# Tag Info

2

If you represent the wave function $\psi(x)$ with it's fourier transform, \begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\ \psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq \end{eqnarray*} (where p and q are almost like "dummy" momenta), ...

2

If you don't want the answer you could look in the wiki Group Velocity. In particular: One derivation of the formula for group velocity is as follows. Consider a wave packet as a function of position $x$ and time $t$: $α(x,t)$. Let $A(k)$ be its Fourier transform at time $t=0$: $\alpha(x,0)= \int_{-\infty}^\infty dk \, A(k) e^{ikx}$, By the superposition ...

0

The solution of this problem is in principle very easy. Since $$z^{4} + 4\left(\mu + 1\right)z^{2} + 8\lambda\Delta z + 4\left(\mu^{2} + \Delta^{2} - V^{2}\right) = 0\text{,}$$ is a fourth order equation I can solved this equation by the hand or with Mathematica/Maple and get 4 solutions. The interesting case is that with two purely real and two complex ...

1

In Mathematica, using the formula $\sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases}$ where $u_0=mL^2V_0/(2\hbar^2)$, we can first compute $u_0$: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since $u_0$ is typically on the order of an ...

2

First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such $\Psi(r,t)$, that fulfill this equation. The finite square well 1.0 fm wide means you have a potential $V(r)$ which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting $\Psi(r,t)$ ...

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If we stick to the Problem #2 you mentioned, then yes, there is an energy difference. Since seems there is no degeneracy from symmetry for $\psi(r-a) \pm \psi(r+a)$, unless numerically accident. $$E_{\pm} = \langle \psi(r-a) \pm \psi(r+a) | \hat{H} | \psi(r-a) \pm \psi(r+a) \rangle \tag{1}$$ $+,-$ correspond to $E_G,E_E$ in the Problem #2, ...

1

Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to ...

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In general every linear combinations of a separable solution is still a solution (superposition principle), so you can take the simplest separable solution, put it in a linear combination with arbitrary coefficient (complex numbers, phases) and you obtained a solution of Sc. equation NON-separable.

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The only thing that's really important is the differential eqaution. The situation is, outside the well, in both cases: $\dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi$ Now it's foundamental notice that for bound states E<0 so we can write: $E=-|E|$ and Sc. equation become: $\dfrac{d^2 \psi}{dx^2}= + \frac{2m|E|}{\hbar^2} \psi$ So the usual way ...

1

First, note that $\hat H_f$ has degenerate spectrum: it has equal eigenvalues for states with $\left|k\right\rangle$ and for $\left|-k\right\rangle$. This in turn means that, in particular, the state $\frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right)$ is also an eigenvector of $\hat H_f$. But in position representation it will look ...

0

Firstly, notice that if you take linear combinations of $\psi_f$ you get $\psi_b$ -- so the basis for both Hilbert spaces is the same. Though I'm not sure how rigorous the following statement is, I'm tempted to say that: The wavefunctions do not sense any region outside the potential well, where we have infinite potential. So as far of the states of our ...

4

One purpose of using $\Psi$ rather than just the probability density is to match observation. Dealing only with probability density isn't sufficient. Imagine you can send particles through two adjacent slits toward a detector screen. You'll find an interesting pattern that looks like an interference phenomenon is happening. This is called the double slit ...

1

The solution manual just uses a simple midpoint approximation to evaluate the integral. Doing it the way you did it should be fine as well. You just need to include both terms when you plug the limits of the integral into the antiderivate. I think you just did the $x/2$ term but not the $\sin$ term.

8

Short version: In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ...

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(a) Your answers are correct. (b) Yes,it is simpler to write out the wave packet in the momentum basis. (This is effectively equivalent to working out the three-dimensional Fourier transform of the given Gaussian wave packet in position basis.)

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It is not correct that the probability distribution of $x$ and $p$ are Gaussian in general. Take a simple system of a particle moving in some potential $V(x)$. The probability distribution of $x$ is the square of the wave-function $\Psi(x)$ of the particle, i.e. the probability of finding your particle in $[x,x+dx]$ is $|\Psi(x)|^2dx$. The probability ...

1

I believe this can be attributed to the central limit theorem, which states that a large number of samples from a population with a well-defined variance will follow a gaussian distribution. The key idea is that because of quantum mechanics, we must treat both position and momentum as random variables; the uncertainty principle gives us a relation between ...

0

No, it doesn't collapse to an eigenstate. Collapse to an eigenstate is a picture of an ideal measurement. In general the final state will not be describable by a wave function, because it's not a pure state, it is instead a mixed state. See this question, which is about inexact measurements. Position eigenstate in position representation is $\langle ... 8 What is your Hilbert space? In$L^2(\mathbb R)$your eigenfunction would have infinite norm. If you dealt instead with a bounded set$L^2([a,b])$, your operator would not be Hermitian unless you impose suitable boundary conditions to discard boundary terms. These boundary conditions, however, would rule out your candidate eigenvector! 0 No. Position operator does not have normalizable eigen-functions ($\delta(x-x_0)\$ is not normalizable). The closest thing one can do in this formalism is to contract the wave function to some sharp peak with non-zero width and finite height, based on the accuracy of the measurement. With continuous space, particle cannot be in an "eigenstate of position", ...

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It is due to the nature of quantum mechanics. In classical regime, the lowest possible energy is zero. But in QM, the lowest state(ground state) still has energy. Quantum nature is wave-nature. In CM, you can pinpoint a location of an object but in QM, it is a distribution (probability density). What you can only do is find the smallest distribution not a ...

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As very often in these sorts of proof, we just need to use a completeness relation: We have, \begin{align} \langle p | \psi _0 \rangle & = \langle p | \int dq |q \rangle \langle q | \psi _0 \rangle \\ & = \int dq e ^{ - i p q } \langle q | \psi _0 \rangle \\ & = \frac{1}{ \sqrt{ a \sqrt{ \pi }}} \int dq e ^{ - i p q } e ^{ - q^2 / 4 a ^2 ...

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