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## Hot answers tagged schroedinger-equation

4

The Schrödinger equation is only correct in the non-relativistic limit $v << c$, for particles without spin. The correct equation for spinless (=spin $0$) particles is the Klein-Gordon equation, which reduces in the non-relativistic limit to the Schrödinger equation. If we want to talk about spin $\frac{1}{2}$, the correct, relativistic equation is ...

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I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...

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A state $\psi$ corresponds to an energy $E$ if: $$H\psi = E\psi$$ Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then $$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$ A linear combination of states with the same energy value again has the same energy value. Now consider $\psi = c_1\psi_1 + ... 2 The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ... 2 Suppose$\psi$satisfies the (dimensionless) time-dependent Schrödinger equation: $$i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi$$ It will also satisfy the conjugate equation: $$-i\frac{\partial\psi^*}{\partial t}=-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^*$$ Now consider how the normalisation changes over time: ... 1 The key here is that E must exceed the minimum of$V(x)$. If one has a delta function well then the minimum is$V_0 = -\infty$while for a delta function barrier the minimum is$V_0 = 0$Hence for the delta function well, one can have bound states with$E<0$or one can have scattering states with$E>0$. For the delta function barrier the necessary ... 1 Comments to the question (v2): Normally we assume that$-\infty< x_1 < x_2<\infty$are the turning points in the 1D potential well. This means $$\forall x\in ]x_1,x_2[:~~ E ~>~ V(x).$$ Hence OP's third last equation $$0 ~<~ \int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} ~=~ -\frac{\pi}{2}\hbar~<~0$$ can never be fulfilled. The correct WKB ... 1 The Dirac notation is simply an alternative to vector notation. Certainly there are PDEs describing the quantum state of a lone particle with spin and they are: The Pauli equation (see Wiki page of this name) was historically the first, and here the quantum state is two$\mathcal{L}^2(\mathbb{R}^3, \mathbb{R})$functions of space and time. The two ... 1 The Dirac equation is mentioned in other answers as PDE describing spin. As you ask "what would Schrodingers equation (or some kind of generalization that allows for you to include spin) look like?", the following may be relevant. Yes, the Dirac equation adequately describes spin. However, it is actually a system of four partial differential equations for ... 1 Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance$\Psi(x)=\sum_nc_n\psi_n(x)$is a general quantum state for a particle in a box, where$\psi_n(x)\$ are the energy ...

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It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...

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First off, there are some very misleading answers given above. Introductory quantum courses fail to properly discuss "time." It is a parameter, not an observable. E(operator)=ih(bar) d/dt has no meaning. That operator simply discribes the time evolution of a wavefuntion that is complex. So it does not describe a physical observable. I know this might be hard ...

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