Timeline for Electron in free space and Schrodinger's equation
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2018 at 17:06 | comment | added | Francesco Bernardini | But if you consider the other constraints related to the physical problem you are considering (i.e. a free electron, so a system whose momentum has to be conserved as no external force is acting on it) you must impose that the solutions to the Schroedinger equation be solutions alsonof the additional eigenvalue problem $P\psi=p\psi$ where $P=-i\hbar\frac{\partial}{\partial x}$ is the differential operator that represents the momentum in QM; so you end up selecting only the energies that allow such solutions to appear; | |
| Dec 27, 2018 at 17:02 | comment | added | Francesco Bernardini | Hi, sorry for delay; you can consider my answer in the following way: the general problem of finding the eigenvalues/eigenfunctions of a differential operator (the Hamiltonian and its related 1-D Schroedinger equation $H\psi=E\psi$) would give you all possible energies (i.e. both positive and negative). | |
| Dec 20, 2018 at 16:19 | comment | added | Anwesa Roy | $$\psi_{x}(x)=A\exp\left[ i\sqrt{\frac{2m\tilde E}{\hbar^2}}\right]x +B\exp\left[- i\sqrt{\frac{2m\tilde E}{\hbar^2}}\right]x$$ | |
| Dec 20, 2018 at 16:19 | comment | added | Anwesa Roy | $$ \frac{\partial^2 \psi_{x}(x)}{\partial x^2} +\frac{2m}{\hbar^2} \tilde E\psi_{x}(x)=0$$ | |
| Dec 20, 2018 at 16:19 | comment | added | Anwesa Roy | $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_{x}(x)}{\partial x^2}- \tilde E\psi_{x}(x)=0$$ | |
| Dec 20, 2018 at 16:18 | comment | added | Anwesa Roy | Since $V$ is constant: $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_{x}(x)}{\partial x^2}+(V-E)\psi_{x}(x)=0$$ | |
| Dec 20, 2018 at 16:18 | comment | added | Anwesa Roy | $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_{x}(x)}{\partial x^2}+(V(x)-E)\psi_{x}(x)=0$$ | |
| Dec 20, 2018 at 16:18 | comment | added | Anwesa Roy | The 1-D Schrodinger's wave equation: $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_{x}(x)}{\partial x^2}+V(x)\psi_{ x}(x)=E\psi_{x}(x)$$ | |
| Dec 20, 2018 at 16:17 | comment | added | Anwesa Roy | Since I am from an electronics background, I don't have much idea about the momentum operator. Would you please explain the rest of the portion in a simpler way. I mean I am acquainted with the one-dimensional Schrodinger's wave equation. So if you could explain in terms of 1-D Schrodinger's wave equation, then I might understand. | |
| Dec 20, 2018 at 16:17 | comment | added | Anwesa Roy | Hi, I have understood your explanation up to this portion: $$\phi(x)=C_0\exp\left[\pm i\sqrt{\frac{2m\tilde E}{\hbar^2}}\right]x$$ I will be posting a few comments continuously in which I have worked out the steps. | |
| Dec 19, 2018 at 22:02 | history | answered | Francesco Bernardini | CC BY-SA 4.0 |