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7

You can show this by using perturbation theory (only for suitable small changes in the potential). When you assume, that $\tilde{V}(x) = V(x) + c$ with $c > 0$, then you can write your problem als perturbation: If the unperturbated hamiltonian $\hat{\mathrm H}$ has eigenstates $ | \Psi_n \rangle $ with discrete energies, then perturbation-theory states ...


4

Let there be given a selfadjoint$^1$ operator $H^{(0)}$ and a (semi)positive operator $V\geq 0$ on a Hilbert space ${\cal H}$. Let the basis of normalized eigenvectors for $H^{(0)}$ be $(|i^{(0)}\rangle)_{i\in I}$ with corresponding eigenvalues $(E^{(0)}_i)_{i\in I}$ ordered such that $$ \forall i,j ~\in~I:\quad i~\leq~j\quad\Rightarrow \quad E^{(0)}_i~\leq~...


2

$T$ and $R$ are transmission and reflection coefficients for waves. They refer to the probability that an incident wave will penetrate the barrier and continue propagating infinitely far. Physically, you should think of it as sending a constant sine wave in from the far left and looking to see what amplitude of constant sine wave you get at the far right. ...


2

Consider $\tilde{V}(x)=V(x)+\Delta(x)$ where $\Delta(x)>0,\forall x$. Within 1st order perturbation, $E_n=E_n^0+\langle \psi_n|\Delta|\psi_n\rangle=E_n^0+\iint dx_1dx_2\psi_n^*(x_1)\delta(x_1-x_2)\Delta(x_2)\psi_n(x_2)=E_n^0+\int dx|\psi_n(x)|^2\Delta(x)>E_n^0$


2

$$A\sin(k_iL)=De^{-qL}$$ $$Ak_i\cos(k_iL)=-Dqe^{-qL}$$ $$k_i\cot(k_iL)=-q$$ Insert the values for $k_i$ and $q$: $$[2m(V_1-E_i)/\hbar^2]^{1/2}\cot[2m(V_1-E_i)/\hbar]^{1/2}L=-[2mE_i/\hbar^2]^{1/2}$$ The allowed energy levels ($E_i<V_2$) for bound states can be determined by numerical solution of that equation. But particles with energy $E>V_2$ can ...


2

Qualitatively, the wave functions of the bound states in a triangular potential well like the one you described, look like this: For $x<-a$, $\psi=0$ because of the infinite potential in that region. Where the wave function crosses the potential line, quantum tunnelling occurs and $\psi \to 0$. For particle energies above $V_0$, no bound states can ...


1

Even with a delta function potential, continuity of the wave function is still required. (Please see comment from ACuriousMind below on this). The derivative of the wavefunction is obviously not continuous, however. You can find the discontinuity by integration about the delta function from +s to -s, where s is a small parameter. You then let s go to 0 ...


1

Generically, the answer is "no". The Zeeman effect is the splitting of degenerate spectral lines in the presence of a static magnetic field. As the field strength increases, some lines move to higher energies and some lines move to lower energies. Example of the splitting of the $5s$ orbitals of Rubidium: (Graph created by: Danski14. Image used under ...


1

Your normalization condition, $$\int_0^\infty\vert{}R\vert{}r^2\mathrm{d}r\int_0^\pi\int_0^{2\pi}\vert{}Y\vert\sin\theta\mathrm{d}\theta\mathrm{d}\phi = 1,$$ can be easily translated to $C_R C_Y = 1$ where $$C_R \equiv \int_0^\infty\vert{}R\vert{}r^2\mathrm{d}r, \quad C_Y \equiv \int_0^\pi\int_0^{2\pi}\vert{}Y\vert\sin\theta\mathrm{d}\theta\mathrm{d}\phi.$$ ...



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