# WKB Approximation

Also called the Semiclassical Approximation for Quantum Mechanics.

$$S(\vec{x},t)=\pm\int^x dx' \sqrt{2m(E-V(x'))}-Et = W(x)-Et$$.

### Continuity of One-Dimension

$$\frac{\partial}{\partial t}\mathcal{P} + \frac{\partial}{\partial x}j = 0$$.

$$j=\frac{\mathcal{P}}{m}\frac{\partial S}{\partial x}$$.

#### Stationary Case

$$\frac{\partial \mathcal{P}}{\partial t} = 0$$.

So, $$\frac{\partial j}{\partial x} = 0$$. Hence, $$\frac{\partial}{\partial x}\left[\mathcal{P}\frac{\partial S}{\partial x}\right] = 0$$. So, $$\rho\frac{d W}{dx}$$ is a constant, $$\mathcal{C}$$. Then, $$\sqrt{\mathcal{P}} = \frac{\mathcal{C}}{\sqrt{E-V(x)}}$$, so $$\mathcal{P}=\frac{\mathcal{C}^2}{\sqrt{E-V(x)}}$$.

Thus, we are able to get a wavefunction without any solving: $$\Psi = \frac{\mathcal{C}}{\sqrt{E-V(x)}}\exp(\pm\frac{i}{\hbar}\int^xdx' \sqrt{2m(E-V(x'))}-\frac{i}{\hbar}Et)$$. Wentzel-Kramers-Brillouin solution. Thus, $$\mathcal{P}\sim\frac{1}{v_{classical}}$$.

## WKB Criteria

$$\hbar\left|\frac{d^2}{dx^2}W\right|\ll\left|\frac{dW}{dx}\right|^2$$. So, $$\hbar\frac{2m\left|\frac{dV}{dx}\right|}{2\sqrt{2m(E-V(x))}}\ll|2m(E-V(x))|$$. Hence, $$\frac{\hbar}{\sqrt{2m(E-V(x))}}=\frac{1}{k}=\frac{\lambda_B}{2\pi}\ll \frac{2(E-V(x))}{\left|\frac{dV}{dx}\right|}$$, where $$k$$ is the wavenumber, related to the momentum by $$\hbar$$. This gives us a deBrolie wavelength to compare.

So, WKB is valid in the low wavelength cases.

### Problem Areas

Turning points pose problem areas for WKB, but they can be solved as linear wavefunction pieces to splice the region solutions together.

### Small Example

Finite Well but with rounded edges, energy such that it is a bound state.

For regions I, III: $$\Psi=\frac{\mathcal{C}}{\sqrt{2m(V(x)-E)}}\exp(\pm \frac{1}{\hbar}\int^x \sqrt{2m(V(x')-E)})dx'$$

For regions II: $$\Psi=\frac{\mathcal{C}}{\sqrt{2m(E-V(x))}}\exp(\pm \frac{i}{\hbar}\int^x \sqrt{2m(E-V(x'))})dx'$$

For turning point solutions to splice together the regions: $$\Psi = Ax+B$$ $$-\frac{\hbar^2}{2m}\Psi''=(E-V(x))\Psi(x) \Rightarrow$$ Linear solutions.

Alternatively, $$-\frac{\hbar^2}{2m}\Psi''+\frac{dV}{dx}|_{x_{1,2}}(x-x_{1,2})\Psi=0$$. Gives a Bessel equation - Airy function.

## Energies

Rigid $$\Rightarrow$$ $$V\to\infty$$.

$$\int_{x_1}^{x_2}dx\sqrt{2m(E_n-V(x))} = \pi\hbar g_n$$.

### Case 1 - No Rigid Walls

$$g_n=n+1/2$$

### Case 2 - 1 Rigid Wall

$$g_n=n+3/4$$

### Case 3 - 2 Rigid Walls

$$g_n=n+1$$

### Notes

If $$n$$ is large, each $$g_n$$ is approximately equal.

Also, as $$n$$ increases, we get closer to classical typically.

### Examples

#### Energies of QMHO

$$E=\frac{m\omega^2 x_{1}^2}{2}$$. $$\int_{-x_1}^{x_1}dx\sqrt{2m(\frac{m\omega^2 x_1^2}{2}-\frac{m\omega^2x^2}{2})}=\frac{\pi E}{\omega}=\pi\hbar(n+1/2)$$. $$x_1=\pm\sqrt{\frac{2E}{m\omega^2}}$$. WKB: $$E=\hbar\omega(n+1/2)$$

#### Bouncing Neutrons

Hard floor. So, the potential is infinite in the negative region and a linear potential at $$x\gt 0$$.

#### Tunneling

Use WKB to find the wavefunction for a more complicated potential barrier, use typical solutions (if possible) for outside propogators.

$$T\approx \exp(-2\gamma), \gamma=\frac{1}{\hbar}\int_{x_1}^{x_2}\sqrt{2m(V(x)-E)}dx$$

Created: 2023-06-25 Sun 02:28

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