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The matrix exponential for any Pauli matrix using the general formula see general formula. This procedure gives, $$\exp[-iHt] = I_{2\times2}\cos \nu_{F}t - i {\bf\sigma}\cdot\left(q-By\hat x\right) \sin \nu_{F}t$$ It then follows that, the state vector at time t $\Psi(r,t) = \exp[-iHt] \Psi (r,0) $ From this, the average value of the angular momentum ...


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The answer is yes: Obviously it is possible to solve radially symmetric Schroedinger equations, including the specialized treatment (radial component only) that you ask about. However, you will need a well-defined problem to meaningfully begin something as rigorous as an analytical (or numerical) treatment. The differential equation you have given for the ...


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Usually one requires, that the propability density $\varrho = |\Psi(x)|^2$ is integrable over the whole range (suchs that it can be normalzied). In this sense $\Psi(x) = \exp(kx)$ is just as divergent as $\Psi(x) = \exp(ikx)$, since the limit $$\lim_{\epsilon \rightarrow \infty } \int_{-\epsilon}^\epsilon |\Psi(x)|^2 \ dx$$ does not exists in both cases. ...


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I am confused about why there is this difference here? What about $E$ being less than $V_o$ that we need to look if $exp(kx)$ diverges or not? Why was this not taken into account in the first case when $E>V_o$, i.e doesn't $C×exp(ikx)$ also diverge in the first case when $E>V_o$? The other answers have explored the question of why there are ...


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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} ...


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 ...


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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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Use a power series $ R(r) = \sum_{n=0}^\infty a_n r^n$. Doing so, you'll find an equation which connets the coefficients $a_n$, $a_{n+1}$ and $a_{n+2}$. By looking at the behaviour for big $n$, one can conclude that the series has to stop at some point $n=N$ (or else it will grow like $\exp()$ and prevent normalization). This $N$ is a new quantum number. For ...


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 ...


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

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 ...


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I don't want to spoil the beauty of 3D QM for you especially if you want to explore it by your own. However I thought I would give you some advise on how to approach to the transition from 1D to 3D. Considering 1D QM first of all think about what it means to be on the x-axis. Who is to decide which axis is x and which axis is the y or z axis for that ...


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

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: $$ ...


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 ...


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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, $$\frac{d^2f}{dx^2} = -k^2f$$ $$\frac{d}{dx}\left(\frac{df}{dx} \right) = -k^2f$$ ...


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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. ...


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The second and the third energy level act like an "almost" degeneracy and therefore it doesn't work. One of the conditions for perturbation theory to work is that the matrix elements of the perturbation can't be bigger than the spacing between the energy levels, which is the case here. A solution would be to introduce a real degeneracy for this "almost" ...


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First I'll describe the (to me) cleanest and clearest example, then we'll extend it. You have a wavepacket that in addition to a complex scalar field also specifies at every point a spin vector/plane. Since we want to describe a measurement we need a Hamiltonian that described the interaction with the device and the system, in our case the device is a ...


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Derivations of the Born rule have been proposed, but they have all been criticized for invoking circular reasoning as CuriosOne mentioned in the comments. You can read a review of the arguments by Huw Price here. Zurek has invoked the fact that due to decoherence, observers are always competely entangled with the environment and then you can reason based on ...


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Thus, should be governed by the Schrodinger's equation. Schroedinger's equation is useful for description of atomic systems. Measurement of such systems involves macroscopic bodies. It is out of place to apply Schroedinger's equation to these macroscopic bodies; classical description is simpler, more accurate and more useful. Why would you describe ...


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According to the Many Worlds Interpretation, the measurement device, and by extension everyone who reads it, gets entangled with the particle in exactly the way the Schrödinger equation predicts. The Born rule tells you how subjectively likely you are to be in a given universe. It's basically just a really bizarre anthropic principle. Feel free to interpret ...


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, ...


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$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\timex}[1]{\mathrm{e}^{\mathrm{i} #1 t}}\newcommand{\ntimex}[1]{\mathrm{e}^{-\mathrm{i} #1 t}}$The interaction picture split of the Schrödinger picture Hamiltonian $$ H_S = H_{0,S} + H_{\text{int}_S} $$ has nothing to do with the Hamiltonians themselves being time (in)dependent. In fact, the choice of ...


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

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 ...


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Look again carefully at the way you have written your spin operator components (up and down together?)and compare them to any q .m textbook spin components. Maybe it's just a typo in the way you have written the question, New to this myself but if u and d are orthogonal this would make each spin component equal zero, right? So I think question is ...


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Your question is misconceived. The wavefunction does not collapse. Rather, when you measure a system, each of the possible outcomes of that measurement happens. Each outcome is associated with a different version of the measurement apparatus. Those different versions of the measurement apparatus can't interact with one another or exchange information. The ...


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

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

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 ...


0

Well, if I'm right, then this problem has no solution - at least no finite analytical one. And btw. the problem is more general than one might think. Look at the upper right quadrant ($x\gt0$ and $y\gt0$). Then the boundary conditions are symmetry boundary conditions (scalar product of normal vector and gradient of the solution is zero at $x=0$ and $y=0$). ...


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 = ...


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

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 ...


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 ...



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