# Tag Info

## Hot answers tagged wavefunction

10

In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting ...

9

"More generally, can the potential term in the Schrödinger equation be dependent on the wavefunction itself?" The answer is negative: The resulting Schroedinger-like equation would turn out to be non-linear. It would not be associated with a unitary time evolutor (the self-adjoint generator, the Hamiltonian operator, would not be defined) against some ...

4

First of all, let me note that it is misleading to say that we're not even dealing with a particle We are still dealing with a particle, but the state of the particle at a given moment in time is no longer described by a pair $(x(t),p(t))$ of position and momentum in a classical phase space, it is instead a state $|\psi(t)\rangle$ in a Hilbert space. ...

4

The time evolution operator of a quantum system is (in units with $\hbar = 1$) $$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$ and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you) $\{\lvert \psi_E \rangle\}$ with $H ... 4 The expectation value (of position) represents the average value (position) for the particle (it has units of length in this case) which is different from the actual location of the particle (also units of length). For example, take an electron on a hydrogen atom; the expectation value for all energy levels is at the nucleus even though many of the energy ... 4 Expectation value is a different concept from probability. In fact, you can have an expectation value of energy, angular momentum, etc., not just for position. An expectation value of an observable for a given state$\Psi$is the average value of a large number of measurements of that observable, assuming each measurement is made on the same state$\Psi$. ... 4 Let$\Omega\subseteq \mathbb{R}^n$; then$\int_\Omega \lvert\psi(x)\rvert^2dx$, for a normalized function$\psi\in L^2(\mathbb{R}^n)$gives the probability that the particle is in the region of space$\Omega$, but does not give any further information on its position. If you want to obtain a quantitative information on the latter (within the limits of ... 3 It is not wrong, all three normalization conditions are natural and they don't contradict each other because, in fact, the first equation is nothing else than the product of the following two equations! Just substitute your formula for$\psi$,$\psi = R Y$, to the first equation. The only mistake you have to fix to show that the first equation becomes the ... 3 Well, it's just terminology: A (wave)function$\psi(x_1,\dots,x_N)$is symmetric in$i,j$iff$\psi(x_1,\dots,x_i,\dots,x_j,\dots,x_N) = \psi(x_1,\dots,x_j,\dots,x_i,\dots,x_N)$antisymmetric in$i,j$iff$\psi(x_1,\dots,x_i,\dots,x_j,\dots,x_N) = -\psi(x_1,\dots,x_j,\dots,x_i,\dots,x_N)$fully symmetric iff symmetric in all$i,j$fully antisymmetric ... 3 The wavefunction returns a complex number whose modulus-squared is a probability density and whose phase is related to the probability current, i.e., where probability is flowing to. If you write in the form $$\Psi({\mathbf x},t) = \sqrt{\rho({\mathbf x},t)}\exp\left(i\frac{S({\mathbf x},t)}{\hbar}\right)\text{,}$$ where$\rho(\mathbf{x},t)\geq0$and ... 2 If this were computer science, we might say$\psi$takes a$d$-tuple of reals ($r$) and another real ($t$) and returns a complex number with the attached unit of$L^{-d/2}$in$d$dimensions (with$L$being the unit of length).1 If you want any more of an interpretation, well then you've already given it:$\psi(r,t)$is the thing such that$\int_R\ \lvert ...

2

Just like $j=1/2$ (spin-half) particles have two-complex-component wave functions, spinors, particles with spin $j$ have $(2j+1)$-dimensional wave functions describing the spin degrees of freedom. The dimension is what it is because $j_z$ always goes from $-j$ to $+j$ with the spacing equal to one. All the transformation rules under rotations may be ...

2

I have 5 bags labelled 1 to 5, and I have randomly dropped the letters A to J into the bags. You choose a letter at random and you win as many Francs as the number on the bag containing your letter. If I have distributed the letters evenly, then there should be 2 letters in each bag, so we could say that ψ(bagnumber) = ψ = sqrt(2). But if we want ...

2

The general form, equivalent to eqn (5) but for outside the well, is $\Psi_o = A_o \dfrac{e^{-k_o r}}{\sqrt{r}} + B_o \dfrac{e^{k_o r}}{\sqrt{r}}$ But just like we know $\Psi_i$ is not infinite at the origin, we also know that $\Psi_o$ doesn't go to infinity at large values of r. So we know $B_o$ must be 0.

1

In the Copenhagen interpretation the collapse of the wavefunction is non-deterministic. Following the collapse the wavefunction evolves in a deterministic manner until the next collapse. However the Copenhagen interpretation gives no account of what the collapse process is and by what mechanism it occurs, and this has led to dissatisfaction with it in some ...

1

The superposition principle of quantum mechanics is not destroyed by quantum (hamiltonian) unitary evolution operator $U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$ as per @ACuriousMind's answer. Event if you dont know of the evolution operator in terms of the hamiltonian (which can be derived easily from the Schrodiger equation), still the fact that the ...

1

A free quark wavefunction does not exist. Instead, inside nucleons quarks are relativistic and asymptotically free, which means that they only behave like individual particles for sufficiently large momentum/energy transfer. Imagine a classical solid state analog: if you apply a very small, slowly acting force to a single atom of a crystal, the force will ...

1

First of all, not every Hamiltonian admits a basis of eigenvectors. An operator has to be either compact or with compact resolvent to admit a basis of eigenvectors. With those type of operators, eigenvalues can accumulate (i.e. being distinct, but getting arbitrarily close) only at zero (compact) or infinity (compact resolvent). A discrete eigenvalue can ...

1

Evidently this is a homework problem. One way to approach the integral is to look at the symmetry of the integrand. To get started: the integrand has two factors. one of them, $y$, is odd under inversion of the axis. Once you have the symmetry of the integrand, you can make quick progress with the integral. Make a sketch of the integrand. That will ...

1

I will answer in layman terms as from your age in your profile I would not expect a very strong background in the necessary mathematics. When we have a function f(x) it returns a value at the point x, a real number. If it is the potential, V(r)=1/r we are able to calculate the potential and solve simple problems or enter the potential in complicated ...

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