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Except in very elementary examples (single particles), the QM wave function has nothing to do with a wave (apart from the historical origin). For a system consisting of $N>1$ particles, the wave function is a function in configuration space (with 3N variables), not one in 3-space (whose coordinates are positions $x$ with 3 components). This can be read ...

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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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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 ...

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You are right, there is no actual wave that the, say, electron turns into whenever it chooses. The only property that forms wave-like images is the probability density function which is given by $$|\Psi ( x , t) |^2$$ This behaviour of the p.d.f. can be plotted as it resembles what we call a wave. Here are some examples from 1D, 2D and 3D potential wells ...

3

There is a bit more structure to be had in the problem, and I would recommend that you take advantage of it. In particular, you know that $\phi$ is really a linear function of $x$, or otherwise it is not a gaussian; and that the real and imaginary parts of its coefficient have distinct physical meaning, which you can get by taking appropriate expectation ...

3

I) The solution to the time-dependent Schrödinger equation (TDSE) is $$\tag{A} \Psi(t_f) ~=~ U(t_f,t_i) \Psi(t_i),$$ where $$\tag{B} U(t_f,t_i)~=~T\exp\left[-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~H(t)\right]$$ is the unitary evolution operator. II) The evolution operator $U(t_2,t_1)$ has the group-property $$\tag{C} U(t_3,t_1)~=~U(t_3,t_2)U(t_2,t_1), ... 3 If you represent the wave function \psi(x) with it's fourier transform, \begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\ \psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq \end{eqnarray*} (where p and q are almost like "dummy" momenta), ... 2 If you don't want the answer you could look in the wiki Group Velocity. In particular: One derivation of the formula for group velocity is as follows. Consider a wave packet as a function of position x and time t: α(x,t). Let A(k) be its Fourier transform at time t=0: \alpha(x,0)= \int_{-\infty}^\infty dk \, A(k) e^{ikx}, By the superposition ... 2 First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such \Psi(r,t), that fulfill this equation. The finite square well 1.0 fm wide means you have a potential V(r) which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting \Psi(r,t) ... 2 The identification that  ψ(x)≡⟨x|ψ⟩ is completely correct, and this is the way to treat wavefunctions in 'grown-up' quantum mechanics. In short, So basicaly wave function is continuous coefficient in the expansion of |ψ⟩ in the basis eigenkets |x⟩, right? Yes, and But this can be true for any state of the system only if the ... 2 I have been told that it is just a misnomer calling it an actual wave, and there is no wave there, its just that the wave function satisfies the general wave differential equation in 3d space. Now is this true? Is there really no wave? The confusion comes from extending our intuitive view of classical waves, as in water, and sound and light, to the ... 1 The equation$$\partial _{t}\psi (t)=-iH\psi (t)$$acting in a Hilbert space with H self-adjoint has the general solution$$\psi (t)=\exp [-iH(t-t_{0})]\psi (t_{0}),$$by Stone's theorem (https://en.wikipedia.org/wiki/Stone\%27s_theorem_on_one-parameter% _unitary_groups). In case H=H(t) depends on t matters change and time ordering becomes ... 1 In Mathematica, using the formula \sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases} where u_0=mL^2V_0/(2\hbar^2), we can first compute u_0: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since u_0 is typically on the order of an ... 1 First, note that \hat H_f has degenerate spectrum: it has equal eigenvalues for states with \left|k\right\rangle and for \left|-k\right\rangle. This in turn means that, in particular, the state \frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right) is also an eigenvector of \hat H_f. But in position representation it will look ... 1 The solution manual just uses a simple midpoint approximation to evaluate the integral. Doing it the way you did it should be fine as well. You just need to include both terms when you plug the limits of the integral into the antiderivate. I think you just did the x/2 term but not the \sin term. 1 Well, without the delta potential the wave function is$$\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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