A particle of mass m is confined to a one-dimensional square...

A particle of mass $m$ is confined to a one-dimensional square well with finite walls. That is, a potential $V(x)=0$ for $-a \leq x \leq+a$, and $V(x)=V_{0}=\eta\left(\hbar^{2} / 2 m a^{2}\right)$ otherwise. You are to find the bound-state energy eigenvalues as $E=\varepsilon V_{0}$ along with their wave functions. a. Set the problem up with a wave function $A e^{\alpha x}$ for $x \leq-a, D e^{-a x}$ for $x \geq+a$, and $B e^{i k x}+C e^{-i k x}$ inside the well. Match the boundary conditions at $x=\pm a$ and show that $k$ and $\alpha$ must satisfy $z=\pm z^{*}$ where $z \equiv e^{\text {iak }}(k-i \alpha) .$ Proceed to find a purely real or purely imaginary expression for $z$ in terms of $k$ and $\alpha$. b. Find the wave functions for the two choices of $z$ and show that the purely real (imaginary) choice leads to a wave function that is even (odd) under the exchange $x \rightarrow-x$. c. Find a transcendental equation for each of the two wave functions relating $\eta$ and $\varepsilon$. Show that even a very shallow well $(\eta \rightarrow 0)$ has at least one solution for the even wave function, but you are not guaranteed any solution for an odd wave function. d. For $\eta=10$, find all the energy eigenvalues and plot their normalized wave functions.

A particle of mass $m$ is confined to a one-dimensional square well with finite walls. That is, a potential $V(x)=0$ for $-a \leq x \leq+a$, and $V(x)=V_{0}=\eta\left(\hbar^{2} / 2 m a^{2}\right)$ otherwise. You are to find the bound-state energy eigenvalues as $E=\varepsilon V_{0}$ along with their wave functions.
a. Set the problem up with a wave function $A e^{\alpha x}$ for $x \leq-a, D e^{-a x}$ for $x \geq+a$, and $B e^{i k x}+C e^{-i k x}$ inside the well. Match the boundary conditions at $x=\pm a$ and show that $k$ and $\alpha$ must satisfy $z=\pm z^{*}$ where $z \equiv e^{\text {iak }}(k-i \alpha) .$ Proceed to find a purely real or purely imaginary expression for $z$ in terms of $k$ and $\alpha$.
b. Find the wave functions for the two choices of $z$ and show that the purely real (imaginary) choice leads to a wave function that is even (odd) under the exchange $x \rightarrow-x$.
c. Find a transcendental equation for each of the two wave functions relating $\eta$ and $\varepsilon$. Show that even a very shallow well $(\eta \rightarrow 0)$ has at least one solution for the even wave function, but you are not guaranteed any solution for an odd wave function.
d. For $\eta=10$, find all the energy eigenvalues and plot their normalized wave functions.

Key Concepts

Boundary Condition Matching

In quantum mechanics, the continuity of the wavefunction and its derivative at the boundaries of different potential regions is essential. For the finite potential well, these conditions at x = ±a ensure that the solutions in the interior and exterior regions are seamlessly connected. This matching process yields equations that determine the allowed energy eigenvalues and the constants in the wavefunction.

Transcendental Equations in Quantum Mechanics

The matching of boundary conditions in the finite potential well typically leads to transcendental equations that relate the parameters of the wavefunction (such as the wave number k and decay constant ?) with the energy eigenvalues. These equations generally cannot be solved using simple algebraic techniques and often require numerical methods to determine the allowed energy levels. They illustrate how quantization arises in systems with nontrivial potential barriers.

Finite Potential Well

This concept refers to a quantum system in which a particle is confined to a region in space by potential barriers that are finite in height. Unlike the infinite well, the finite potential well allows the possibility of quantum tunneling, meaning that even though the particle is bound within the well, its wavefunction penetrates into the classically forbidden region outside. This model is fundamental in studying confined quantum systems and the discretization of energy levels.

Schrödinger Equation for Bound States

The problem relies on solving the time-independent Schrödinger equation for bound states, where the energy of the particle is less than the potential barrier height. This involves obtaining solutions in different regions of the well—oscillatory inside the well (where the potential is zero) and exponentially decaying outside (where the potential is finite)—and applying the respective differential equations for these regions.

Parity of Wave Functions

The symmetry of the potential in a one-dimensional finite well allows the classification of the eigenfunctions into even and odd parity solutions. Even parity (symmetric) solutions are represented by wavefunctions that mirror themselves about the origin, while odd parity (antisymmetric) solutions change sign upon reflection. This classification simplifies the analysis and leads to separate transcendental relationships for each case.

Transcript

00:01 Okay, so one solution of this short -danger's time -independent equation can be written as this is z -zi and this is equal to x into x y of y and z of z.

00:23 And substituting this expression into the short injures equation and dividing by this on both side of the equation we have that one over x into x partial square x into x over partial x square plus one over y into y times partial square y into y divided by partial y square plus 1 over z into z times partial square z times z over partial z square it is equal to minus 8 pi square m over h squared times e right so this yields the partial square x over x divided by partial x square plus e sub x x into x is equal to zero and this is equal to this is equal to partial square y into y divided by partial y square plus e y y y into y this is equal to 0 and partial square z times z over partial with respect to z square plus e z in z into z is equal to zero right and we are we are x e y and e z are constants and this satisfies that e x plus e y plus ez is equal to 8 pi square m over 8 square times e right and by solving the above three equations we have that x or in terms of x is equal to cx e power a aorta square root e e x time x right and and plus d x e power minus aorta square root e x time x right and y of y is equal to c sub y power iota square root e sub y times y plus d sub y power e power a iota square root e.

03:19 Of y times y and this is z into z is equal to c z e power iota square root e z times z plus d z e power minus iota square root e z times z right and finally this finally putting the values we have that this is equal to c x e power iota square root e sub x times x plus d sub x e power minus iota x x times x and this is c y e power iota square root ey times y times y times x right and plus okay so this is at times c yota square root ey times y plus d, y, e power, minus iota square root, ey times y.

04:47 So we are just putting the values here, right? and here this is times cz, e power, iota square root, ez times z plus dze power minus iota square root, ez, e power, minus iota square root, e z times z right and here uh here uh these uh x and d x c y d y and c y and c y and c d g are constants right these are constants okay and uh by boundary conditions this i is equal to zero at x is equal to 0 and it gives c sub x is equal to minus d sub x and by the boundary condition si is equal to uh zero right and y is equal to zero it goes c y is equal to minus d sub y and by boundary conditions si is equal to zero at uh z is equal to zero so it gives us c z is equal to minus d z so we have that this is equal to this is equal to cx c y c z and this is e power iota square root x power times x minus e power minus aota square root uh e y this is e x right so this is e x times x and this is e power aota square root e sub y times y minus e y power minus iota square root e sub y times y right and this is times e power iota square root uh e power z times z minus e power minus iota square root uh e sub z time z right and finally this is equal to uh this is equal to finally this is equal to a sine square root x x x x times x so this is x right sign up square root e sub y times y and sign up square root e sub z times z right and using the above expression this expression for the boundary condition which is this is equal to 0 at x is equal to l x it gives x x is equal to l pi over l x whole square right and here here by the boundary conditions the above boundary conditions which is x is equal to l z and this gives us e z is equal to n pi over l sub z whole square right and we are l r n is equal to 0 1 2 and up to so on so we have l pi over l x whole square plus r pi over l sub y whole square plus n pi over l sub y whole square plus n pi over l sub z whole square this is equal to 8 pi square times m over it square times e so e is equal to h square over eight times m into l square or l sub x square plus r square over l sub y square plus n square over l sub z square and this is equal to so when let when l x is equal to l y and this is equal to l y and this is equal to l z and this is equal to l.

09:23 So e will be equal to each square over 8 times l square and this is l square plus r square plus n square and if e is equal to e0 par l is equal to 1, r is equal to n is equal to 0...

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