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3

The Wikipedia page appears to be a little confused on this point. It claims to provide the Schrodinger Equation for a charged particle in an electric field. One might infer from the wording that the equation to follow is for a fixed, static electric field set up by fixed external charges. What actually follows, however, is the Schrodinger equation ...

2

The reduced mass is used where we are modelling a two (or more) body system by a single body. It isn't just used in quantum mechanics - you'll find it used in classical mechanics as well. For example, suppose we are modelling a hydrogen atom. We normally think of the electron as orbiting the proton, but actually both orbit their mutual centre of mass. This ...

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One of many good reasons to know it is that you can also use the value you get for a metal, along with the dependence on the number of the energy level (usually "n") and the fact that electrons, as fermions, "pile up" in terms of their energy states, to get an approximation of the energy distribution of electrons in metals. I would second the notion given by ...

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As you yourself have illustrated, you gain an understanding of how the energy changes with respect to the parameter $a$. For fixed $a$ you may argue that this ground state energy is just an irrelevant constant. But if $a$ is not fixed (e.g., you cite two different values of $a$ in your example) then this formula does tell you how the energy varies with $a$. ...

0

First, the $k$ in the first region is incorrect. Check your signs. You should have $$k_{I} = \sqrt{2M(E-V_o)}/\hbar.$$ Second, your solution to the second region will have sine and cosine solutions with $$k_{II} = \sqrt{2ME}/\hbar.$$ Real exponential solutions will occur in each region if $E<V_o$ in region one and/or $E<0$ in region II. Look at ...

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What you're missing is that you're interested in $E < 0$. Such states are "bound" in the square well. The equation: $\frac{d^2 f}{dx^2} ~=~ - k^2 ~ f(x)$ is solved by $f(x) ~=~ A e^{i k x} + B e^{-i k x}$. If $k^2 < 0$ then you get another $i$ in this picture and you switch from sinusoidal behavior to plus/minus exponential behavior.

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To include magnetic field into non-relativistic Schrödinger's equation, you can consider Pauli's phenomenological equation, which is a non-relativistic approximation of Dirac equation. It includes both spin and vector potential. If you drop spin-dependent term from it, you'll get an equation for charged spinless particle in magnetic field. For a spinless ...

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In the Schrödinger equation you can introduce, in principle, whatever form of potential you like. All the question is whether it allows a physical solution. About a particle in the magnetic field you can very well use the Schrödinger equation in which you introduce the interaction term $mB$, where $m$ is the magnetic dipole of the particle and $B$ the ...

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We cannot know where the particle is with certainty. The particle, in general, does not have a definite location to know. Under certain conditions, we can know and predict future probability distributions. The evolution of the state is determined by the Hamiltonian (in the Schrodinger picture). The problem is that we don't know the Hamiltonian ...

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Correct, expect just don't say "on average," just we will find the particle at different locations more often than others in accordance with the probability distribution described by a wave function. Incorrect, you may always calculate the evolution of an initial state if you know the effective Hamiltonian or effective Action the state is subject to. ...

2

Schroedinger's equations may have both normalizable and non-normalizable solutions. The function $$\psi_k(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}. \tag{2}$$ is a solution of the free-particle Schroedinger equation for any real $k$ and $\omega = |k|/c$. As a rule, if the equation has a class of solutions whose members are functions ...

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The other answers are right. But you might like some insight on why the plane wave is not normalizable, yet still useful. The general solution is a superposition of components of the form $\psi(x,t) = A e^{i(kx-\omega t)} + B e^{i(-kx-\omega t)}$. Each component will have a different $k$ and $\omega$. A single component is a uniform distribution over all ...

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The solutions of type $e^{i(kx - \omega t)}$ (so-called Fourier components) are not normalizable to 1. Though, they are widely used in quantum theory. They are normalized (quite improper expression) to $\delta$ Dirac, $$\frac {1}{2\pi} \int_{-\infty}^{\infty} e^{i(k'x - \omega' t)}e^{-i(kx - \omega t)} dx = \delta (k - k') e^{i(\omega - \omega')}. \tag{i}$$ ...

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Plane wave solutions to the Schrödinger equation are not normalizable because they extend to infinity with a constant amplitude. Any physical particle will be constrained to a finite space, though (at least to the visible universe), so you need to look at superpositions of plane waves. This means that your starting point is $$\psi(x, 0) = \int \mathrm dk ... 1 There is nothing wrong with this, as the solution to this equation doesn't live in the Hilbert space. That is to say that there is no eigenvector solution to the free particle equation since the Hamiltonian has only a continuous part in its spectrum. The best you can do if you want to stick with the Hilbert space is to find a sequence of vectors that ... 0 It's just an Ansatz and as long as it fulfills the given homogeneous ordinary differantial equation it's OK to use it. In the end you'll get the same results with the more general formula you gave because at one point (e.g. when inserting initial or boundary conditions) your B turns out to be 0. The most general (correct me if I'm wrong) Ansatz would be ... 0 I post this answer to check my understanding. Imagine a wavefunction in 1 dimensions with a known energy and momentum it's wavefunction will be:$$\Psi(x, t) = e^{i(kx-\omega t)} = e^{i(px-E t)/\hbar}$$With some calculus and algebra you can derive the momentum operator and get this:$$-i\hbar \partial_x \Psi = p \Psi There $-i\hbar \partial_x$ is ...

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The propagator does satisfy the Schrodinger equation for most values of x and t... The easiest way to show this is to let $\hbar=m=\omega=1$. Also, we can work with $\sqrt{2\pi}e^{i\pi/4}K\to K$ instead of $K$ to clean up the mess a little further. Further, let $x_f\to x$ and $x_0 \to 0$. Then let: $K=fe^{ig}$ with $f=\frac{1}{\sqrt{\sin(t)}}$ and ...

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Except that they don't have the same eigenvalue. To understand it in terms of differential operators; divide the first equation by $1/r^2$, this gives you an eigenvalue equation for the total hamiltonian operator $H = -h^2\nabla^2/2m + V$, whose eigenvalue is clearly $E$. Then we divide $\nabla^2 = \hat{R} + 1/r^2 \hat{\Omega}$ in your notation (into radial ...

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If you consider the complex logarithm, then $\log z=\log|z|+i\arg z$. (plus integer multiples of $2\pi i$ for different branches.) Now if $z\in \mathrm{R}$ and $z<0$, then $\log z = \log(-z) + i \pi$. So the action $S$ has just changed by an additive constant. However, adding something to your action does not change the physics described by it, since the ...

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