Solid State Physics: Essential Concepts 1st Edition by David Snoke

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Solid State Physics: Essential Concepts 1st Edition by David Snoke

النسخة 1الرقم المعياري الدولي: 978-0805386646

Solid State Physics: Essential Concepts 1st Edition by David Snoke

النسخة 1الرقم المعياري الدولي: 978-0805386646

تمرين 1

التوضيح

موثّق

According to the prescription of the text, we assume that the solutions in the two regions take the form

\Psi_{1}(x)=A_{1} \sin (k x)+B_{1} \cos (k x),

and

\Psi_{2}(x)=A_{2} e^{\kappa x}+B_{2} e^{-\kappa x} .

There is no need to write down the solution for

\Psi_{3}(x)

, since the complete solution is either symmetric, in which case the derivative of

\Psi_{2}(x)

vanishes at

x=0

, or antisymmetric, therefore,

\Psi_{2}(x)

itself vanishes at

x=0

.
The boundary conditions of the problem are

\begin{gathered}\Psi_{1}\left(x=-a-\frac{b}{2}\right)=0 \longrightarrow A_{1} \sin \left(k\left(-a-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-a-\frac{b}{2}\right)\right)=0 \\\Psi_{1}\left(x=-\frac{b}{2}\right)=\Psi_{2}\left(x=-\frac{b}{2}\right) \longrightarrow A_{1} \sin \left(k\left(-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-\frac{b}{2}\right)\right)=A_{2} e^{-\kappa \frac{b}{2}}+B_{2} e^{\kappa \frac{b}{2}} \\\left.\frac{\Psi_{1}}{d x}\right|_{x=-\frac{b}{2}}=\left.\frac{\Psi_{2}}{d x}\right|_{x=-\frac{b}{2}} \longrightarrow A_{1} k \cos \left(k\left(-\frac{b}{2}\right)\right)-B_{1} k \sin \left(k\left(-\frac{b}{2}\right)\right)=A_{2} \kappa e^{\kappa \frac{b}{2}}-B_{2} \kappa e^{-\kappa \frac{b}{2}},\end{gathered}

and finally,

\Psi_{2}(x=0)=0 \longrightarrow A_{2}+B_{2}=0

i.e.,

A_{2}=-B_{2}

for the antisymmetric case, or

\left.\frac{\Psi_{2}}{d x}\right|_{x=0}=0 \longrightarrow A_{2} \kappa-B_{2} \kappa=0

i.e.,

A_{2}=B_{2}

for the symmetric case. Regarding the amplitudes

A_{1}, B_{1}, A_{2}

and

B_{2}

as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution,

the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation

\kappa \sin a k \cosh \frac{\kappa b}{2}+k \cos a k \sinh \frac{\kappa b}{2}=0

for the antisymmetric case, and

\kappa \sin a k \sinh \frac{\kappa b}{2}+k \cos a k \cosh \frac{\kappa b}{2}=0

for the symmetric one. Dividing by

\cos a k

and

\cosh \frac{b \kappa}{2}

, or

\sinh \frac{b \kappa}{2}

, respectively, we get

\begin{aligned}\frac{\tan a k}{k} & =-\frac{\tanh \frac{b \kappa}{2}}{\kappa} \\\frac{\tan a k}{k} & =-\frac{\operatorname{coth} \frac{b \kappa}{2}}{\kappa}\end{aligned}

for the antisymmetric and symmetric case, respectively.
We have an additional equation, which links

k

and

\kappa

. Namely,

\begin{aligned}\frac{\hbar^{2} k^{2}}{2 m} & =E \\\frac{\hbar^{2} \kappa^{2}}{2 m} & =V_{0}-E,\end{aligned}

i.e.,

\kappa=\sqrt{\frac{2 m V_{0}}{\hbar^{2}}-k^{2}}

Then the two equations above, Eqs. (10-11), lead to two equations for

k

, where

a, b

and

2 m V_{0} / \hbar^{2}

play the role of parameters. Let us note that in Eq. (14),

\kappa

becomes imaginary, when

\hbar^{2} k^{2} / 2 m>V_{0}

. This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle's energy is larger than the "confining" potential, i.e., the particle is not bound in that region.
The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1 , where

a=1,2 m V_{0} / \hbar^{2}=100

, and

b=0.1

, or

b=0.02

. We plotted only

k>0

, since the equations are invariant under the transformation

k \leftrightarrow-k

. By trying various values for

b

, we notice that as we increase

b

, the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when

b \rightarrow 0

, the right hand side of Eq. (11) tends to

-\infty

, which means that the first solution will be at

k=\pi / 2

. At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of

\pi

. This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more.
$According to the prescription of the text, we assume that the solutions in the two regions take the form \Psi_{1}(x)=A_{1} \sin (k x)+B_{1} \cos (k x), and \Psi_{2}(x)=A_{2} e^{\kappa x}+B_{2} e^{-\kappa x} . There is no need to write down the solution for \Psi_{3}(x) , since the complete solution is either symmetric, in which case the derivative of \Psi_{2}(x) vanishes at x=0 , or antisymmetric, therefore, \Psi_{2}(x) itself vanishes at x=0 . The boundary conditions of the problem are \begin{gathered} \Psi_{1}\left(x=-a-\frac{b}{2}\right)=0 \longrightarrow A_{1} \sin \left(k\left(-a-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-a-\frac{b}{2}\right)\right)=0 \\ \Psi_{1}\left(x=-\frac{b}{2}\right)=\Psi_{2}\left(x=-\frac{b}{2}\right) \longrightarrow A_{1} \sin \left(k\left(-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-\frac{b}{2}\right)\right)=A_{2} e^{-\kappa \frac{b}{2}}+B_{2} e^{\kappa \frac{b}{2}} \\ \left.\frac{\Psi_{1}}{d x}\right|_{x=-\frac{b}{2}}=\left.\frac{\Psi_{2}}{d x}\right|_{x=-\frac{b}{2}} \longrightarrow A_{1} k \cos \left(k\left(-\frac{b}{2}\right)\right)-B_{1} k \sin \left(k\left(-\frac{b}{2}\right)\right)=A_{2} \kappa e^{\kappa \frac{b}{2}}-B_{2} \kappa e^{-\kappa \frac{b}{2}}, \end{gathered} and finally, \Psi_{2}(x=0)=0 \longrightarrow A_{2}+B_{2}=0 i.e., A_{2}=-B_{2} for the antisymmetric case, or \left.\frac{\Psi_{2}}{d x}\right|_{x=0}=0 \longrightarrow A_{2} \kappa-B_{2} \kappa=0 i.e., A_{2}=B_{2} for the symmetric case. Regarding the amplitudes A_{1}, B_{1}, A_{2} and B_{2} as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution, the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation \kappa \sin a k \cosh \frac{\kappa b}{2}+k \cos a k \sinh \frac{\kappa b}{2}=0 for the antisymmetric case, and \kappa \sin a k \sinh \frac{\kappa b}{2}+k \cos a k \cosh \frac{\kappa b}{2}=0 for the symmetric one. Dividing by \cos a k and \cosh \frac{b \kappa}{2} , or \sinh \frac{b \kappa}{2} , respectively, we get \begin{aligned} \frac{\tan a k}{k} & =-\frac{\tanh \frac{b \kappa}{2}}{\kappa} \\ \frac{\tan a k}{k} & =-\frac{\operatorname{coth} \frac{b \kappa}{2}}{\kappa} \end{aligned} for the antisymmetric and symmetric case, respectively. We have an additional equation, which links k and \kappa . Namely, \begin{aligned} \frac{\hbar^{2} k^{2}}{2 m} & =E \\ \frac{\hbar^{2} \kappa^{2}}{2 m} & =V_{0}-E, \end{aligned} i.e., \kappa=\sqrt{\frac{2 m V_{0}}{\hbar^{2}}-k^{2}} Then the two equations above, Eqs. (10-11), lead to two equations for k , where a, b and 2 m V_{0} / \hbar^{2} play the role of parameters. Let us note that in Eq. (14), \kappa becomes imaginary, when \hbar^{2} k^{2} / 2 m>V_{0} . This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle's energy is larger than the confining potential, i.e., the particle is not bound in that region. The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1 , where a=1,2 m V_{0} / \hbar^{2}=100 , and b=0.1 , or b=0.02 . We plotted only k>0 , since the equations are invariant under the transformation k \leftrightarrow-k . By trying various values for b , we notice that as we increase b , the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when b \rightarrow 0 , the right hand side of Eq. (11) tends to -\infty , which means that the first solution will be at k=\pi / 2 . At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of \pi . This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more. Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for a=1, b=0.1 , and 2 m V_{0} / \hbar=100 on the left hand side, and for a=1, b=0.02 , and 2 m V_{0} / \hbar^{2}=100 on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for b=0.1 on the left hand side. Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width, b . We see from Fig. 1.1 that both of these roots are in the interval [\pi / 2, \pi] , and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for b \rightarrow 0 , we indeed have a factor of 2 in the values of the wave number, while for b \rightarrow \infty , the two energies will virtually be the same. This behavior can be understood, if we notice that as b \rightarrow \infty , the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled. Once we have the value of k , we can solve for A_{1}, B_{1}, A_{2} and B_{2} , which give the wavefunctions. Two typical solutions are shown in Fig. 1.3 , for a=1,2 m V_{0} / \hbar=100 , and b=0.1 , or b=0.3 . Mathematica source codes$
Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for

a=1, b=0.1

, and

2 m V_{0} / \hbar=100

on the left hand side, and for

a=1, b=0.02

, and

2 m V_{0} / \hbar^{2}=100

on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for

b=0.1

on the left hand side.
Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width,

b

. We see from Fig. 1.1 that both of these roots are in the interval

[\pi / 2, \pi]

, and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for

b \rightarrow 0

, we indeed have a factor of 2 in the values of the wave number, while for

b \rightarrow \infty

, the two energies will virtually be the same. This behavior can be understood, if we notice that as

b \rightarrow \infty

, the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled.
Once we have the value of

k

, we can solve for

A_{1}, B_{1}, A_{2}

and

B_{2}

, which give the wavefunctions. Two typical solutions are shown in Fig. 1.3 , for

a=1,2 m V_{0} / \hbar=100

, and

b=0.1

, or

b=0.3

.
Mathematica source codes
$According to the prescription of the text, we assume that the solutions in the two regions take the form \Psi_{1}(x)=A_{1} \sin (k x)+B_{1} \cos (k x), and \Psi_{2}(x)=A_{2} e^{\kappa x}+B_{2} e^{-\kappa x} . There is no need to write down the solution for \Psi_{3}(x) , since the complete solution is either symmetric, in which case the derivative of \Psi_{2}(x) vanishes at x=0 , or antisymmetric, therefore, \Psi_{2}(x) itself vanishes at x=0 . The boundary conditions of the problem are \begin{gathered} \Psi_{1}\left(x=-a-\frac{b}{2}\right)=0 \longrightarrow A_{1} \sin \left(k\left(-a-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-a-\frac{b}{2}\right)\right)=0 \\ \Psi_{1}\left(x=-\frac{b}{2}\right)=\Psi_{2}\left(x=-\frac{b}{2}\right) \longrightarrow A_{1} \sin \left(k\left(-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-\frac{b}{2}\right)\right)=A_{2} e^{-\kappa \frac{b}{2}}+B_{2} e^{\kappa \frac{b}{2}} \\ \left.\frac{\Psi_{1}}{d x}\right|_{x=-\frac{b}{2}}=\left.\frac{\Psi_{2}}{d x}\right|_{x=-\frac{b}{2}} \longrightarrow A_{1} k \cos \left(k\left(-\frac{b}{2}\right)\right)-B_{1} k \sin \left(k\left(-\frac{b}{2}\right)\right)=A_{2} \kappa e^{\kappa \frac{b}{2}}-B_{2} \kappa e^{-\kappa \frac{b}{2}}, \end{gathered} and finally, \Psi_{2}(x=0)=0 \longrightarrow A_{2}+B_{2}=0 i.e., A_{2}=-B_{2} for the antisymmetric case, or \left.\frac{\Psi_{2}}{d x}\right|_{x=0}=0 \longrightarrow A_{2} \kappa-B_{2} \kappa=0 i.e., A_{2}=B_{2} for the symmetric case. Regarding the amplitudes A_{1}, B_{1}, A_{2} and B_{2} as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution, the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation \kappa \sin a k \cosh \frac{\kappa b}{2}+k \cos a k \sinh \frac{\kappa b}{2}=0 for the antisymmetric case, and \kappa \sin a k \sinh \frac{\kappa b}{2}+k \cos a k \cosh \frac{\kappa b}{2}=0 for the symmetric one. Dividing by \cos a k and \cosh \frac{b \kappa}{2} , or \sinh \frac{b \kappa}{2} , respectively, we get \begin{aligned} \frac{\tan a k}{k} & =-\frac{\tanh \frac{b \kappa}{2}}{\kappa} \\ \frac{\tan a k}{k} & =-\frac{\operatorname{coth} \frac{b \kappa}{2}}{\kappa} \end{aligned} for the antisymmetric and symmetric case, respectively. We have an additional equation, which links k and \kappa . Namely, \begin{aligned} \frac{\hbar^{2} k^{2}}{2 m} & =E \\ \frac{\hbar^{2} \kappa^{2}}{2 m} & =V_{0}-E, \end{aligned} i.e., \kappa=\sqrt{\frac{2 m V_{0}}{\hbar^{2}}-k^{2}} Then the two equations above, Eqs. (10-11), lead to two equations for k , where a, b and 2 m V_{0} / \hbar^{2} play the role of parameters. Let us note that in Eq. (14), \kappa becomes imaginary, when \hbar^{2} k^{2} / 2 m>V_{0} . This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle's energy is larger than the confining potential, i.e., the particle is not bound in that region. The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1 , where a=1,2 m V_{0} / \hbar^{2}=100 , and b=0.1 , or b=0.02 . We plotted only k>0 , since the equations are invariant under the transformation k \leftrightarrow-k . By trying various values for b , we notice that as we increase b , the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when b \rightarrow 0 , the right hand side of Eq. (11) tends to -\infty , which means that the first solution will be at k=\pi / 2 . At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of \pi . This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more. Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for a=1, b=0.1 , and 2 m V_{0} / \hbar=100 on the left hand side, and for a=1, b=0.02 , and 2 m V_{0} / \hbar^{2}=100 on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for b=0.1 on the left hand side. Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width, b . We see from Fig. 1.1 that both of these roots are in the interval [\pi / 2, \pi] , and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for b \rightarrow 0 , we indeed have a factor of 2 in the values of the wave number, while for b \rightarrow \infty , the two energies will virtually be the same. This behavior can be understood, if we notice that as b \rightarrow \infty , the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled. Once we have the value of k , we can solve for A_{1}, B_{1}, A_{2} and B_{2} , which give the wavefunctions. Two typical solutions are shown in Fig. 1.3 , for a=1,2 m V_{0} / \hbar=100 , and b=0.1 , or b=0.3 . Mathematica source codes$ $According to the prescription of the text, we assume that the solutions in the two regions take the form \Psi_{1}(x)=A_{1} \sin (k x)+B_{1} \cos (k x), and \Psi_{2}(x)=A_{2} e^{\kappa x}+B_{2} e^{-\kappa x} . There is no need to write down the solution for \Psi_{3}(x) , since the complete solution is either symmetric, in which case the derivative of \Psi_{2}(x) vanishes at x=0 , or antisymmetric, therefore, \Psi_{2}(x) itself vanishes at x=0 . The boundary conditions of the problem are \begin{gathered} \Psi_{1}\left(x=-a-\frac{b}{2}\right)=0 \longrightarrow A_{1} \sin \left(k\left(-a-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-a-\frac{b}{2}\right)\right)=0 \\ \Psi_{1}\left(x=-\frac{b}{2}\right)=\Psi_{2}\left(x=-\frac{b}{2}\right) \longrightarrow A_{1} \sin \left(k\left(-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-\frac{b}{2}\right)\right)=A_{2} e^{-\kappa \frac{b}{2}}+B_{2} e^{\kappa \frac{b}{2}} \\ \left.\frac{\Psi_{1}}{d x}\right|_{x=-\frac{b}{2}}=\left.\frac{\Psi_{2}}{d x}\right|_{x=-\frac{b}{2}} \longrightarrow A_{1} k \cos \left(k\left(-\frac{b}{2}\right)\right)-B_{1} k \sin \left(k\left(-\frac{b}{2}\right)\right)=A_{2} \kappa e^{\kappa \frac{b}{2}}-B_{2} \kappa e^{-\kappa \frac{b}{2}}, \end{gathered} and finally, \Psi_{2}(x=0)=0 \longrightarrow A_{2}+B_{2}=0 i.e., A_{2}=-B_{2} for the antisymmetric case, or \left.\frac{\Psi_{2}}{d x}\right|_{x=0}=0 \longrightarrow A_{2} \kappa-B_{2} \kappa=0 i.e., A_{2}=B_{2} for the symmetric case. Regarding the amplitudes A_{1}, B_{1}, A_{2} and B_{2} as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution, the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation \kappa \sin a k \cosh \frac{\kappa b}{2}+k \cos a k \sinh \frac{\kappa b}{2}=0 for the antisymmetric case, and \kappa \sin a k \sinh \frac{\kappa b}{2}+k \cos a k \cosh \frac{\kappa b}{2}=0 for the symmetric one. Dividing by \cos a k and \cosh \frac{b \kappa}{2} , or \sinh \frac{b \kappa}{2} , respectively, we get \begin{aligned} \frac{\tan a k}{k} & =-\frac{\tanh \frac{b \kappa}{2}}{\kappa} \\ \frac{\tan a k}{k} & =-\frac{\operatorname{coth} \frac{b \kappa}{2}}{\kappa} \end{aligned} for the antisymmetric and symmetric case, respectively. We have an additional equation, which links k and \kappa . Namely, \begin{aligned} \frac{\hbar^{2} k^{2}}{2 m} & =E \\ \frac{\hbar^{2} \kappa^{2}}{2 m} & =V_{0}-E, \end{aligned} i.e., \kappa=\sqrt{\frac{2 m V_{0}}{\hbar^{2}}-k^{2}} Then the two equations above, Eqs. (10-11), lead to two equations for k , where a, b and 2 m V_{0} / \hbar^{2} play the role of parameters. Let us note that in Eq. (14), \kappa becomes imaginary, when \hbar^{2} k^{2} / 2 m>V_{0} . This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle's energy is larger than the confining potential, i.e., the particle is not bound in that region. The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1 , where a=1,2 m V_{0} / \hbar^{2}=100 , and b=0.1 , or b=0.02 . We plotted only k>0 , since the equations are invariant under the transformation k \leftrightarrow-k . By trying various values for b , we notice that as we increase b , the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when b \rightarrow 0 , the right hand side of Eq. (11) tends to -\infty , which means that the first solution will be at k=\pi / 2 . At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of \pi . This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more. Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for a=1, b=0.1 , and 2 m V_{0} / \hbar=100 on the left hand side, and for a=1, b=0.02 , and 2 m V_{0} / \hbar^{2}=100 on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for b=0.1 on the left hand side. Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width, b . We see from Fig. 1.1 that both of these roots are in the interval [\pi / 2, \pi] , and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for b \rightarrow 0 , we indeed have a factor of 2 in the values of the wave number, while for b \rightarrow \infty , the two energies will virtually be the same. This behavior can be understood, if we notice that as b \rightarrow \infty , the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled. Once we have the value of k , we can solve for A_{1}, B_{1}, A_{2} and B_{2} , which give the wavefunctions. Two typical solutions are shown in Fig. 1.3 , for a=1,2 m V_{0} / \hbar=100 , and b=0.1 , or b=0.3 . Mathematica source codes$ $According to the prescription of the text, we assume that the solutions in the two regions take the form \Psi_{1}(x)=A_{1} \sin (k x)+B_{1} \cos (k x), and \Psi_{2}(x)=A_{2} e^{\kappa x}+B_{2} e^{-\kappa x} . There is no need to write down the solution for \Psi_{3}(x) , since the complete solution is either symmetric, in which case the derivative of \Psi_{2}(x) vanishes at x=0 , or antisymmetric, therefore, \Psi_{2}(x) itself vanishes at x=0 . The boundary conditions of the problem are \begin{gathered} \Psi_{1}\left(x=-a-\frac{b}{2}\right)=0 \longrightarrow A_{1} \sin \left(k\left(-a-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-a-\frac{b}{2}\right)\right)=0 \\ \Psi_{1}\left(x=-\frac{b}{2}\right)=\Psi_{2}\left(x=-\frac{b}{2}\right) \longrightarrow A_{1} \sin \left(k\left(-\frac{b}{2}\right)\right)+B_{1} \cos \left(k\left(-\frac{b}{2}\right)\right)=A_{2} e^{-\kappa \frac{b}{2}}+B_{2} e^{\kappa \frac{b}{2}} \\ \left.\frac{\Psi_{1}}{d x}\right|_{x=-\frac{b}{2}}=\left.\frac{\Psi_{2}}{d x}\right|_{x=-\frac{b}{2}} \longrightarrow A_{1} k \cos \left(k\left(-\frac{b}{2}\right)\right)-B_{1} k \sin \left(k\left(-\frac{b}{2}\right)\right)=A_{2} \kappa e^{\kappa \frac{b}{2}}-B_{2} \kappa e^{-\kappa \frac{b}{2}}, \end{gathered} and finally, \Psi_{2}(x=0)=0 \longrightarrow A_{2}+B_{2}=0 i.e., A_{2}=-B_{2} for the antisymmetric case, or \left.\frac{\Psi_{2}}{d x}\right|_{x=0}=0 \longrightarrow A_{2} \kappa-B_{2} \kappa=0 i.e., A_{2}=B_{2} for the symmetric case. Regarding the amplitudes A_{1}, B_{1}, A_{2} and B_{2} as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution, the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation \kappa \sin a k \cosh \frac{\kappa b}{2}+k \cos a k \sinh \frac{\kappa b}{2}=0 for the antisymmetric case, and \kappa \sin a k \sinh \frac{\kappa b}{2}+k \cos a k \cosh \frac{\kappa b}{2}=0 for the symmetric one. Dividing by \cos a k and \cosh \frac{b \kappa}{2} , or \sinh \frac{b \kappa}{2} , respectively, we get \begin{aligned} \frac{\tan a k}{k} & =-\frac{\tanh \frac{b \kappa}{2}}{\kappa} \\ \frac{\tan a k}{k} & =-\frac{\operatorname{coth} \frac{b \kappa}{2}}{\kappa} \end{aligned} for the antisymmetric and symmetric case, respectively. We have an additional equation, which links k and \kappa . Namely, \begin{aligned} \frac{\hbar^{2} k^{2}}{2 m} & =E \\ \frac{\hbar^{2} \kappa^{2}}{2 m} & =V_{0}-E, \end{aligned} i.e., \kappa=\sqrt{\frac{2 m V_{0}}{\hbar^{2}}-k^{2}} Then the two equations above, Eqs. (10-11), lead to two equations for k , where a, b and 2 m V_{0} / \hbar^{2} play the role of parameters. Let us note that in Eq. (14), \kappa becomes imaginary, when \hbar^{2} k^{2} / 2 m>V_{0} . This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle's energy is larger than the confining potential, i.e., the particle is not bound in that region. The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1 , where a=1,2 m V_{0} / \hbar^{2}=100 , and b=0.1 , or b=0.02 . We plotted only k>0 , since the equations are invariant under the transformation k \leftrightarrow-k . By trying various values for b , we notice that as we increase b , the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when b \rightarrow 0 , the right hand side of Eq. (11) tends to -\infty , which means that the first solution will be at k=\pi / 2 . At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of \pi . This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more. Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for a=1, b=0.1 , and 2 m V_{0} / \hbar=100 on the left hand side, and for a=1, b=0.02 , and 2 m V_{0} / \hbar^{2}=100 on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for b=0.1 on the left hand side. Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width, b . We see from Fig. 1.1 that both of these roots are in the interval [\pi / 2, \pi] , and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for b \rightarrow 0 , we indeed have a factor of 2 in the values of the wave number, while for b \rightarrow \infty , the two energies will virtually be the same. This behavior can be understood, if we notice that as b \rightarrow \infty , the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled. Once we have the value of k , we can solve for A_{1}, B_{1}, A_{2} and B_{2} , which give the wavefunctions. Two typical solutions are shown in Fig. 1.3 , for a=1,2 m V_{0} / \hbar=100 , and b=0.1 , or b=0.3 . Mathematica source codes$