New answers tagged

0

Actually of course this is true for eigenstates (or stationary solutions) of the Schrodinger equation, and others states for which $\langle x\rangle $ is time independent. Since $\langle p\rangle$ captures the time dependence of $\langle x\rangle$ then it follows that $\langle p\rangle=0$ as well. If you start with a wavefunction of the form $$ \Psi(x,0)=\...


0

Consider the particle like a prisoner in a cell. The unfortunate prisoner can pace up and down all day and all night long but he can't escape. Effectively his average velocity (as vector) is zero and thus also his average momentum (as a vector). This does not mean the prisoner does not move however.


0

Not via uncertainty principle but via Ehrenfest theorem -- If the particle is stuck at the bottom of the well, that means its expected position is constant, so $\frac{d}{dt}\langle x\rangle = 0$. Ehrenfest theorem says $ \frac{d}{dt}<x> = \frac{<p>}{m}$ so $\langle p \rangle$ must be zero.


7

The non-relativistic behavior of antiparticles can be understood with the Schrodinger equation. For example, anti-hydrogen is approximated by the Schrodinger equation to the same accuracy as hydrogen is. This is often never mentioned in an introductory Quantum Mechanics course. But the relationship between particles and antiparticles can only be understood ...


21

The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ where the electrons are denoted 1 and 2, and $r_i$ is the distance to the nucleus at the origin and $r_{12}$ the distance between the electrons. Since the ...


0

I am answering with these quotes(and also my comment to probably_someone) Since opposite charges attract via a basic electromagnetic force, the negatively-charged electrons orbiting the nucleus and the positively-charged protons in the nucleus attract each other. Also, an electron positioned between two nuclei will be attracted to both of them. Thus, the ...


2

For a particle to have a definite energy, it must be in an energy eigenstate of the Hamiltonian that describes it. But this in no way means that our system always has to be in one of the energy eigenstates. Take the simple case of particle in a box. The only condition on the wavefunctions is that it is $0$ at and outside the boundaries of the box and that ...


0

You are asking why the formation of covalent bonds between two (or more) atoms causes attraction (bonding) between them. Now it is very important to understand that the more correct expression is the sharing of electron orbitals that causes this attraction (overlap is not even needed, when there is only one electron being shared). The sharing (between the ...


2

When you put two atoms close enough together that both nuclei are important to the behavior of both sets of electrons, you have significantly changed the potential that an individual electron is moving within. This potential generates wavefunctions that are quite different from the wavefunctions in an isolated atom, with different energies and very different ...


2

In this "overlap" explanation, the overlap is not an overlap between two electrons. You get a $\sigma$ bond in the H2+ molecule, where there is only one electron. There are two reasons why you get a lower energy in a covalent bond. One is that the kinetic energy is lower than in the unbound state, and the other is that the potential energy is lower. The ...


3

The reason that we take it to be the energy is that this is closely related to the classical energy, when one follows the standard rules and conventions of "quantizing" the classical quantities. In a similar manner, you can take any classical quantity $A$, and then from the classical concept of the Hamiltonian (which is the classical energy $E_K+V$ in ...


8

This is essentially the definition of energy in quantum mechanics -- it is $\hbar$ times the rate of change of phase. That's one of the fundamentally new ideas in quantum mechanics, so it can't be derived from anything you already know classically. If that's not very satisfying, we can say the same thing in more steps. The energy of an energy eigenstate is ...


2

The Schrödinger equation can be quantized canonically to get the Schrödinger field. The latter is a quantum field theory that has the Schrödinger equation as its operator equation of motion. Here is an outline of how that works. The Schrödinger equation is $$H\psi(r,t) = i\frac{\partial}{\partial t} \psi(r,t).$$ where $H$ is the ''first quantized'' ...


8

Exactly the same Schrodinger equation as in quantum mechanics, only with a different Hamiltonian. The QFT Hamiltonian is $$ \hat{\mathcal{H}} = \int d^3 x \, \hat{T}^{00}, $$ where $\hat{T}_{\mu \nu}$ is the quantization of the classical stress-energy tensor $T_{\mu \nu}$ (strictly speaking, it is not required to mathematically exist; only its integral ...


2

You could look at Schrödinger's original paper where he introduces the equation. It's actually very nicely written. E.Schrödinger, An Undulatory Theory of the Mechanics of Atoms and Molecules, Physical Review (1926) Vol. 28, No. 6 pp. 1049-1070 As people have pointed out, you need to make some assumptions to derive the equation. Schrödinger's approach was ...


0

Suppose you concluded after seeing the double slit experiment that the position of a particle is in a (linear) superposition of all positions: $$|{\psi}\rangle = \sum_i \psi_i |x_i\rangle \xrightarrow[\text{cont. limit}]{} \int \mathrm{dx}\ \psi(x) |x\rangle$$ such that the absolute square of $\psi(x)$ gives the probability distribution of finding the ...


0

Partial differential equations are derived from basic principles of physics, such as conservation of energy or quantization of energy. They are not axioms. I prefer starting with the Hamiltonian and the principle of least action via calculus of variations, which is indeed axiomatic. PDE have an infinite number of solutions. The physically reasonable ones ...


1

The other solution isn't square-integrable and so it doesn't correspond to a physical state.


2

Strict definition of "plane wave" means that the phase of the wave is constant along parallel planes, and subsequently the phase velocity is perpendicular to these planes. It doesn't matter if the wave represents classical light of quantum electrons. Mathematically it means that the phase velocity of the wave is constant and along single direction. Phase of ...


5

So indeed, a free electron is described by plane wave $$\Psi(r) =A e^{ikr}$$ where $A$ is constant. This means the probability to observe the electron will be equal to $|\Psi(r)|^2=A^2$ for every $r$, i.e. the probability to observe a free electron in space is everywhere the same. Now if we take into account a periodic potential $U(r)$ we indeed get a Bloch ...


25

A bit of a different perspective than other answers: I was once in a strange physics class as an undergraduate, where an old 90 year old professor would mumble to himself while drawing terribly on a tablet connected to a projector. Everyone would get A's by default so no one would pay attention, in fact some days I would be the only one to show up, but this ...


4

Equations aren't 'derived' in a fully rigorous way in physics as the derivation always uses physics in some or all of its key steps. Also physicists have access to tools which mathematicians do not have access to because they do not require full rigour in their derivation: Feyman path integrals are a prime example. As an example, in the derivation of the ...


5

Yes, there is. In order to describe emission (or absorption) we need to include in our description the thing that is emitted or absorbed, namely the electromagnetic field. While a complete description of the coupling of the electron to the EM field is quite complicated, at the energies and setups involved in most experiments or to explain visible phenomena, ...


5

Start with the classical nonrelativistic energy expression. Make the De Broglie assumption that matter, not just light, can be described by waves. As a consequence identify E with $\frac{\hbar} {i} \partial_t $ and similar for P. There you have the Schrödinger equation.


30

Although knzhou's answer makes a good point stressing the possibility that what is taken as a starting point at introductory level could become a consequence of a more fundamental principle, I think that there is a key point which should be stressed more clearly. In physics, whatever conceptual tool we develop has to be rooted in, and its motivation comes ...


164

A derivation means a series of logical steps that starts with some assumptions, and ends up at the result you want. Just about anything can be "derived", as long as you vary what the assumptions are. So when people say "X can't be derived", they mean "at your current level of understanding, there's no way to derive X that sheds more light on why X is true, ...


0

The problem is that you defined what your system is, concluded a discrete spectrum, then assume it has a definite energy prior to measurement, and on top of that it is not an allowed energy. This (The system, not the error) can be realized with an analogy. A guitar string strapped on a guitar. Playing is a superposition of standing waves, and slightly and ...


2

I guess we can try using semi-classical approximation. Notice that in one 'period' the particle moves from $x=0$ to $x=a$ and back. This gives a phase shift $\gamma = -\pi$ since there are two smooth turning points in such an orbit. Use the Bohr-Sommerfeld equation ($n \in \mathbb{N} = \{0,1,2,...\}$): $$\frac{1}{\hbar}\oint p_x dx = 2\pi n - \gamma = 2\pi(...


0

I think you may have a factor of 2 error for the $E$ and the power of $M$, here is my derivation, I use $\theta$ instead of $\phi$.(we may double check) In polar coordinates, the Del operator $\nabla^2$ is defined as: $$ \begin{align} \nabla^2 &=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{1}{r^{2}} \frac{\...


4

Just a small addition to ZeroTheHero's answer, I think there's a more instructive reason to label the wavefunctions by $e^{in\phi}$ rather than sines and cosines. As you have correctly shown, in this problem (similar to the particle in the box), the energies are given by $$E_n = \frac{h^2 n^2}{2m R^2}.$$ However, what is different is that unlike the particle ...


2

In principle one can use any normalized linear combination of $\cos(n\phi)$ and $\sin(n\phi)$ but the solutions for this special case are usually written in terms of exponentials: $$ \psi_n(\phi)=\frac{1}{\sqrt{2\pi}}e^{in\phi}\, ,\qquad -\infty\le n \le \infty\, . \tag{1} $$ Since $\psi_{\pm n}(\phi)$ are degenerate, an arbitrary linear combination $$ ...


2

Can we write down the solutions to this potential in closed form? Yes, you can, though they will likely depend on the eigenvalue, and this will likely be the solution of a transcendental equation that can only be solved numerically. (Note that there's nothing strange about that $-$ you get the same behaviour with the finite square well.) Why aren't ...


14

I see the fatal flaw here. Inside the well, your solution must be a solution to the HO, BUT there are two solutions to the HO. There is the one we're all familiar with, a hermite polynomial times a gaussian, but $e^{x^2}$ is also a valid solution. For the HO, we discard that solution because it isn't normalizeable at the boundaries. In the finite well ...


2

Any superposition/linear combination of monochromatic left- and right-moving exponential waves is of course equivalent to a superposition/linear combination of sine and cosine waves, as OP demonstrated above. However from the point of view of a 1D scattering experiment where wavepackets enter from (and leave to) spatial infinity, the travelling exponential ...


0

If the spatial part were to be multiplied by (for example) $\cos \left( \dfrac{\hbar k^2}{2m}t \right)$ this would be true. In this case you get a standing wave. But multiplying by $e^{-i\frac{\hbar k^2}{2m}t}$ yields a travelling wave.


1

In the description provided by QM, a particle in a harmonic potential does not oscillate around any point. There is no trajectory, no amplitude and no oscillation at all. The only thing you can say is that there is a gaussian-like probability density with its maximum at the minimum of the parabolic potential. If at some time the harmonic potential would ...


1

If the particle is at an eigenstate of the harmonic oscillator, then its wavefunction is characterized by zero expectation value of the position $\langle n | x | n \rangle = 0$. For the ground state the wavefunction is a Gaussian. Then removing the harmonic potential will lead, as you said, to the particle keep having a zero expectation value of the position ...


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