New answers tagged

2

The Schrödinger equation is but a differential equation, so, then, solvable by one of the most effective iterative approximation techniques for such, namely WP:Asymptotic analysis; WP:Asymptotic expansion; WP:Asymptotic theory; etc... Asymptotic analysis consists of either starting from a base solution and improving it iteratively as a series, or else, ...


2

The wavefunction associated to a free particle of momentum $p$ is $$ \psi_p(x) = \mathrm{e}^{\mathrm{i}px/\hbar}$$ which is obviously just a plane wave with wavelength $h/p$, fully compatible with the deBroglie relation. Strictly speaking, this function itself is not an admissible wavefunction because it is not square-integrable and hence ...


1

It is true that two electrons can't have identical quantum numbers, but spin itself is a quantum number. That is, the state with quantum numbers 111 can hold two electrons: one of spin up and one of spin down. So when you are finding the ground state, for example, find the four lowest energy eigenstates (ignoring spin), and the ground state of eight ...


1

Quantum mechanics is a bit more than the Schrödinger equation. In particular, it says that all states evolve in time as given by the Schrödinger equation - there is the Hamiltonian operator $H$ and every time-dependent state $\lvert\psi(t)\rangle$ fulfills $\partial_t\lvert\psi(t)\rangle = -\mathrm{i}H\lvert\psi(t)\rangle$. In contrast, the potential ...


0

The main point is that the wave function is postulated to contain the complete physical information of the system, in the sense that all there is to know is contained in the wavefunction. This is why Schrödinger's equation is an eigenvalue equation: this way the physical state at a later time is completely determined by the state at the current time: ...


2

This subtlety is related to the fact that the momentum operator $\hat{P}$ (unlike the Hamiltonian $\hat{H}=\frac{\hat{P}^2}{2m}$) has no eigenfunctions compatible with the Dirichlet boundary conditions, and $\hat{P}$ is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, ...


1

Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...


10

What you call an operator theory is usually called the Heisenberg picture of quantum mechanics. What you call a wave function theory is usually called the Schroedinger picture of quantum mechanics. It is well-known that for every quantum mechanical model, the Heisenberg picture and the Schroedinger picture are fully equivalent through a dual description, ...


0

If you are interested in an path integral with the action: $$ \mathcal{S}= \int_{t_0}^{t_1} dt \langle \Phi(t) | i \hbar\partial / \partial t - \hat{H}(t) | \Phi(t) \rangle \tag{1} $$ then $\Phi(k, t)=\langle k| \Phi(t) \rangle$ is now an operator or a mude variable inside the path integral. The bridge between the operator and path integral linguage is: ...


1

Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2.


1

In general, what we use for periodic boundary conditions is defined by $Y(0)=Y(L)$ and $\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition $Y(0) = Y(L)$ I believe is not sufficient to impose conditions on $k$. This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the ...


0

For the strict boundary conditions, the -n solution is linearly dependent with the n solution: it's the same wavefunction multiplied by -1. $\psi_{-n}(x)=\sin(-n\frac\pi Lx) = -\sin(n\frac\pi Lx)=-\psi_{n}(x)$


1

I personally find most notation in the Heisenberg picture totally unsatisfactory for the exact same reason as you, its hard to look at the notation and not want to see a time dependent operator as a function from time to a matrix (or to an operator on Hilbert Space, or such). But it's similar to thermodynamics, a thermodynamic variable could be written as a ...


0

First off, the pictures are equivalent, because you can always map one picture to the other. For example, go to back from Heisenberg to Schrodinger operators (in the case of constant $H$) use $$A_S(t) = e^{iHt} A_H(t) e^{-iHt}.$$ Since we can always recover the Schrodinger operator time dependency, there's no information lost. You already know that a total ...


1

Let's make things a little more fun. How can we use the conservation of energy equation to derive Schrödinger equation in QM? Let's say we know the system's Hilbert space $\mathcal H$ and we know how to define a Hamiltonian $H:\mathcal H \rightarrow \mathcal H$ whose average value $\langle \psi | H | \psi\rangle$ provides the average energy in state ...


3

The Schrödinger equation cannot be derived from classical physics. There are various consistency checks and motivations, such as its consistency with conservation of energy, but it is not derived from those considerations. However, that the Schrödinger equation conserves energy is built in when one knows that the Hamiltonian is the energy operator since $$ ...


2

Quantum mechanics is not derived from classical mechanics or energy conservation, but there are "jumping off points" in classical mechanics that may serve to answer your question. If you study classical mechanics at a sufficiently advanced level you will discover the Hamiltonian formalism. The Hamiltonian for an isolated system with only conservative ...


2

This is more of a comment than an answer, but I can't fit this into the amount of characters; Writing a quick bit of code, it looks to me like there's not much wrong with the method: The numerical and the analytical solution go on top of one another. N = 256 T = 256*128 L = 1. dt = 0.000001 x = linspace(0., N-1, N)*L/N psix = exp(1j*2*pi*x) psik = ...



Top 50 recent answers are included