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2

Yes, this looks correct, except that the energy of the state $|2\rangle$ in $|\psi(t)\rangle$ should be $E_2$. I also don't think you want hats on the energies $E_2$ and $E_3$. Such hats are usually used in basic quantum mechanics to indicate operators, bur $E_i$ are the eigenvalues of the operator $\hat{H}$, and hence just ordinary numbers. Since $\hat{H}$ ...

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Here is my attempt at an answer, following the suggestion of @Lagerbaer. We first subtitute the Fourier Transform for $\psi_{LP}(k)$, $$\psi_{LP}(k)=\int dxe^{-ikx}\psi_{LP}(x),$$ and get \begin{multline} \int dxe^{-ikx}i\frac{d}{dt}\psi_{LP}(x)=\int ...

1

When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate $x$ must match that at coordinate $x+L$, so we have: $$\Psi(x)=\Psi(x+L)$$ In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be ...

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Some broadly applicable background might be in order, since I remember this aspect of quantum mechanics not being stressed enough in most courses. [What follows is very good to know, and very broadly applicable, but may be considered overkill for this particular problem. Caveat lector.] What the OP lays out is exactly the motivation for finding how an ...

1

For a free particle, the energy/momentum eigenstates are of the form $e^{i k x}$. Going over to that basis is essentially doing a Fourier transform. Once you do that, you'll have the wavefunction in the momentum basis. After that, time-evolving that should be simple. Hint: The fourier transform of a Gaussian is another Gaussian, but the width inverts, in ...

3

Within the superposition of the ground and the first excited state, the wavefuncion oscillates between "hump at left" and "hump at right". Maybe you are asked to find the half-period of these oscillations?

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Yes, I believe you have to think of it as if it were a semiclassical problem; you evaluate with QM the mean square velocity $\left< v^2 \right>$ of the particle, then calculate its square root; this should give you an estimate of the typical velocity of the particle. Once you have it, you divide the length of the well by it and find the time it takes ...

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In my view, the important question to answer here is a special case of the more general question Given a space $M$, what are the physically allowable wavefunctions for a particle moving on $M$? Aside from issues of smoothness wavefunctions (which can be tricky; consider the Dirac delta potential well on the real line for example), as far as I can tell ...

2

Your solution is valid. It has zero kinetic energy. It doesn't necessarily have zero energy. It can have any potential energy you'd like. Just because your particle is "freely moving," that doesn't mean the potential is zero. You could have $V(x)=k$ for any constant $k$. The value of $k$ is not observable and has no physical significance. In general there ...

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1) In general, $\psi(\vec{r},t) = {\sf U}(t,0) \psi(\vec{r},0)$, where ${\sf U}(t,0)$ is the time-evolution operator (a unitary matrix). 2) Given your superposition state at initial time, after time $t$ the wave function would look like $$\psi(r,\theta,\phi,t) = A \left( 2R_{10}Y_{00} e^{-iE_1 t/\hbar} + 4 R_{21}Y_{1,-1} e^{-iE_2 t/\hbar} \right)$$ where ...

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I am not sure if you are looking for this, but you can define a Lagrangian in such a way that the L-EOM (equation of motion) is the SchrÃ¶dinger equation. $\cal{L}=\Psi^{t}(i\frac{\partial}{\partial t}+\nabla^2/2m)\Psi$ $\frac{\partial\cal{L}}{\partial\Psi^t}=0$ The second term of the Lagrange-equation (derivative with respect to $\partial_{\mu}\Psi^t$) is ...

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I want to elaborate on John Rennie's answer. The Schrodinger equation for a free particle is ($\hbar=1$): $$i\frac{\partial}{\partial t}\psi=-\frac{1}{2m}\frac{\partial^2}{\partial x^2}\psi.$$ It is a first-order differential equation in variable $t$. To solve it, you should specify initial data, say, $\psi(t=0)$. At this point, you should be aware that ...

4

When you solve the Schrodinger equation for a free particle you get a family of solutions of the form $\Psi(x,t) = A e^{i(kx - \omega t)}$ and all superpositions of these functions. So just solving the Schrodinger equation doesn't give you a solution for a specific particle. For that you need to specify the initial conditions. If you take the solution to be ...

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The problems you have been encountered are related to that fact that you try to calculate probability of some unphysical situation. Quantum mechanics can give you probability of an outcome from some experiment. This wave functions does not contain any information (restrictions) concerning the way how you are going to measure it and what you are going to ...

3

First of all you should recall that Schroedinger equation is an Eigenvalue equation. If you are unfamiliar with eigenvalue equations you should consult any math book or course as soon as possible. Answer 1 (my apologies, I will use my own notation, as this is mainly copy-paste from my old notes): First define constants x_0 = ...

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Time-dependent Schrodinger equation is an elliptic PDE if the Hamiltonian is time-independent.

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The time-independent Schrodinger equation is mainly useful for describing standing waves. It has serious shortcomings when used to describing traveling waves. If you have an example like a constant potential, then there are only traveling-wave solutions, and the time-independent Schrodinger equation may be the wrong tool for the job. Physically, the ...

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Picking up on your comment that plane waves are not renormalisable: Only infinite plane waves are not renormalisable, and an infinite plane wave is not physically realistic simply because we can't make, or indeed even observe) infinite objects. Any plane wave we can observe will be finite and therefore normalisable. An infinite plane wave represents an ...

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In my opinion,if a potential function is authentic, the solution of the time independent Schrodinger equation will have physics meaning. You say a one dimensional universe with constant potential,in fact ,this kind of potential is not existed. I think the plane wave cannot be normalized is a reflection of this.

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