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From the time-independ shr$\ddot{o}$dinger equation: $$ V(x) - E = \frac{1}{\Psi(x)}\frac{\hbar^2}{2m}\frac{d^2\Psi(x)}{dx^2}. $$

A sinusoidal wave means that a sine wave profile has already established on the string. Probably, you are thinking about the initial stage of building up the wave by an oscillating drive in one end of …

Since you structure was sandwiched by two infinite potential barrier at both ends, therefore strictly speaking, they are all bond states. If classified these states from the barrier height, you may lo …

In three waves, you may class them into two group of equal amplitude. Group one: $$ \frac{1}{2} e^{ik_0x}\,\, \text{ and } \,\,\frac{1}{2} e^{i\left(k_0+\frac{\Delta k}{2}\right) x}; $$ Group two: $ …

The relation of the first derivative of the wave function is meant to keep the flux of propability constant across the interface. Therefore, if the effectice masses are different across a heterojuncti …

The properties of a physical system depends on the boundary condition as well as the governing equations. All eigen vlaues are results from a giving boundary condition. Therefore, it is to be kept in …

The corresponding exponent of wavefunction is $q = \sqrt{\frac{2 m E}{\hbar^2}}$: Region 1: $x < -a$ $$ \Psi_1(x) = A \exp(qx). $$ Region 2: $-a < x < a$ $$ \Psi_2(x) = B \exp(qx) + C \exp(-qx); … $$ Region 3: $x > a$ $$ \Psi_3(x) = D \exp(-qx); $$ Then we match boundary conditions at $x= -a$: wavefunction continum $\Psi_1(-a) = \Psi_2(-a) $ $$ A \exp(-qa) = B \exp(-qa) + C \exp(qa); \tag …

A choice be made for all eigenvalues to be either even or odd ? The analysis in group theory can classify how many different sysmmetry styles will be in the eigen functions according to the symmeytry …

Starting from your equation: $${\Psi^*(x)}\,(d^2\Psi(x)/dx^2)-(d^2{\Psi^*(x)}/dx^2)\,\Psi(x)=0$$ You are just one-step away from the answer: \begin{align} 0 &= {\Psi^*(x)}\,(d^2\Psi(x)/dx^2)-(d^2{\Psi …

As Al Brown suggested, the current operator is equivalent to \begin{align} \vec j(\vec r, t) &= \frac{1}{2} \left\{\psi^*(\vec r, t) \frac{\hbar}{m i}\mathbf \nabla \psi(\vec r, t) - \psi(\vec r, t …