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Chapter 5: Problem 5

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 \(W>\pi / 2\)

Short Answer

Expert verified
Odd-parity bound states require \(W > \pi/2\).

Step by step solution

01

Define the square potential well

The potential well is defined by the function \(V(x)\) such that \(V(x) = 0\) for \(|x| < a\) and \(V(x) = V_0\) for \(|x| \ge a\). Bound states correspond to energies \(E < V_0\). The particle is free inside the well \(|x| < a\) and must satisfy Schr枚dinger's equation.
02

Write Schr枚dinger's equation solutions inside and outside the well

Inside the well \(|x| < a\), where \(V(x) = 0\), the wave function \(\psi(x)\) satisfies the free particle equation: \(-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi\). This gives the form \(\psi(x) = A \sin(kx) + B \cos(kx)\) with \(k^2 = \frac{2mE}{\hbar^2}\). Outside the well \(|x| \ge a\), the solution is evanescent: \(\psi(x) = C e^{-\kappa x}\) or \(\psi(x) = D e^{\kappa x}\) with \(\kappa^2 = \frac{2m(V_0 - E)}{\hbar^2}\).
03

Apply boundary and odd-parity conditions

For odd-parity states, \(\psi(x) = -\psi(-x)\), leading to \(\psi(0) = 0\). This implies that \(B = 0\), giving \(\psi(x) = A \sin(kx)\). At \(x = a\), the continuity of \(\psi(x)\) and \(\psi'(x)\) across the boundary requires \(A \sin(ka) = C e^{-\kappa a} \) and \(kA \cos(ka) = -\kappa C e^{-\kappa a}\). Dividing these gives \(\cot(ka) = -\frac{\kappa}{k}\).
04

Relate parameters to the given transcendental equation

Substituting \(\kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}\) and \(k = \sqrt{\frac{2mE}{\hbar^2}}\) into the continuity condition \(\cot(ka) = -\frac{\kappa}{k}\), manipulate using \(W = \sqrt{\frac{2mV_0a^2}{\hbar^2}}\) until you achieve \(\cot(ka) = -\sqrt{\frac{W^2}{(ka)^2} - 1}\).
05

Show that odd-parity states require \(W > \pi/2\)

To find the condition for an odd parity bound state to exist, consider the behavior of \(\cot(x)\): it is undefined at integer multiples of \(\pi\) and must match \(-\sqrt{\frac{W^2}{x^2} - 1}\), which itself is real and negative. At \(ka = \pi/2\), \(\cot(x)\) is zero, hence \(-\sqrt{\frac{W^2}{(\pi/2)^2} - 1}\) must also be zero, implying \(W = \pi/2\) is a cutoff.

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Key Concepts

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

square potential well
A square potential well is a foundational concept in quantum mechanics, often used to simplify complex systems into a more manageable form. It describes a system where a particle is confined to a region, a well, of finite width, with zero potential energy inside and a constant positive potential outside. This model helps illustrate quantum principles like wave functions and quantized energy levels. Zero potential energy within the well means the particle behaves like a free particle, leading to solutions of Schr枚dinger's equation that resemble unrestricted waves. Outside the well, the potential energy boost means the wave function decays rapidly, as the particle is less likely to be found in these regions. Square potential wells are useful for understanding bound states and quantum confinement, showing how particles are limited in their energy states and available space.
bound state energies
In quantum mechanics, bound state energies refer to the discrete energy levels that a quantum particle can occupy within a potential well. For a finite square potential well, these energies are less than the potential barrier height, ensuring the particle remains trapped in the well. Bound states exhibit quantization due to the wave nature of particles; only specific energies allow the wave function to match boundary conditions and decay properly outside the well. The presence of these quantized energies is a key departure from classical mechanics, where particles could possess any energy. Understanding bound states is crucial for exploring how particles behave under constraints, especially in atomic and molecular systems where electrons exhibit similar quantized behavior.
odd-parity stationary states
Odd-parity stationary states in a quantum system like a square potential well are configurations where the wave function changes sign when the spatial coordinates are inverted. Mathematically, this means \[ \psi(x) = -\psi(-x) \] Odd parity implies symmetry properties that simplify the analysis of wave functions and the determination of energy states. In the square well scenario, odd-parity solutions ensure that the wave function is zero at the center of the well. This results from boundary conditions demanding that the probability of finding the particle split evenly about the center. Identifying odd-parity states helps in categorizing energy levels and understanding quantum systems with symmetrical potentials.
Schr枚dinger equation solutions
The Schr枚dinger equation is central to quantum mechanics, defining how quantum systems evolve over time. Solving it for a system like a square potential well reveals the allowed wave functions and corresponding energy levels. Inside the well, where potential energy is zero, the Schr枚dinger equation resembles that of a free particle, leading to sine and cosine solutions. Outside the well, the exponentially decaying solutions reflect the decreasing probability of finding the particle outside the region of confinement. Applying both boundary conditions and specific properties like parity, we determine the discrete energy levels or quantized states numerically or analytically. By studying these solutions, we can predict behaviors in more complex quantum systems, as the square potential model serves as a foundation for understanding fundamental quantum phenomena.

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Most popular questions from this chapter

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

A particle is confined by the potential well $$ V(x)= \begin{cases}0 & \text { for }|x|

\({ }^U\) decays by \(\alpha\) emission with a mean lifetime of \(6.4\) Gyr. Take the nucleus to have a diameter \(\sim 10^{-14} \mathrm{~m}\) and suppose that the \(\alpha\) particle has been bouncing around within it at speed \(\sim c / 3\). Modelling the potential barrier that confines the \(\alpha\) particle to be a square one of height \(V_{0}\) and width \(2 a\), give an order-of-magnitude estimate of \(W=\left(2 m V_{0} a^{2} / \hbar^{2}\right)^{1 / 2}\) Given that the energy released by the decay is \(\sim 4 \mathrm{MeV}\) and the atomic number of uranium is \(Z=92\), estimate the width of the barrier through which the \(\alpha\) particle has to tunnel. Hence give a very rough estimate of the barrier's typical height. Outline numerical work that would lead to an improved estimate of the structure of the barrier.

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.

This problem is about the coupling of ammonia molecules to electromagnetic waves in an ammonia maser. Let \(|+\rangle\) be the state in which the \(\mathrm{N}\) atom lies above the plane of the \(\mathrm{H}\) atoms and \(|-\rangle\) be the state in which the \(\mathrm{N}\) lies below the plane. Then when there is an oscillating electric field \(\mathcal{E} \cos \omega t\) directed perpendicular to the plane of the hydrogen atoms, the Hamiltonian in the \(|\pm\rangle\) basis becomes $$ H=\left(\begin{array}{cc} \bar{E}+q \mathcal{E} s \cos \omega t & -A \\ -A & \bar{E}-q \mathcal{E} s \cos \omega t \end{array}\right) $$ Transform this Hamiltonian from the \(|\pm\rangle\) basis to the basis provided by the states of well-defined parity \(|\mathrm{e}\rangle\) and \(|\mathrm{o}\rangle\) (where \(|\mathrm{e}\rangle=(|+\rangle+|-\rangle) / \sqrt{2}\), etc). Writing $$ |\psi\rangle=a_{\mathrm{e}}(t) \mathrm{e}^{-\mathrm{i} E_{e} t / \hbar}|\mathrm{e}\rangle+a_{\mathrm{o}}(t) \mathrm{e}^{-\mathrm{i} E_{0} t / \hbar}|\mathrm{o}\rangle $$ show that the equations of motion of the expansion coefficients are $$ \begin{aligned} &\frac{\mathrm{d} a_{\mathrm{e}}}{\mathrm{d} t}=-\mathrm{i} \Omega a_{\mathrm{o}}(t)\left(\mathrm{e}^{\mathrm{i}\left(\omega-\omega_{0}\right) t}+\mathrm{e}^{-\mathrm{i}\left(\omega+\omega_{0}\right) t}\right) \\ &\frac{\mathrm{d} a_{\mathrm{o}}}{\mathrm{d} t}=-\mathrm{i} \Omega a_{\mathrm{e}}(t)\left(\mathrm{e}^{\mathrm{i}\left(\omega+\omega_{0}\right) t}+\mathrm{e}^{-\mathrm{i}\left(\omega-\omega_{0}\right) t}\right) \end{aligned} $$ where \(\Omega \equiv q \mathcal{E} s / 2 \hbar\) and \(\omega_{0}=\left(E_{o}-E_{e}\right) / \hbar .\) Explain why in the case of a maser the exponentials involving \(\omega+\omega_{0}\) can be neglected so the equations of motion become $$ \frac{\mathrm{d} a_{\mathrm{e}}}{\mathrm{d} t}=-\mathrm{i} \Omega a_{\circ}(t) \mathrm{e}^{\mathrm{i}\left(\omega-\omega_{0}\right) t}, \quad \frac{\mathrm{d} a_{\mathrm{o}}}{\mathrm{d} t}=-\mathrm{i} \Omega a_{\mathrm{e}}(t) \mathrm{e}^{-\mathrm{i}\left(\omega-\omega_{0}\right) t} $$ Solve the equations by multiplying the first equation by \(\mathrm{e}^{-\mathrm{i}\left(\omega-\omega_{0}\right) t}\) and differentiating the result. Explain how the solution describes the decay of a population of molecules that are initially all in the higher energy level. Compare your solution to the result of setting \(\omega=\omega_{0}\) in \((5.86)\).
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