Question
Evaluate quantitatively the standing wave solutions for the square well potential of section 4 in the case $E>V_0$. Use these solutions to verify the qualitative conclusions reached at the end of that section.
Step 1
The square well potential is defined as: \[ V(x) = \begin{cases} 0 & \text{for } |x| < a \\ V_0 & \text{for } |x| \geq a \end{cases} \] where \( V_0 \) is the potential outside the well and \( a \) is the width of the well. Step 2: Set up the Schrödinger Show more…
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-?? (Section 7.8) Consider the problem of a particle in finite square well of depth $U_{0}$, as described in Section 7.8. To solve this problem, use a coordinate system centered on the box, with the left edge of the well at $x=-a / 2$ and the right edge at $x=+a / 2 .$ From the symmetry of the potential, one can argue that the stationary states should have symmetric probability distributions, that is, $|\psi(x)|^{2}=|\psi(-x)|^{2}$. This implies that solutions are either symmetric and satisfy $\psi(x)=\psi(-x)$ or are antisymmetric and satisfy $\psi(x)=-\psi(-x)$. Thus, with our choice of coordinates, the solutions within the well are either of the form $\cos (k x)$ (for $n=1,3,5, \ldots)$ or of the form $\sin (k x)$ (for $n=2,4,6, \ldots)$. (a) For the case of the symmetric solutions, derive an equation relating $k$ and $\alpha$. [Hint: Using the boundary conditions that both $\psi(x)$ and $\psi^{\prime}(x)$ are continuous at $x=-a / 2$, you can produce two equations that relate $k, \alpha$, and the arbitrary coefficients $A$ and $G$ in Section 7.8. By dividing these equations, you can eliminate the coefficients. The final equation you produce is a transcendental equation relating $E$ and $U_{0}$; it cannot be solved for $E$ using elementary methods.] (b) Show that the equation derived in (a) produces the correct stationary-state energies in the limit $U_{0} \rightarrow \infty$. (c) Using numerical techniques, find the ground-state energy $E_{1}$ of a square well of depth $U_{0}=3 E_{1} .[$ Hint $:$ A simple trialand-error search for the solution works surprisingly well.]
The Schrödinger Equation in One Dimension
The Time-Dependent Schrödinger Equation
Consider the case of a symmetrical double well, such as the one $*$ pictured in Figure $8.13 .$ We are interested in bound states with $E<V(0)$. (a) Write down the WKB wave functions in regions (i) $x>x_{2}$, (ii) $x_{1}<x<x_{2}$, and (iii) $0<x<x_{1}$. Impose the appropriate connection formulas at $x_{1}$ and $x_{2}$ (this has already been done, in Equation $8.46$, for $x_{2} ;$ you will have to work out $x_{1}$ for yourself), to show that $$ \psi(x) \cong\left\{\begin{array}{l} \frac{D}{\sqrt{|p(x)|}} \exp \left[-\frac{1}{\hbar} \int_{x_{2}}^{x}\left|p\left(x^{\prime}\right)\right| d x^{\prime}\right] \\ \frac{2 D}{\sqrt{p}(x)} \sin \left[\frac{1}{\hbar} \int_{x}^{x_{2}} p\left(x^{\prime}\right) d x^{\prime}+\frac{\pi}{4}\right] \\ \frac{D}{\sqrt{|p(x)|}}\left[2 \cos \theta e^{\frac{1}{\hbar} \int_{v}^{r_{1}}\left|p\left(x^{\prime}\right)\right| d x^{\prime}}+\sin \theta e^{\left.-\frac{1}{\hbar} \int_{v}^{x_{1}}\left|p\left(x^{\prime}\right)\right| d x^{\prime}\right]}\right. \end{array}\right. $$ where $$ \theta \equiv \frac{1}{\hbar} \int_{x_{1}}^{x_{2}} p(x) d x $$ (b) Because $V(x)$ is symmetric, we need only consider even $(+)$ and odd $(-)$ wave functions. In the former case $\psi^{\prime}(0)=0$, and in the latter case $\psi(0)=0 .$ Show that this leads to the following quantization condition: $$ \tan \theta=\pm 2 e^{\phi} $$ where $$ \phi \equiv \frac{1}{h} \int_{-x_{1}}^{x_{1}}\left|p\left(x^{\prime}\right)\right| d x^{\prime} $$ Equation $8.59$ determines the (approximate) allowed energies (note that $E$ comes into $x_{1}$ and $x_{2}$, so $\theta$ and $\phi$ are both functions of $E$ ). (c) We are particularly interested in a high and/or broad central barrier, in which case $\phi$ is large, and $e^{\phi}$ is huge. Equation $8.59$ then tells us that $\theta$ must be very close to a half-integer multiple of $\pi .$ With this in mind, write $\theta=$ $(n+1 / 2) \pi+\epsilon$, where $|\epsilon| \ll 1$, and show that the quantization condition becomes $$ \theta \cong\left(n+\frac{1}{2}\right) \pi \mp \frac{1}{2} e^{-\phi} $$ (d) Suppose each well is a parabola: ${ }^{16}$ $$ V(x)=\left\{\begin{array}{ll} \frac{1}{2} m \omega^{2}(x+a)^{2}, & \text { if } x<0 \\ \frac{1}{2} m \omega^{2}(x-a)^{2} . & \text { if } x>0 \end{array}\right. $$
The WKB Approximation
The Connection Formulas
- (a) Derive an approximate expression for the number of bound states in a finite square well of depth $U_{0}$. To do this, make the assumption that the energy of the highest state of the finite well is close to the corresponding energy of the infinite well. This state is the one that has an energy $E$ near the top of the well, that is, $E=U_{0}$. (b) Compute the approximate number of bound states for the case of an electron in a well of depth $U_{0}=10 \mathrm{eV}$, and width $a=0.3 \mathrm{~nm}$. These numbers crudely approximate valence electrons in a solid.
The Nonrigid Box
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