctions individually. Quantum mechanics isn't my area, but I'll tie it back to ...

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 ...

0

Here's an example I like of why entanglement doesn't let you violate relativity. Say you have two spaceships moving in opposite directions along a line, with constant velocity. At $t = 0$, they synchronize clocks and entangle two particles. They also decide, at some predetermined time $T$, to measure the spins of the particles (actually, ship 1 will measure ...

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 ...

0

1 & 2) I think that with rotational invariant they mean transformations that do not change the inner product: $$\langle \chi|\chi\rangle \rightarrow \langle \chi'|\chi'\rangle = \langle \chi|U^\dagger U |\chi\rangle$$ which is clearly preserved if $U$ is unitary. This should also answer question 2, because a true symmetry ...

0

1) I think that they mean a rotation with a unitary matrix. But $\hat U$ will be unitary not for all $\hat A$ : you need that $\hat A$ is hermitian (this is a constraint on $\hat A$). 2) If $\hat A$ is not such that these products are invariant, then the matrix $\hat U$ is not unitary, and this is a problem. All that gives constraints on $\hat A$. 3) All ...

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 ...

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 ...

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

In Heisenberg picture states (kets) do not depend on time. But the so-called effective wave function here must clearly depend on time, seeing how it is defined. Think of nonrelatvistic QM, where the value $\psi\left(\mathbf{x},t\right)$ of a wave function at some point in space and time does not depend on whether you're working in Schrödinger or Heisenberg ...

0

I myself overlooked it too but wikipedia actually happens to have a great such list at https://en.wikipedia.org/wiki/Quantum_gravity#Points_of_tension There are other points of tension between quantum mechanics and general relativity. First, classical general relativity breaks down at singularities, and quantum mechanics becomes inconsistent ...

7

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 ...

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 ...

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, ...

0

The point $(11)$ is not correct, by doing $2$ successive "exchanges", you may have a global phase factor, such as $\psi'(x,y) = e^{i\alpha}\psi(x,y)$. The two wave functions describe the same physical state. The correct considerations are topological, inside of considering a discrete operation, consider a continuous operation, so that it is equivalent to ...

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 ...

0

The general two-particle state will look like $\displaystyle \int dp_1 dp_2 \psi(p_1,p_2) a^\dagger_{p_1} a^\dagger_{p_2}| 0\rangle$ Here $\psi(p_1,p_2)$ is the momentum-space wavefunction. Since the creation operators commute, only the symmetric part matters, so we may as well take $\psi(p_1,p_2)=\psi(p_2,p_1)$ (there would be a minus sign if they were ...

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 ...

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

As operators, you have $[L_{\vec e}, L_{\vec f}] = i L_{\vec e \wedge \vec f}$ (in units $\hbar=1$)

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 ...

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 ...

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 ...

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 ...

0

The spectra are actually always continuous if the resolution in frequency of the spectral analyzer and sensitivity in intensity of the detector is high enough. The spectrum of radiation emitted by hot atomic gases and molecules is called discrete because it has very sharp spectral peaks and looks as if formed of lines. But they are actually peaks with ...

1

Well, you are not yet done! You should also fix $A,B,C,D$ by imposing that the wavefunction is $C^1$ crossing the discontinuities of the potential. Following that way you obtain that only an overall arbitrary constant remains. You have thus a function like this $k\psi_E$, where $k\neq 0$ is any complex number, can be fixed arbitrarily (without imposing ...

0

Maybe a way for you to be not confused is to imagine a time dependence. For instance, let suppose three times $t_i, t, t_f$ with $t_i < t < t_f$. One may suppose that the particle is in the initial state $|A\rangle$ at time $t_i$, is in the final state $|z\rangle$ at time $t_f$, and, at the intermediary time $t$ is in one of the $2$ states ...

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 ...

1

This is wholly analogous to the evanescent optical field that arises in the classically (i.e. computed by raytracing) forbidden region beyond a totally internally reflecting interface between two optical mediums. I analyse this situation in my answer here and there is also a great plot of the situation in Ruslan's answer here. Let's think of a 1D barrier ...

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: ... 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

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, ...

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 ...

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 ...

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: ...

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