23

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 ...


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 ...


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 ...


4

This is a very interesting question, and to answer it I am going to consider the following example. Suppose, as you said, that we have a source that, somehow (there are different techniques which allow us to obtain entangled photons such as employing nonlinear cristals) generates a pair of photons that are in the following global state \begin{equation} |\...


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 ...


4

You shouldn't do a Fourier in the time coordinate just in the position coordinates (i.e. x y z). And you know by simple calculations that $\hat{\Delta \psi}$ is just $-k^2\psi(t, \vec{k})$ where $\vec{k}$ is (x, y, z) after the Fourier transform. And the by substituting this to the equation you get a simple ode for $\hat{\psi}(t, \vec{k})$ because $\hat{\...


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 ...


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

Energy can have many different units e.g. joules and electron volts are commonly used. We tend to choose whatever units are convenient for our calculation. In this case the energy units being used are units of $\hbar\omega$ (or equivalently $h\nu$) so the ground state energy is $\tfrac12$, the first excited state is $\tfrac32$, the second excited state $\...


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 ...


2

The only possible completely antisymmetric wave function for three spins $1/2$ is identical zero. From three spin variables $s_1, s_2, s_3$, each being equal $1/2$ or $-1/2$, at least two have same value. The antisymmetry of wave function leads to its zero value in this case.


2

Measurements are represented by operators, and collapse is described as producing an eigenstate of that operator. The measurement you're doing is represented by a projection operator. In the situation you describe, the wavefunction after collapse will have had one region projected down to zero, but the other region will still all be there. The particle is ...


2

No. Parallel corresponds to $$S_{z1}S_{z2}|\psi\rangle=\left(\frac{\hbar}{2}\right)^2|\psi\rangle$$ And anti-parallel corresponds to $$S_{z1}S_{z2}|\psi\rangle=0$$ Thus $|\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\rangle$ are parallel states and the other two are antiparallel.


2

Is it correct to say that particle is oscillating ? It is not the particle that is oscillating,but the wavefunction is sinusoidal, due to the nature of wave equations. By the postulates of quantum mechanics, the meaning of the wavefunction is, when complex conjugate squared, to give the probability of finding a particle at (x,y,z,t). It is the probability ...


1

I can name two intuitive reasons. Firstly if the oposite is true, $E-V<0$, then the solution becomes a decaying exponential. So in this classicaly forbidden region the chance of finding the particle drops of very quickly. Secondly the term $E-V$ is the kinetic energy operator: it tells you how quickly a particle moves (under some circumstances). For a ...


1

What we need is two things. Uncertainty principle and superposition of states. If the state has a fixed momentum, then it’s position is completely undefined. This can be visualised as follows: Here the wavelength is defined, thus momentum is defined. But the position is completely undefined. This has to be the case as dictated by the uncertainty principle. ...


1

I think the confusion is resolved by realizing that these are actually different $n$'s. Symmetry, conserved quantities, and degeneracy We can solve the time-independent Schrödinger equation (TISE) for a particle on a ring of radius $R$ directly, but it's worth thinking about the symmetry of the situation first. Assuming that the ring is oriented so ...


1

What you've done seems fine, it's just that you added an extra conjugate by accident. Using your notation, if $$F[\psi, \psi^*] = \int \psi^* \psi \, d \vec{r}$$ then we have $$F[\psi + t h, \psi^* + t h^*] = \int (\psi^* + t h^*) (\psi + t h) \, d \vec{r}.$$ Taking the derivative with respect to $t$ at $t = 0$ gives $$\frac{\partial F[\psi + t h, \psi^* + ...


1

So the question is basically about how entanglement works. There's probably a gazillion answers for that question by now, but there's no harm done in addressing it again in the context of the current question. First, let's clear up some possible misconceptions that may be implied in the question. Photons carry many degrees of freedom. Therefore, they could ...


1

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


1

I know about the wave function describing the uncertainty of two 'dependent' properties at the quantum level. Not really. The wave function $\psi(r)$ is obtained by solving the Schrödinger equation (SE) of the quantum system at hand. The wave function contains all the information of the quantum system, information that is obtained from it by treating $\...


1

The state of the electron in the Hydrogen atom is defined by 3 quantum numebrs - its energy level $n$, its angular momentum $l$ and the projection of the angular momentum on the $z$-axis $m$. The energy levels get the values of positive integers $n=1,2,3,\ldots$, the angular momentum of non-negative integers but is restricted to be smaller than the energy ...


1

Let's start from the basic physical reasoning. The idea is that we want to find the general form of the wavefunction that has the symmetry of the problem. Let's say that we have a one-dimensional lattice with lattice spacing $a$. So, a wavefunction that shares the symmetry of the lattice would be invariant under translations by $na$ where $n\in\mathbb{Z}$. ...


1

This is actually the fourier transformation. Normally you see $$f(\vec{x})=N\sum_{\vec{k}}e^{i\vec{k}\vec{x}}f(\vec{k})$$ where N is a normalization parameter. This in terms of linear algebra changes your basis from eigenvalues of the position operator to eigenvalues of the momentum operator. The same thing happens here, but instead of functions you ...


1

Okay here I'll try to give an introductory summary of time evolution. Let's assume we have an arbitrary initial state $\psi(x,0)$ and a Hamiltonian operator $\hat H$. If we know the set of eigenfunctions of $\hat H$, we can expand this arbitrary $\psi(x,0)$ as a linear combination of them. The easiest case is when we have a bound state and thus only ...


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