Question 705052

SCQHARD

A particle of energy $E$ is scattered off a one-dimensional potential $\lambda\delta(x)$ , where $\lambda$ is a real positive constant, with a transmission amplitude $t_+$ . In a different experiment, the same particle is scattered off another one-dimensional potential $-\lambda\delta(x)$ , with a transmission amplitude $t_-$ . In the limit $E \to 0$ , the phase difference between $t_+$ and $t_-$ is

(A)

$\pi/2$

(B)

$\pi$

(C)

$0$

(D)

$3\pi/2$

Detailed Solution

For potential $\lambda\delta(x)$ , $t_+ = \frac{1}{1 + \frac{m\lambda}{i\hbar^2k}}$ . For $-\lambda\delta(x)$ , $t_- = \frac{1}{1 - \frac{m\lambda}{i\hbar^2k}}$ . In the limit $k \to 0$ ( $E \to 0$ ), the ratio $\frac{t_+}{t_-} \to -1 = e^{i\pi}$ . Thus the phase difference is $\pi$ .

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