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4

No, you can't do that. First of all because it doesn't make much sense to talk about the wave function corresponding to a spatial dimension. A wave function $\psi(x_1,\dots,x_n,t)$ gives you the probability amplitude of the system being at the position $(x_1,\dots,x_n)$ at the time $t$. It must by definition be dependent on all the coordinates necessary to ...


3

The evolution immediately after the measurement occurs according to the Schrödinger equation, but now with a known initial state. In general, in order to be able to "repeat" a measurement you have to prepare the state again. For example, say I have a particle in a harmonic oscillator potential prepared in the state $$|\psi\rangle = ...


3

There is a different interpretation of Schroedinger's equation than the one in terms of Born's rule or Copenhagen interpretation. It is called bohmian mechanics or DeBroglie-Bohm-theory or pilot-wave theory. The general idea is to set the wave-function as $\Psi(t, \vec{x}) = R(t,\vec{x}) \cdot \exp(i\ \frac{S(t,\vec{x})}{\hbar})$, which is no restriction. ...


3

When solving differential equations we often us a technique called ansatz. The word ansatz is German for guess. In other words we guess at a possible solution then feed it back into the differential equation to see if it works. In this case we guess that the solution is $\Psi(x) = e^{ikx}$. If we take the second differential of this we get: $$ ...


3

Probably one of the first things one should do after arriving at a solution to an ODE is stick it back in to see if it satisfies the original equation. In your case, you obtain $$ \left(4k^4x^2\Psi-2k^2\Psi\right)+k^2\Psi\neq0 $$ where the term in the parenthesis is the result of $\frac{d^2}{dx^2}\Psi$. So clearly you went wrong somewhere. I'd wager that it ...


3

Can someone explain why the time-independent Schrödinger equation isn't an eom? The TISE is an eigenvalue equation due to applying separation of variables to the TDSE; it is an equation for the spatial function alone. Can someone explain in what sense exactly is the time-dependent Schrödinger equation an equation of motion? A Lagrangian ...


2

In bra-ket notation, we have $$\hat H |\psi(t)\rangle = i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle $$ The Hermitian conjugate of this equation is $$\langle\psi(t)|\hat H^\dagger = -i\hbar \frac{\partial}{\partial t} \langle\psi(t)|$$ But $\hat H$ is self-adjoint thus $$\langle\psi(t)|\hat H = -i\hbar \frac{\partial}{\partial t} ...


2

$$\frac{\partial}{\partial t}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i\hbar kx}\left(\frac{\partial}{\partial t}e^{-\frac{1}{2}\frac{\hbar^2k^2}{m}t/\hbar}\right) dk$$ $$\frac{\partial}{\partial x}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) \left(\frac{\partial}{\partial x}e^{i\hbar ...


2

The general solution to the differential equation you have given is $Ae^{ikx}+Be^{-ikx}$, which you can verify by substituting back in. (This is a common method, putting in some parametric form of a potential solution and determining the values of the parameters after substituting the trial solution). A plane wave has the form $\Psi(x,t)=Ae^{i(kx-\omega ...


2

I'll tell you a very powerful way to think about this problem. I myself would not find the eigenvectors like Chris2807 did. I'd proceed as such: The general solution you refer to is $\exp(-\frac{i t}{\hbar}\hat{H})|\varphi_0\rangle$. So, you seem to be aware that this is $\exp(-\frac{i t}{\hbar}C B \hat{S}_y)|\varphi_0\rangle$. It is useful to think of ...


1

Author of this text first splits the Hamiltonian up as $$H=H_0+H_{int}$$ The time dependence of operators is governed by $H_0$ while the time dependence of states is governed by $H_{int}$. (Question, does this say that the Hamiltonian is broken into a time dependent $H_0$ and a time independent $H_{int}$ parts?) No, not necessarily, ...


1

What you're asking for is impossible. The energy eigenvalues are given by the zeros of the determinant (as you surmised) and cannot be related to the zeros of the airy function. $$Ai(\zeta_a)Bi(\zeta_0)-Ai(\zeta_0)Bi(\zeta_a)=0$$ For the odd solution (and with derivatives on $Ai$ for the even solution). Also be careful, your $z$ is not dimensionless as it ...


1

Indeed, there is no getting around having to solve this problem numerically. Departing from the determinant method originally posted, perhaps an alternative approach here would be to solve the system using only the energy. We start with the full form of our points of interest taking the odd wavefunctions as an example: $$z(0) = ...


1

The quantum numbers serve to enumerate the solutions to the time-independent Schrodinger equation (Energy eigenstates) which are also eigenstates of the angular momentum (squared) operator and have a definite component of angular momentum in some direction labeled $ z $. When solving the problem, it just turns out to be handy to label the solutions with the ...


1

The answer to your question comes from math, not from physics. Consider the wave equation in 1+1 dimensions: $$\partial^{2}_{x}\phi(x,t) - \partial^{2}_{t}\phi = 0$$ One obvious way to solve this is to convert the PDE in two variables into two ODEs in one variable. This is done by assuming that the solution can be written as some sum $$\phi(x,t) = ...


1

Hints: Consider the potential $$\tag{1} V_{\rm cf}(r)~:=~\frac{\hbar^2}{2m} \frac{\ell(\ell+1)}{r^2}. $$ For which radius $r$ is the potential (1) smallest? If it smallest for large $r$, it is reasonable to call it a centrifugal (rather than a centripetal) potential. Alternatively, what is the sign of the corresponding force $$\tag{2} F_{\rm ...


1

I will get you started with the first bit and finding the eigenvectors to the Hamiltonian, so you are right in that this Hamiltonian is proportional to the $\hat{S}_y$ operator so you need to find the eigenvectors of the $\hat{S}_y$ operator and they will also be eigenvectors of the full Hamiltonian. So we want some states $\vert y_+\rangle$ and $\vert ...


1

Do you mean a particle with quantum behaviour, or classical behaviour? If it's classical you'll need to update it's position based on its initial velocity, and include bounces off walls, which are most easily modelled as elastic collisions which preserve the magnitude of its velocity at walls. If quantum, you will need to know its energy and the size of the ...


1

Assuming $\Psi$ is the wave function. The Schrodinger equation biing, $$ \frac{\mathrm{d^{2}\Psi } }{\mathrm{d} x^{_{2}}}+k^{2}\Psi =0 $$ This is a standard solution of differential equations. If fou have done simple harmonic oscillators, it appears there. In particilar, $$\frac{d^2f}{dx^2} = -k^2f$$ $$\frac{d}{dx}\left(\frac{df}{dx} \right) = -k^2f$$ ...


1

Your solution is right. What you get verifying it is that $\psi$ is also an eigenfunction of the momentum operator, which means $$\hat p\psi=p\psi,$$ where $\hat p=-i\hbar\nabla$ is momentum operator, and $p$ is its eigenvalue. Now, applying $\hat p$ twice and dividing by $2m$, you can get $$\frac1{2m}\hat p^2\psi=\frac1{2m}p^2\psi,$$ which is just ...


1

I'm glad I can finally answer this. Numerov's method as described on Wikipedia is not how you want to proceed here. To give you an idea of how to proceed, let's start with a simplified version of the method. What I'm going to do is to just naively discretize the differential operator, like so: $$ \frac{d^2}{dx^2}\psi \approx \frac{\psi(x-d)-2\psi(x) ...


1

The proper constructions resembling quantum mechanic's formalism does exist in classical mechanics, but it goes a bit beyond lagrangian formalism. In classical mechanics, you can represent a system by a phase space with points corresponding to states of the system. Now, functions over that phase space form a symplectic Lie algebra together with the Poisson ...



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