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Chapter 5: Problem 9

A free particle of energy \(E\) approaches a square, one-dimensional potential well of depth \(V_{0}\) and width \(2 a\). Show that the probability of being reflected by the well vanishes when \(K a=n \pi / 2\), where \(n\) is an integer and \(K=\left(2 m\left(E+V_{0}\right) / \hbar^{2}\right)^{1 / 2}\). Explain this phenomenon in physical terms.

Short Answer

Expert verified
Reflection vanishes when \( Ka = n\pi/2 \), ensuring complete transmission without reflection.

Step by step solution

01

Understanding the Scenario

We are given a potential well scenario where a free particle approaches a square one-dimensional potential well. The well has depth \( V_0 \) and width \( 2a \). The problem requires us to find conditions under which the probability of reflection is zero. We consider a particle approaching with energy \( E \), with \( K = \sqrt{\frac{2m(E + V_0)}{\hbar^2}} \).
02

Establish the Reflection Condition

A particle will be completely transmitted through the potential well without reflection if the wave function and its derivative are continuous. For no reflection, the phase condition must satisfy a certain criterion related to wave matching inside the well.
03

Condition for Vanishing Reflection

For reflection to vanish, the wave function must constructively interfere inside the well. Mathematically, this translates to the condition when the phase shift \( Ka \), which results from the well perturbation, is a multiple of \( \pi \). This is expressed as \( Ka = n\pi/2 \) where \( n \) is an integer.
04

Physical Explanation

The condition \( Ka = n\pi/2 \) ensures that the wave function inside the potential well matches perfectly, allowing for constructive interference and complete transmission. This happens because the integral number of wavelengths fit perfectly inside the well, leading to a standing wave that does not reflect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Reflection
The probability of reflection describes how likely it is for a wave, such as a particle in quantum mechanics, to be reflected back by a potential barrier rather than passing through or being absorbed. In a potential well, this is an essential concept as it helps us understand the behavior of particles as they encounter changes in potential energy.
For the potential well problem described, reflection can completely vanish under specific conditions. This occurs when there is perfect wave matching inside the well.
  • The wave function's phase shift, denoted by the term \( Ka \), plays a crucial role in determining this probability.
  • When \( Ka = n\pi/2 \), this means that an integral number of half-wavelengths fits the potential well exactly.
When this alignment happens, the reflections cancel each other out due to the interference pattern, leading to zero probability of reflection. This results in complete transmission through the potential well.
Wave Function Continuity
In quantum mechanics, the continuity of the wave function and its derivative across a potential well boundary is an essential condition for understanding particle behavior. The wave function must be continuous everywhere, and its derivative must also be continuous to ensure physical reality.
What's at stake here is the seamless progression of the particle's quantum wave through different potential regions.
  • For the particle encountering a potential well, certain parameters must be aligned to achieve full continuity.
  • It ensures that there is no abrupt change in the wave function, facilitating the transition with either transmission or reflection.
By satisfying the condition \( Ka = n\pi/2 \), the well becomes perfectly matched for continuity of the wave function and its derivative, allowing for uninterrupted passage, hence vanishing reflection.
Constructive Interference
Constructive interference is a phenomenon where two or more waves superpose to form a wave of greater amplitude. In the context of the potential well, constructive interference occurs when the waves inside the well align perfectly with the dimensions of the well, leading to amplified wave activity that promotes complete transmission without reflection.
Here's how it works in our scenario:
  • The key to constructive interference is setting the phase condition such that the waves inside the potential well reinforce each other.
  • Mathematically, this is achieved when \( Ka = n\pi/2 \).

When this condition holds, the standing wave inside the well resonates with the well dimensions, preventing any wave from bouncing back. Essentially, the well acts like a perfect trap for a standing wave, resulting in seamless passage through the well for the particle.

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Most popular questions from this chapter

Show that the energies of bound, odd-parity stationary states of the square potential well $$ V(x)= \begin{cases}0 & \text { for }|x|0 & \text { otherwise }\end{cases} $$ are governed by \(\cot (k a)=-\sqrt{\frac{W^{2}}{(k a)^{2}}-1}\) where \(\quad W \equiv \sqrt{\frac{2 m V_{0} a^{2}}{\hbar^{2}}}\) and \(k^{2}=2 m E / \hbar^{2}\) \((5.77)\) Show that for a bound odd-parity state to exist, we require \(W>\pi / 2\)

We consider the scattering of free particles of mass \(m\) that move in one dimension in the potential \(V(x)=-W \delta(x)\), with \(W>0\). (a) For a well of finite depth \(V_{0}\) and width \(2 a\) the condition on the phases \(\phi\) and \(\phi^{\prime}\) of the even- and odd-parity wavefunctions \(\psi \propto \sin (k x+\phi)\), etc., for free particles are $$ \tan (k a+\phi)=-\frac{k}{K} \cot (K a), \quad \tan \left(k a+\phi^{\prime}\right)=-\frac{k}{K} \tan (K a) $$ Show that in the limit \(a \rightarrow 0, V_{0}=W / 2 a \rightarrow \infty\) we have \(\tan \phi \rightarrow\) \(-\hbar^{2} k / m W\) and \(\phi^{\prime} \rightarrow 0\). Hence obtain the scattering cross-section by the \(\delta\)-function potential $$ \sigma=\frac{2}{1+(\hbar k / m W)^{2}} $$ (b) Re-derive the equation above for \(\phi\) by requiring that \(\psi=\sin (k|x|+\phi)\) satisfy the TISE. Convince yourself that \(\psi=\sin (k x)\) is also consistent with the TISE.

Show that the correctly normalised wavefunction of a particle trapped by the potential \(V(x)=-V_{\delta} \delta(x)\) is \(\psi(x)=\sqrt{K} e^{-K|x|}\), where \(K=\) \(m V_{\delta} / \hbar^{2}\). Show that although this wavefunction makes it certain that a measurement of \(x\) will find the particle outside the well where its kinetic energy is negative, the expectation value of its kinetic energy \(\left\langle E_{K}\right\rangle=\frac{1}{2} m V_{\delta}^{2} / \hbar^{2}\) is in fact positive. Reconcile this apparent paradox as follows: (i) Show that for a narrow, deep potential well of depth \(V_{0}\) and half- width \(a\), with \(2 V_{0} a=V_{\delta}, k a \simeq W \equiv\left(2 m V_{0} a^{2} / \hbar^{2}\right)^{1 / 2}\), while \(K a \simeq W^{2}\). (ii) Hence show that the contribution from inside the well to \(\left\langle E_{K}\right\rangle\) is \(|\psi(0)|^{2} V_{\delta}\) regardless of the value of \(a .\) Explain physically what is happening as we send \(a \rightarrow 0\).

Give an example of a potential in which there is a complete set of bound stationary states of well-defined parity, and an alternative complete set of bound stationary states that are not eigenkets of the parity operator. Hint: modify the potential discussed apropos \(\mathrm{NH}_{3}\).

An electron moves along an infinite chain of potential wells. For sufficiently low energies we can assume that the set \(\\{|n\rangle\\}\) is complete, where \(|n\rangle\) is the state of definitely being in the \(n^{\text {th }}\) well. By analogy with our analysis of the \(\mathrm{NH}_{3}\) molecule we assume that for all \(n\) the only non-vanishing matrix elements of the Hamiltonian are \(\mathcal{E} \equiv\langle n|H| n\rangle\) and \(A \equiv\langle n \pm 1|H| n\rangle\). Give physical interpretations of the numbers \(A\) and \(\mathcal{E}\). Explain why we can write $$ H=\sum_{n=-\infty}^{\infty} \mathcal{E}|n\rangle\langle n|+A(|n\rangle\langle n+1|+| n+1\rangle\langle n|) $$ Writing an energy eigenket \(|E\rangle=\sum_{n} a_{n}|n\rangle\) show that $$ a_{m}(E-\mathcal{E})-A\left(a_{m+1}+a_{m-1}\right)=0 $$ Obtain solutions of these equations in which \(a_{m} \propto \mathrm{e}^{\mathrm{i} k m}\) and thus find the corresponding energies \(E_{k}\). Why is there an upper limit on the values of \(k\) that need be considered? Initially the electron is in the state $$ |\psi\rangle=\frac{1}{\sqrt{2}}\left(\left|E_{k}\right\rangle+\left|E_{k+\Delta}\right\rangle\right) $$ where \(0
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