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4

Of course, now we've adopted Schrödinger equation as a postulate: it is true. However, Schrödinger derived the equation from previous knowledge. Schrödinger thought his equation from Hamilton-Jacobi formalism. If you take the classical limit in that equation you'll find the Hamilton-Jacobi equation. You can also read the original Schrödinger papers in ...


4

The Schrodinger equation is an approximation because it ignores relativistic effects and it ignores spin. However, aside from these limits it applies to any system and not just electrons. The trouble is that the Schrodinger equation, and indeed most partial differential equations, are impossible to solve except in a few special cases. Since we have no ...


2

The probability of just one of your atoms getting through the wall is very low, multiply that probability (which is way lower than 1) by the probabilty that all of your atoms go through, you get a number so small that the universe would almost certainly have ended long before you got through the wall. A 70 Kg human body has approximately 7 X $10^{27}$ ...


2

I don't want to be too much precise, but the Schrödinger equation ($i \dot{\psi}= H\psi$ to avoid confusion) has at most an unique solution under very general assumptions on the Hamiltonian operator $H$, even if you see it as a liner equation in the more general setting of Banach spaces. In particular, it is not a priori necessary that $H$ is self-adjoint ...


2

The standard procedure is the following: starting from $ \langle nlm|\,\partial^2_z\,| n'l'm'\rangle $ insert the identity operator with respect to the position basis $$ 1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}| $$ to have $$ \int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ...


1

This is perhaps best seen via the dispersion relation after doing a Fourier transformation to the energy-momentum representation. The time derivative $\partial_0$ corresponds (up to a factor of $i$) to multiplication with energy $E$ (which is the total energy minus the rest energy that was subtracted in eq. (2).). In the non-relativistic (NR) limit, we are ...


1

Solving time-dependent SE as danielsmw mentioned above good starting point. $$ i\hbar\frac{\partial\psi}{\partial t}=H\psi $$ $$ \frac{d\psi}{\psi}=\frac{H}{i\hbar}dt $$ $$ log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0} $$ suppose $\psi=|\alpha,t>$ and $\psi_{0}=|\alpha, t_{0}>$ $$ \psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0 $$ Under ...


1

As Tom pointed out, I made a mistake with assuming that the operator PA is -ih(∂A/∂x) when it should be -ih(∂A/∂x+A∂/∂x). This means that there is a (1-α) coefficient for the AP operator.


1

If I understand the question. Assume you have a quantum oscillator, then if the system is an eigenvector of the hamiltonian, then a series of measurements will give a series of results reflecting eigenvalues of the system. The appearance of these eigenvalues must obey at a number of many measurements the probabilities of each eigenvalue to appear. It's ...



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