Homework 3 Key

Q1

Write the time-independent and time-dependent Schrödinger equations for each the following systems:

1. A particle of mass \(m\) moving in 3-D space with the potential energy given by \(V(x, y, z) = -\dfrac {(x^2 + y^2 + z^2)}{3}\).

General form of time-dependent Schrödinger equation:

\[{\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\Psi(\vec{r},t)}={ i\hbar\dfrac{\partial}{\partial t}\Psi(\vec{r},t)}\tag{1}\]

or

\[{\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})e^{-iEt / \hbar}}={ i\hbar\dfrac{\partial}{\partial t}\psi(\vec{r})e^{-iEt / \hbar}}\tag{2}\]

General form of time-independent Schrödinger equation:

\[{ \left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{3}\]

Plug in 3-D potential \(V(x, y, z) = -\dfrac {(x^2 + y^2 + z^2)}{3}\) in both equations (\(2\) and \(3\)):

\[{ \left[-\dfrac{\hbar^2}{2m}\nabla^2+(-\dfrac {(x^2 + y^2 + z^2)}{3}))\right]\psi(\vec{r})e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(\vec{r})e^{-iEt / \hbar}}\tag{4}\]

\[{ \left[-\dfrac{\hbar^2}{2m}\nabla^2+(-\dfrac {(x^2 + y^2 + z^2)}{3})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{5}\]

Write \(\nabla^2\) in 3-D:

\[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2})+(-\dfrac {(x^2 + y^2 + z^2)}{3})\right]\psi(\vec{r})e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(\vec{r})e^{-iEt / \hbar}}\tag{6}\]

\[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2})+(-\dfrac {(x^2 + y^2 + z^2)}{3})\right]\psi(\vec{r}) = E\psi(\vec{r})}\tag{7}\]

And finally change the functions to 3-D:

\[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2})+(-\dfrac {(x^2 + y^2 + z^2)}{3})\right]\psi(x,y,z)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x,y,z)e^{-iEt / \hbar}}\tag{8}\]

\[{ \left[-\dfrac{\hbar^2}{2m}(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2})+(-\dfrac {(x^2 + y^2 + z^2)}{3})\right]\psi(x,y,z) = E\psi(x,y,z)}\tag{9}\]

b. A particle of mass \(mu\) moving in 1-D space with a potential energy given by \(V(x)=1/2 k (x-x_o)^2\) where \(x_o\) and \(k\) are constants.

Time-dependent Schrödinger equation:

\[{ \left[-\dfrac{\hbar^2}{2mu}\dfrac{\partial^2}{\partial x^2}+\dfrac {1}{2}(x-x_0)^2\right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\tag{10}\]

Time-independent Schrödinger equation:

\[{ \left[-\dfrac{\hbar^2}{2mu}\dfrac{\partial^2}{\partial x^2}+\dfrac {1}{2}(x-x_0)^2\right]\psi(x) = E\psi(x)}\tag{11}\]

c. A particle of mass \(m\) within a 1-D box with infinitely high walls and of length \(L\).

Box with infinitely high walls has the following potential:

Region I. \(V(x)=\infty, x<0\)

Region II. \(V(x)=0, 0\leqslant x\leqslant L\)

Region III. \(V(x)=\infty, x>L\)

Time-dependent Schrödinger equation:

Region I. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+\infty \right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\]

(Notation: You can write the equation with \(V(x)=\infty\), although it is not very strict from mathematical perspective, OR you can state that there is no solutions for this region without writing the equation)

Region II. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\tag{10}\]

Region III.

\[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+\infty \right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\]

(Notation: You can write the equation with \(V(x)=\infty\), although it is not very strict from mathematical perspective, OR you can state that there is no solutions for this region without writing the equation)

Time-independent Schrödinger equation:

Region I. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} +\infty \right]\psi(x) = E\psi(x)}\]

Region II. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\right]\psi(x) = E\psi(x)}\tag{11}\]

Region III. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} +\infty \right]\psi(x) = E\psi(x)}\]

(Notation: You can write the equation with \(V(x)=\infty\) for Regions I and III, although it is not very strict from mathematical perspective, OR you can state that there is no solutions for these regions without writing the equation)

(Notation: The solutions exist only inside the box with infinitely high walls, i.e. where \(0\leqslant x\leqslant L\). For the box with finite walls the solution is not confined to the box, but exists at any x, i.e. the particle "penetrates" the finite walls. The solution inside the walls is different from that inside the box.)

d. A particle of mass \(m\) within a 1-D box with finite high walls of length \(L\) and height \(B\).

Box with infinitely high walls has the following potential:

Region I. \(V(x)=B, x<0\)

Region II. \(V(x)=0, 0 \leqslant x \leqslant L\)

Region III. \(V(x)=B, x>L\)

Time-dependent Schrödinger equation:

Region I. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+B\right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\tag{12.1}\]

Region II. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\tag{12.2}\]

Region III. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+B\right]\psi(x)e^{-iEt / \hbar}=i\hbar\dfrac{\partial}{\partial t}\psi(x)e^{-iEt / \hbar}}\tag{12.3}\]

Time-independent Schrödinger equation:

Region I. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+B\right]\psi(x) = E\psi(x)}\tag{13.1}\]

Region II. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\right]\psi(x) = E\psi(x)}\tag{13.2}\]

Region III. \[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+B\right]\psi(x) = E\psi(x)}\tag{13.3}\]

Q2

Which of the following operators are linear?

An operator \(\hat{A}\) is linear if

\[ \hat{A}[c_1f_1(x)+c_2f_2(x)]= c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \tag{14}\]

and the operator is nonlinear if

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] \neq c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \tag{15}\]

1. \( \displaystyle \hat{A} = 2x^2 \frac{d^2}{dx^2} \)

Start with the left side of \((14)\) or \((15)\), simplify the expression and check if you get the right side of \((14)\) or \((15)\):

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= 2x^2 \frac{d^2}{dx^2}[c_1f_1(x)+c_2f_2(x)]=2x^2\frac{d^2}{dx^2}c_1f_1(x)+2x^2\frac{d^2}{dx^2} c_2f_2(x)=c_1 2x^2\frac{d^2}{dx^2}f_1(x)+c_2 2x^2 \frac{d^2}{dx^2}f_2(x)=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

2. \( \displaystyle \hat{A} f(x) = f(x) \)

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= \hat{A}c_1f_1(x)+\hat{A}c_2f_2(x) =c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \)

You get the equality sign between left and right side of \(14\) or \((15)\), therefore, \(\hat{A}\) is linear.

3. \( \displaystyle \hat{A} f(x,y) = \dfrac{\partial f(x,y)}{\partial x} + \dfrac{\partial f(x,y)}{\partial y} \) [sum of two partial derivatives f(x,y)]

\( \hat{A}[c_1f_1(x,y)+c_2f_2(x,y)]= [\dfrac{\partial}{\partial x} + \dfrac{\partial}{\partial y}][c_1f_1(x,y)+c_2f_2(x,y)] = [\dfrac{\partial}{\partial x} + \dfrac{\partial}{\partial y}]c_1f_1(x,y) + [\dfrac{\partial}{\partial x} + \dfrac{\partial}{\partial y}]c_2f_2(x,y) = c_1\hat{A}f_1(x,y)+c_2\hat{A}f_2(x,y) \)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

4. \( \hat{A} f(x) = \sqrt{f(x)}\) [square f(x)]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= \sqrt{c_1f_1(x)+c_2f_2(x)} \neq c_1\sqrt {f_1(x)}+c_2\sqrt {f_2(x)} = c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the inequality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is nonlinear.

5. \( \hat{A} f(x) =f^∗(x)\) [form the complex conjugate of f(x)]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= [c_1f_1(x)+c_2f_2(x)]^*= c_1^*f_1^*(x)+c_2^*f_2^*(x)={if \ c_1 \ and \ c_2 \ are \ real}=c_1f_1^*(x)+c_2f_2^*(x)=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

(In case c₁ and c₂ are complex: \( \hat{A}[c_1f_1(x)+c_2f_2(x)]= [c_1f_1(x)+c_2f_2(x)]^*= c_1^*f_1^*(x)+c_2^*f_2^*(x) \neq c_1f_1^*(x)+c_2f_2^*(x)=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\). You get the inequality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is nonlinear. )

6. \( \hat{A} = \left( \begin{array}{cc} a & b \\ d & e \end{array} \right) \)

Matrices possess distributive property:

\(A(B+C)=AB+AC\)

If \( A = \hat{A} =\left( \begin{array}{cc} a & b \\ d & e \end{array} \right) \), \(c_1f_1=B\) and \(c_2f_2=C\) the distributive property above becomes:

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= \hat{A}[c_1f_1(x)]+\hat{A}[c_2f_2(x)]\)

According to the rules of multiplication of matrices by scalar, \(c_1\) and \(c_2\) could be pulled out to the front:

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= \hat{A}[c_1f_1(x)]+\hat{A}[c_2f_2(x)]=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

7. \( \hat{H} = \left( \begin{array}{cc} 1 & \delta \\ \delta & 1 \end{array} \right) \)

Matrices possess distributive property:

\(A(B+C)=AB+AC\)

If \( A = \hat{H} = \left( \begin{array}{cc} 1 & \delta \\ \delta & 1 \end{array} \right) \), \(c_1f_1=B\) and \(c_2f_2=C\) the distributive property above becomes:

\( \hat{H} [c_1f_1(x)+c_2f_2(x)]= \hat{H} [c_1f_1(x)]+\hat{H}[c_2f_2(x)]\)

According to the rules of multiplication of matrices by scalar, \(c_1\) and \(c_2\) could be pulled out to the front:

\( \hat{H}[c_1f_1(x)+c_2f_2(x)]= \hat{H}[c_1f_1(x)]+\hat{H}[c_2f_2(x)]=c_1\hat{H}f_1(x)+c_2\hat{H}f_2(x)\)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

8. \( \displaystyle \hat{A} f(x) = \int_0^x f(x')dx' \) [Integrating from 0 to x of f(x)]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]=\int_0^x [c_1f_1(x')+c_2f_2(x')] dx'=\int_0^x c_1f_1(x') dx' +\int_0^x c_2f_2(x')dx'=c_1\int_0^x f_1(x') dx' +c_2\int_0^x f_2(x')dx' = c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

9. \( \hat{A} f(x) = 0 \) [multiply f(x) by zero]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= 0 \times [c_1f_1(x)+c_2f_2(x)] = c_1 \times 0 \times f_1(x)+c_2\times 0 \times f_2(x)=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

10. \( \hat{A} f(x) =[f(x)]^{−1} \) [take the reciprocal of f(x)]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]=\dfrac{1}{c_1f_1(x)+c_2f_2(x)} \neq c_1\dfrac{1}{f_1(x)}+c_2\dfrac{1}{f_2(x)} = c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the inequality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is nonlinear.

11. \( \hat{A} f(x)=f(0)\) [evaluate f(x) at x=0]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= c_1f_1(0)+c_2f_2(0)=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the equality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is linear.

12. \( \hat{A} f(x) =\log_{10} f(x) \) [take the log of f(x)]

For this operator it is easier to start with the right side of \((14)\) or \((15)\), simplify the expression and check if you get the left side of \((14)\) or \((15)\):

\( c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)=c_1\log_{10}f_1(x)+c_2\log_{10}f_2(x)=\log_{10}[f_1(x)]^{c_1}+\log_{10}[f_2(x)]^{c_2}= \log_{10}([f_1(x)]^{c_1} \times [f_2(x)]^{c_2}) \neq \log_{10}[c_1f_1(x)+c_2f_2(x)]=\hat{A}[c_1f_1(x)+c_2f_2(x)] \)

You get the inequality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is nonlinear.

13. \( \hat{A} f(x) =\sin ⁡f(x) \) [take the sin of f(x)]

\( \hat{A}[c_1f_1(x)+c_2f_2(x)]= \sin (c_1f_1(x)+c_2f_2(x)) = \sin (c_1f_1(x)) \cos (c_2f_2(x)) +\cos (c_1f_1(x)) \sin (c_2f_2(x)) \neq c_1 \sin (f_1(x)) + c_2 \sin (f_2(x))=c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\)

You get the inequality sign between left and right side of \((14)\) or \((15)\), therefore, \(\hat{A}\) is nonlinear.

Q3

Demonstrate that the function \(e^{ikx}\) is an eigenfunction of the kinetic energy operator. What are the corresponding eigenvalues?

General form for kinetic energy operator:

\(\hat{T}=-\dfrac{\hbar^2}{2m}\nabla^2\)

For 1-D case:

\(\hat{T}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\)

Eigenfunctions \(\psi (x)\) satisfy the equation \(\hat{A}\psi (x) = a\psi (x) \) \((16)\) , where \(a\) are eigenvalues of operator \(\hat{A}\) (eigenvalue is always a CONSTANT).

To show if a function is an eigenfunction of an operator we should check if \((16)\) takes place (or \(a=const\) after simplification).

In this case we check for \(\hat{A} = \hat{T}\) and \(\psi (x)=e^{ikx}\) and \((16)\) transforms into:

\(\hat{T} e^{ikx}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} (e^{ikx})=-\dfrac{\hbar^2}{2m} (ik)^2 e^{ikx}= -\dfrac{\hbar^2}{2m} (-k^2) e^{ikx} = \dfrac{\hbar^2 k^2}{2m} e^{ikx} \)

If compare to \((16)\) \(a= \dfrac{\hbar^2 k^2}{2m} = const\), therefore, function \(e^{ikx}\) is an eigenfunction of the kinetic energy operator with eigenvalues \(a= \dfrac{\hbar^2 k^2}{2m} \)

(From physics \(\hbar k = p\) and, therefore, \(a= \dfrac{p^2}{2m}\). This is a classical expression for kinetics energy.)

Q4

Given a particle in a 1D box with infinite high walls in the \(n=5\) state:

1. How many nodes are there (the edges do not count)?

\(N_{nodes}=n-1=5-1=4\)

1. How many antinodes are there?

\(N_{antinodes}=n=5\)

b. What is the probability of finding the particle outside the box?

Since the box has infinite high walls, the probability of finding the particle outside the box is 0. (For finite walls is non-zero)

c. How do the above questions change if:

1. The mass is doubled (i.e., \(2m)\)?

Number of nodes does not depend on mass.

Number of antinodes does not depend on mass.

The probability of finding the particle outside the box does not depend on mass.

2. The box length is doubled (i.e., \(2L\))?

Number of nodes does not depend on the box length.

Number of antinodes does not depend on the box length.

The probability of finding the particle outside the box does not depend on the box length.

3. The quantum number is doubled to \(2n\)?

Number of nodes will become bigger: \(N_{nodes}=2n-1=2 \times 5-1=9\).

Number of antinodes will be doubled: \(N_{antinodes}=2 \times n=10\).

The probability of finding the particle outside the box does not depend on quantum number.

Q5

Consider a particle of mass \(m\) that has energy \( E = 0 \) and a time-independent wave function given by:

\[ \psi(x) = A x^2 e^{-B^2 x^2} \tag{18}\]

\(A \) and \(B\) are constants. Determine the potential energy \( V(x) \) of the particle. (Hint: Use the time-independent Schroedinger equation.) What does this potential energy look like?

Time-independent Schrödinger equation for 1-D case is (here the wavefunction \(\psi(x)\) depends on x only):

\[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x)\right]\psi(x) = E\psi(x)}\]

Plug in \( E = 0 \) and \((18)\) for \(\psi(x)\) :

\[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x)\right] A x^2 e^{-B^2 x^2} = 0}\]

Simplify:

\[{ \left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x) \right] A x^2 e^{-B^2 x^2} = -\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} [A x^2 e^{-B^2 x^2}] +V(x) A x^2 e^{-B^2 x^2} =-\dfrac{\hbar^2}{2m}A\dfrac{\partial^2}{\partial x^2} [x^2 e^{-B^2 x^2}] +V(x) A x^2 e^{-B^2 x^2} = -\dfrac{\hbar^2}{2m}A\dfrac{\partial}{\partial x} [2x e^{-B^2 x^2}+x^2 (-2B^{2}x)e^{-B^2 x^2}]+V(x) A x^2 e^{-B^2 x^2} = \\ -\dfrac{\hbar^2}{2m} A \dfrac{\partial}{\partial x} [2x e^{-B^2 x^2}-2B^2 x^3 e^{-B^2 x^2}]+V(x) A x^2 e^{-B^2 x^2}=-\dfrac{\hbar^2}{2m} A[2 e^{-B^2 x^2}-4B^2x^2e^{-B^2 x^2}-2B^2[3x^2e^{-B^2 x^2}+x^3 (-2B^2x) e^{-B^2 x^2}]+V(x) A x^2 e^{-B^2 x^2} = \\ -\dfrac{\hbar^2}{2m} A e^{-B^2 x^2} [2-4B^2x^2-2B^2(3x^2-2B^2x^4)] +V(x) A x^2 e^{-B^2 x^2} } = A e^{-B^2 x^2} [ -\dfrac{\hbar^2}{2m} [2-4B^2x^2-2B^2(3x^2-2B^2x^4)] +V(x)x^2] = A e^{-B^2 x^2} [ -\dfrac{\hbar^2}{2m} (2-10B^2x^2+4B^4x^4) +V(x)x^2] =0\tag{19}\]

Solve \((19)\) for \(V(x)\):

\(V(x)=\dfrac{\hbar^2}{2m} (\dfrac{2}{x^2}-10B^2+4B^4x^2)\)

Q6

Find following values for a particle of mass, \(m \), for a particle in a 1D box with length \(L\) in the \(n=1\) state and \(n=3\) states.

a. \(\langle x \rangle\)

The general formula for expectation value of the position operator is:

\[\langle \hat{x}\rangle = \int_{0}^{L}\psi_n^*\hat{x}\psi_n dx\]

For particle in a 1D box with length \(L\) the wavefunctions are \(\psi (x) = \sqrt{\frac{2}{L}}\sin(\dfrac{n\pi x}{L})\) where \(n =1,2,...\)

\[\langle \hat{x}\rangle = \int_{0}^{L} x\psi^*(x) \psi(x)dx=\int_{0}^{L}x\dfrac{2}{L}\sin^2{\dfrac{n\pi x}{L}}dx=\dfrac{2}{L}\dfrac{1}{2}\int_{0}^{L}x(1-\cos{\dfrac{2n\pi x}{L}})dx=\dfrac{L}{2}\]

b. \(\langle x^2 \rangle\)

The general formula for expectation value of the position operator squared is:

\[\langle \hat{x^2}\rangle = \int_{0}^{L}\psi_n^*\hat{x^2}\psi_n dx\]

\[\langle \hat{x^2}\rangle = \dfrac{2}{L}\int_{0}^{L}x^2\sin^2\dfrac{n\pi x}{L}dx = \dfrac{L^2}{3} - \dfrac{L^2}{2n^2\pi^2}\]

Again we used the half angle formula and integration by parts to compute this integral.

c. \(\langle p \rangle\)

The general formula for expectation value of the momentum operator is:

\[\langle \hat{p}\rangle = \int_{0}^{L}\psi_n^*\hat{p}\psi_n dx\]

\[\langle \hat{p}\rangle = -i\hbar\dfrac{\pi}{L}\dfrac{d}{dx}\]

\[\langle \hat{p}\rangle = \int_{0}^{L} \sqrt{\dfrac{2}{L}} \sin\dfrac{n\pi x}{L}(-i\hbar)\dfrac{d}{dx} \sqrt{\dfrac{2}{L}} \sin\dfrac{n\pi x}{L} dx = \dfrac{2}{L} (-i\hbar) \int_{0}^{L} \sin\dfrac{n\pi x}{L}\dfrac{d}{dx} \sin\dfrac{n\pi x}{L} dx =\dfrac{2}{L}(-i\hbar) \dfrac{n \pi}{L}\int_{0}^{L}\sin\dfrac{n\pi x}{L}\cos\dfrac{n\pi x}{L}dx = 0\tag{17}\]

The integral in \((17)\) is zero because sin(x) times cos(x) is an odd function and the integral of an odd function over a symmetric interval is zero.

d. \(\langle p^2 \rangle \)

The general formula for expectation value of the momentum operator squared is:

\[\langle \hat{p^2}\rangle = \int_{0}^{L}\psi_n^*\hat{p^2}\psi_n dx\]

\[ \hat{p^2} = -\hbar^2\dfrac{d^2}{dx^2}\]

\[\langle \hat{p^2}\rangle = \int_{0}^{L} \sqrt{\dfrac{2}{L}} \sin\dfrac{n\pi x}{L}(-\hbar^2) \dfrac{d^2}{dx^2} \sqrt{\dfrac{2}{L}} \sin\dfrac{n\pi x}{L} dx = \dfrac{2}{L} (-\hbar^2) \int_{0}^{L} \sin\dfrac{n\pi x}{L} \dfrac{d^2}{dx^2} \sin\dfrac{n\pi x}{L} dx = \dfrac{2}{L}(-\hbar^2) (-\dfrac{n^2 \pi^2}{L^2}) \int_{0}^{L}\sin^2\dfrac{\pi x}{L}dx = \dfrac{2}{L}(-\hbar^2) (-\dfrac{n^2 \pi^2}{L^2}) (\dfrac{1}{2}x-\dfrac{L}{4\pi} \sin\dfrac{2\pi x}{L} )\Biggr\rvert_{0}^{L}= \hbar^2\dfrac{n^2\pi^2}{L^2}\]

Q7

For each of the expectation values in Q6, explain why or why not, the respective values depends on \(n\).

a. The expectation value of the position operator does not depend on \(n\), because the wavefunction is symmetric for all \(n\)'s. Symmetric wavefunctions give expectation values of the position operator in the middle of the range of their argument (here, in the middle of the box).

b. The expectation value of the position operator squared depend on \(n\), because it corresponds to the spread of the probability distribution which depends on the wavefunction shape. The shape is different for different \(n\)'s.

(Notation: For greater \(n\)'s the probability distribution becomes less centered (in contrast to \(n\)=1 with one bump in the center), but has more uniform distribution (\(n \) bumps uniformly distributed along the box), therefore, the spread increases. When \(n=\infty\) the distribution is perfectly uniform and the probability of finding the particle at any x inside the box is the same.)

c. The expectation value of the momentum does not depend on \(n\) and equals 0 for all \(n\)'s. This is the consequence of the symmetry of the wavefunction (the particle "moves to the right" as much as it "moves to the left").

d. The expectation value of the momentum operator squared depends on \(n\), as with expectation value of the position operator squared, the momentum operator squared corresponds to the spread of the distribution, but for momentum. The distribution for momentum changes with \(n\).

(Notation: As \(n\) increases, the total energy of the system \(E\) increases ( \( E_n = \dfrac{h^2 n^2 }{8m_e L^2}\) ) and kinetic energy \(T=E-V\) increases (since \(V=0\) does not change inside the box). Since expectation value of kinetic energy is proportional to the expectation value of the momentum operator squared: \(\langle \hat{T} \rangle=\dfrac{1}{2m} \langle \hat{p^2} \rangle \), no wonder the expectation value of the momentum operator squared \(\langle \hat{p^2} \rangle \) increases with \(n\).)

Q8

The uncertainties of a position and a momentum of a particle (\(\Delta x\)) and (\(\Delta p\)) are defined as

\[ \Delta x = \sqrt{ \langle x^2 \rangle - \langle x \rangle ^2} \]

\[ \Delta p = \sqrt{ \langle p^2 \rangle - \langle p \rangle ^2} \]

For the particle in the box at the ground eigenstate (\(n=1\)) and first excited state (\(n=2\)), what is the uncertainty in the value \( x \)? How would you interpret the results of these calculations?

\[\Delta x = \sqrt{\langle \hat{x^2}\rangle - \langle \hat{x}\rangle^2 }\]

See Q6a and Q6b for finding \(\langle \hat{x^2}\rangle\) and \(\langle \hat{x}\rangle^2\).

\[\Delta x =\sqrt{\dfrac{L^2}{3} - \dfrac{L^2}{2n^2\pi^2} - \dfrac{L^2}{4}}=L\sqrt{\dfrac{1}{12}-\dfrac{1}{2n^2\pi^2} }\]

For n=1:

\[\Delta x =\sqrt{\dfrac{L^2}{3} - \dfrac{L^2}{2n^2\pi^2} - \dfrac{L^2}{4}}=L\sqrt{\dfrac{1}{12}-\dfrac{1}{2 \times 1^2\pi^2} }=L\sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} }={may \ calculate \ if \ you \ like}=0.18L\]

For n=2:

\[\Delta x =\sqrt{\dfrac{L^2}{3} - \dfrac{L^2}{2n^2\pi^2} - \dfrac{L^2}{4}}=L\sqrt{\dfrac{1}{12}-\dfrac{1}{2 \times 2^2\pi^2} }=L\sqrt{\dfrac{1}{12}-\dfrac{1}{8\pi^2} }={may \ calculate \ if \ you \ like}=0.27L\]

Uncertainty in the value \( x \) for particle in a box depends on principal number \(n\), since the spread of the wavefunction (\(\langle x^2 \rangle\)) depends on \(n\).

(Notation: Greater \(n\)'s have less centered and more uniform distribution, therefore, for greater \(n\)'s the spread is greater ("fatter" distribution) and there is more uncertainty in where the particle is.)

b. For the particle in the box at the ground eigenstate (\(n=1\)) and first excited state (\(n=2\)), what is the uncertainty in the value \( p \)? How would you interpret the results of these calculations?

\[ \Delta p = \sqrt{ \langle p^2 \rangle - \langle p \rangle ^2} ={use \ results \ in \ Q6c \ and \ Q6d} = \sqrt{ \langle p^2 \rangle - 0} =\sqrt{ \hbar^2\dfrac{n^2\pi^2}{L^2}}= \hbar\dfrac{n\pi}{L}\]

For n=1:

\[ \Delta p =\hbar\dfrac{\pi}{L}={may \ calculate \ if \ you \ like}=\dfrac{3.31 \times 10^{-34}}{L}\]

For n=2:

\[ \Delta p =\hbar\dfrac{2\pi}{L}={may \ calculate \ if \ you \ like}=\dfrac{6.62 \times 10^{-34}}{L}\]

Uncertainty in the value \( p \) for particle in a box depends on principal number \(n\), since (\(\langle p^2 \rangle\)) depends on \(n\).

c. What is the product for the ground and first excited state: \(\Delta x \Delta p\).

\[\Delta p = \sqrt{\langle \hat{p^2}\rangle - \langle \hat{p}\rangle^2 } \]

\[\Delta x = \sqrt{\langle \hat{x^2}\rangle - \langle \hat{x}\rangle^2 } \]

Use the calculated \(\Delta x\) and \(\Delta p\) for n=1 above:

\[\Delta p\Delta x =\hbar\dfrac{\pi}{L} L\sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} } = \hbar \pi \sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} }=\dfrac{\hbar}{2} 2\pi \sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} }=1.13 \dfrac{\hbar}{2} \]

Use the calculated \(\Delta x\) and \(\Delta p\) for n=2 above:

\[\Delta p\Delta x =\hbar\dfrac{2\pi}{L} L\sqrt{\dfrac{1}{12}-\dfrac{1}{8\pi^2} } = \hbar 2\pi \sqrt{\dfrac{1}{12}-\dfrac{1}{8\pi^2} }= \dfrac{\hbar}{2} 4\pi \sqrt{\dfrac{1}{12}-\dfrac{1}{8\pi^2} } = 3.39 \dfrac{\hbar}{2}\]

d. Does the Heisenberg Uncertainty Principle hold for a particle in each of these states?

As shown above \(\Delta x\) times \(\Delta p\) is 1.13 times greater than \(\dfrac{\hbar}{2}\) for n=1 and 3.39 times greater than \(\dfrac{\hbar}{2}\) for n=2, thus the Heisenberg Uncertainty principle holds.

Q9

What is the expectation value of kinetic energy for a particle in a box of length (\(L\)) in the ground eigenstate (n=1)? What about for the first excited eignestate (n=2). Explain the difference.

Use kinetic energy operator:

\[ \hat{T}= \dfrac{\hat{p^2}}{2m} = \dfrac{1}{2m} \hat{p^2}\]

\[\langle \hat{T}\rangle = \int_{0}^{L}\psi_n^*\hat{T}\psi_n dx = \int_{0}^{L}\psi_n^* \dfrac{1}{2m} \hat{p^2} \psi_n dx = \dfrac{1}{2m} \int_{0}^{L}\psi_n^* \hat{p^2} \psi_n dx={use \ result \ in \ Q6.d}= \dfrac{1}{2m} \hbar^2\dfrac{n^2\pi^2}{L^2}= \hbar^2\dfrac{n^2\pi^2}{2mL^2}\]

For n=1 expectation value of kinetic energy is \(\hbar^2\dfrac{\pi^2}{2mL^2}\).

For n=2 expectation value is of kinetic energy is \(\hbar^2\dfrac{4\pi^2}{2mL^2}\).

The expectation value of kinetic energy is greater for n=2 because the spread of the momentum \(\langle \hat{p^2}\rangle \) is greater.

Q10

What is the most probable position for a particle in a box of length (\(L\)) in the ground eigenstate (n=1)? What about for the first excited eignestate (n=2). Explain the difference.

The most probable positions are where the probability distribution \(\psi^2\) has MAXIMA. These MAXIMA are at the same positions as the EXTREMA (MAXIMA and MINIMA) of the amplitude of the probability \(\psi\). (You can see it in the figures of Graphical method below: extrema of \(\psi\) (left figure) correspond to maxima of \(\psi^2\) (right figure).) For simplicity here we find the EXTREMA (MAXIMA and MINIMA) of the amplitude of the probability \(\psi\).

Mathematically you can do this by setting the its derivative to 0:

\(\dfrac{d}{dx} \psi_n(x) = 0\)

Remember that for particle in a 1D box with length \(L\) the wavefunctions are \(\psi (x) = \sqrt{\frac{2}{L}}\sin(\dfrac{n\pi x}{L})\) where \(n =1,2,...\)

\[\dfrac{d}{dx} [\sqrt{\frac{2}{L}} sin{\dfrac{n \pi x}{L}}]=\sqrt{\frac{2}{L}} \dfrac{d}{dx} sin{\dfrac{n \pi x}{L}}= \sqrt{\frac{2}{L}} \dfrac{n \pi}{L} cos {\dfrac{n \pi x}{L}}=0\tag{18}\]

Solve for x using n=1:

\(cos {\dfrac{\pi x}{L}}=0\) when its argument

\(\dfrac{\pi x}{L}=\dfrac{\pi}{2}(2k+1) \tag{19}\)

where \(k=0,1,...\)

Solve \((19)\) for x:

\(x=\dfrac{L}{2}(2k+1)\) where \(k=0,1,...\)

Out of all \(k\) only \(k=0\) gives solution inside the box, therefore, \(x=\dfrac{L}{2}\).

Solve for x using n=2:

\(cos {\dfrac{2\pi x}{L}}=0\) when its argument

\(\dfrac{2\pi x}{L}=\dfrac{\pi}{2}(2k+1) \tag{20}\)

where \(k=0,1,...\)

Solve \((20)\) for x:

\(x=\dfrac{L}{4}(2k+1)\) where \(k=0,1,...\)

Out of all \(k\) only \(k=0,1\) give solutions inside the box, therefore, \(x=\dfrac{L}{4}\) and \(x=\dfrac{3L}{4}\).

Graphical method:

The easy way is just to note in the right figure that the maximum of \(\psi^2\) (probability distribution) for n=1 occurs in the middle of the box so the most probable position is \(x = \dfrac{L}{2}\)

For n=2 there are two maxima of \(\psi^2\) (probability distribution) as shown in the right figure. These occur at \(x = \dfrac{L}{4}, \dfrac{3L}{4}\).

The difference in most probable position for \(n=1\) and \(n=2\) comes from different shapes of wavefunction at different \(n\).

Q11

Find the following expectation values for a particle with mass, \(m \), in a 3D box with dimensions \(a_1 \), \(a_2 \), \(a_3 \). Assume that the quantum numbers are given by \(n_x\), \(n_y\), and \(n_z\).

The wavefunction for 3D box is

\[\psi (x,y,z)=\sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) = \sqrt{\dfrac{8}{a_1a_2a_3}} \sin(\dfrac{n_x\pi x}{a_1}) \sin(\dfrac{n_y\pi y}{a_2}) sin(\dfrac{n_z\pi z}{a_3}) \]

1. \( \langle x \rangle \)

Since the wavefunction is real, then \(|\psi|^2=\psi^*\psi =\psi^2\), so no need to "sandwich" the variable for calculating the expectation value of position like \(\langle \hat{x}\rangle = \int_{-\infty}^{\infty}\psi^*x \psi dx\) , but rather we could use \(\langle \hat{x}\rangle = \int_{-\infty}^{\infty} x \psi^2 dx\) (this is true only for coordinate! for momentum (like in Q11.4) you have to keep the order unchanged \(\langle \hat{p_x}\rangle = \int_{-\infty}^{\infty}\psi^* \hat{p_x} \psi dx= \int_{-\infty}^{\infty}\psi \hat{p_x} \psi dx = \int_{-\infty}^{\infty}\psi (-i \hbar) \dfrac{\partial}{\partial x} \psi dx\), because the order matters for taking the derivative.)

Transition to 3d coordinates and use finite limits for the integral because the particle exist only inside the box \(0 \leqslant x \leqslant a_1\), \(0 \leqslant y \leqslant a_2\) and \(0 \leqslant z \leqslant a_3\):

\[\langle \hat{x}\rangle = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3}\psi^*x \psi dxdydz= {since \ wavefunction \ is \ real}= \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3}\psi x \psi dxdydz = \int_{0}^{a_1}x \psi^2 dx dydz=\int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} x (\sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) )^2 dxdydz = \\ \int_{0}^{a_2}\int_{0}^{a_3}{\dfrac{4}{a_2a_3}} \sin^2(\dfrac{n_y\pi y}{a_2}) sin^2(\dfrac{n_z\pi z}{a_3}) dydz \int_{0}^{a_1}x (\sqrt{\dfrac{2}{a_1}})^2\sin^2\dfrac{n_x \pi x}{a_1} dx ={use \ Q6a \ result} ={\dfrac{4}{a_2a_3}} \int_{0}^{a_2}\int_{0}^{a_3} \sin^2(\dfrac{n_y\pi y}{a_2}) sin^2(\dfrac{n_z\pi z}{a_3}) dydz \dfrac{a_1}{2} = {\dfrac{4}{a_2a_3}} {\dfrac{a_2a_3}{4}} \dfrac{a_1}{2}=\dfrac{a_1}{2}\]

2. \( \langle y \rangle \)

Analogously to Q11a:

\(\langle \hat{y}\rangle = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} y (\sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}))^2 dxdydz= \dfrac{a_2}{2}\)

3. \( \langle z \rangle \)

Analogously to Q11a:

\(\langle \hat{z}\rangle = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} z (\sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) )^2 dxdydz = \dfrac{a_3}{2} \)

4. \( \langle p_z \rangle \)

Use the result in Q6.c:

\(\langle \hat{p_z}\rangle = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3}\psi^*p_z\psi dxdydz= {since \ wavefunction \ is \ real}=\int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} \psi p_z \psi dxdydz = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} \sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) \hat{p_z} \sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) dxdydz = \\ \dfrac{4}{a_1a_2} \int_{0}^{a_1} \int_{0}^{a_2}\sin^2(\dfrac{n_x\pi x}{a_1}) sin^2(\dfrac{n_y\pi y}{a_2}) dxdy \int_{0}^{a_3} \sqrt{\dfrac{2}{a_3}}\sin \dfrac{n_z \pi z}{a_3} \hat{p_z} \sqrt{\dfrac{2}{a_3}}\sin \dfrac{n_z \pi z}{a_3} dz = \int_{0}^{a_3} \sqrt{\dfrac{2}{a_3}}\sin \dfrac{n_z \pi z}{a_3} (-i \hbar) \dfrac{\partial}{\partial x} \sqrt{\dfrac{2}{a_3}}\sin \dfrac{n_z \pi z}{a_3} dz = \\ (\dfrac{2}{a_3}\dfrac{-i\hbar \pi}{a_3}\int_{0}^{a_3}\sin\dfrac{\pi z}{a_3}\cos\dfrac{\pi z}{a_3}dz )= 0\)

5. \( \langle z^2 \rangle \)

Use the result in Q6.b:

\(\langle \hat{z^2}\rangle = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} \psi^*z^2 \psi dxdydz= {since \ wavefunction \ is \ real}=\int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3}z^2 \psi^2 dxdydz = \int_{0}^{a_1} \int_{0}^{a_2}\int_{0}^{a_3} z^2 (\sqrt{\dfrac{2}{a_1}}\sin(\dfrac{n_x\pi x}{a_1}) \sqrt{\dfrac{2}{a_2}}\sin(\dfrac{n_y\pi y}{a_2}) \sqrt{\dfrac{2}{a_3}}\sin(\dfrac{n_z\pi z}{a_3}) )^2 dxdydz = \\ \int_{0}^{a_1} \int_{0}^{a_2}\dfrac{2}{a_1a_2} \sin^2(\dfrac{n_x\pi x}{a_1}) sin^2(\dfrac{n_y\pi y}{a_2})dxdy \dfrac{2}{a_3}\int_{0}^{a_3}z^2\sin^2\dfrac{n_z\pi z}{a_3}dz = (\dfrac{a_3^2}{3} - \dfrac{a_3^2}{2n_z^2\pi^2})\)

6. \( \langle z \rangle^2 \)

Use the result in Q11.3:

\( \langle z \rangle^2 = (\dfrac{a_3}{2})^2\)

Q12

Consider a particle of mass \(m\) in a one-dimensional box of width \(L\). Calculate the general formula for energy of the transitions between neighboring states. How much energy is required to excite the particle from the \(n=2\) to \(n=3\) state?

\(E_{n+1}- E_n= \dfrac{(n+1)^{2}h^{2}}{8mL^{2}}-\dfrac{n^{2}h^{2}}{8mL^{2}}=\dfrac{(n^2+2n+1)h^{2}}{8mL^{2}}-\dfrac{n^{2}h^{2}}{8mL^{2}}=\dfrac{(2n+1)h^{2}}{8mL^{2}}, n=1,2,.... \tag{21}\)

or

\(E_n- E_{n-1}= \dfrac{n^{2}h^{2}}{8mL^{2}}-\dfrac{(n-1)^{2}h^{2}}{8mL^{2}}=\dfrac{n^{2}h^{2}}{8mL^{2}}-\dfrac{(n^2-2n+1)h^{2}}{8mL^{2}}=\dfrac{(2n-1)h^{2}}{8mL^{2}}, n=2,3,....\)

Using \((21)\) the energy required to excite the particle from the \(n=2\) to \(n=3\) state is \(\dfrac{(2n+1)h^{2}}{8mL^{2}} = \dfrac{(2 \times 2+1)h^{2}}{8mL^{2}} = \dfrac{5h^{2}}{8mL^{2}}\).

*The coefficients may vary since \(\dfrac{h^{2}}{8m}=\dfrac{(2\pi\hbar)^{2}}{8m}=\dfrac{4\pi^{2}\hbar^{2}}{8m}=\dfrac{\pi^{2}\hbar^{2}}{2m}\).

Q13

Consider a particle of mass \(m\) in a two-dimensional square box with sides \(L\). Calculate the four lowest (different!) energies of the system. Write them down in the increasing order with their principal quantum numbers.

The energies of particle of mass \(m\) in a two-dimensional square box with sides \(L\) are

\(E_{n_x,n_y}=\dfrac{h^{2}}{8mL^{2}}({n_x^{2}}+{n_y^{2}})\), \(n_x,n_y=1,2,...\).

The four lowest (different!) energies of the system in the increasing order with their principal quantum numbers:

\(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{1^{2}})=\dfrac{2h^{2}}{8mL^{2}}=\dfrac{h^{2}}{4mL^{2}}\) \(n_x,n_y=1,1\)

\(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{2^{2}})=\dfrac{5h^{2}}{8mL^{2}}\) \(n_x,n_y=1,2\) or \(2,1\) (degenerate state)

\(\dfrac{h^{2}}{8mL^{2}}({2^{2}}+{2^{2}})=\dfrac{8h^{2}}{8mL^{2}}=\dfrac{h^{2}}{mL^{2}}\) \(n_x,n_y=2,2\)

\(\dfrac{h^{2}}{8mL^{2}}({1^{2}}+{3^{2}})=\dfrac{10h^{2}}{8mL^{2}}=\dfrac{5h^{2}}{4mL^{2}}\) \(n_x,n_y=1,3\) or \(3,1\) (degenerate state)

*The coefficients may vary since \(\dfrac{h^{2}}{8m}=\dfrac{(2\pi\hbar)^{2}}{8m}=\dfrac{4\pi^{2}\hbar^{2}}{8m}=\dfrac{\pi^{2}\hbar^{2}}{2m}\).

Q14

For a particle in a one-dimensional box of length \(L\), the second excited state wavefunction (n=3) is

\[\psi_3=\sqrt{\dfrac{2}{L}}\sin{\dfrac{3\pi x}{L}}\]

1. What is the probability that the particle is in the left half of the box?

Graphical method:

The total probability of finding the particle in the box is 1, therefore, the area under \(\psi_3^2\) the curve is 1(right figure). Since the potential is symmetric, probability distribution \(\psi_3^2\) is symmetric as well and the area under the half of \(\psi^2\) curve (shaded area) equals 0.5. Therefore, the probability that the particle is in the left half of the box equals to 0.5.

Another way is to calculate the integral \(\int_{0}^{L/2}\psi_3^*\psi_3 dx\). You will get the same result:

\[P\left(0,\dfrac{L}{2}\right)=\int_{0}^{L/2}\psi_3^*\psi_3 dx= \int_{0}^{L/2}(\sqrt{\dfrac{2}{L}}\sin{\dfrac{3\pi x}{L}})^* \sqrt{\dfrac{2}{L}}\sin{\dfrac{3\pi x}{L}} dx=\dfrac{2}{L}\int_{0}^{L/2}\sin^2{\dfrac{3\pi x}{L}} dx= \dfrac{\dfrac{L}{2}-0}{L}-\dfrac{1}{6\pi}\left({\sin({\dfrac{6\pi (\dfrac{L}{2})}{L})}-\sin({\dfrac{6\pi (0)}{L})}}\right)=\dfrac{1}{2}\]

b. What is the probability that the particle is in the middle third of the box?

Middle third of the box is where \(L/3 \leqslant x \leqslant 2L/3\).

Graphical method:

The shaded area in the right figure equals to 1/3.

Another way is to calculate the integral \(\int_{L/3}^{2L/3}\psi_3^*\psi_3 dx\). You will get the same result:

\[P\left(\dfrac{L}{3},\dfrac{2L}{3}\right)= \int_{L/3}^{2L/3}\psi_3^*\psi_3 dx= \int_{L/3}^{2L/3}(\sqrt{\dfrac{2}{L}}\sin{\dfrac{3\pi x}{L}})^* \sqrt{\dfrac{2}{L}}\sin{\dfrac{3\pi x}{L}} dx=\dfrac{2}{L}\int_{L/3}^{2L/3}\sin^2{\dfrac{3\pi x}{L}} dx= \dfrac{\dfrac{2L}{3}-\dfrac{L}{3}}{L}-\dfrac{1}{6\pi}\left({\sin({\dfrac{6\pi (\dfrac{2L}{3})}{L})}-\sin({\dfrac{6\pi (\dfrac{L}{3}) }{L})}}\right)=\dfrac{1}{3}\]

Q15

For each wavefunction below, (i) normalize the following wavefunction. (ii) sketch a plot of \( |\psi |^2 \) as a function of x.

1. \( \psi(x) = Ae^{-|x|/a_o} \) with \(A,a_o \) all real

Normalize:

\(\int_{-\infty}^{\infty}\psi^*\psi dx= 1\)

\(\int_{-\infty}^{\infty}\psi^*\psi dx= \int_{-\infty}^{\infty} A^* (e^{-|x|/a_o})^* Ae^{-|x|/a_o} dx =A^2 \int_{-\infty}^{\infty} e^{-2|x|/a_o} dx =2A^2 \int_{0}^{\infty} e^{-2x/a_o} dx=2A^2 (-\dfrac{a_o}{2}) e^{-2x/a_o} \Biggr\rvert_{0}^{\infty}= A^2 a_o (1-0)= A^2 a_o=1\)

\(A=\pm \sqrt{\dfrac{1}{a_o}}\)

Normalized wavefunction is \( \psi(x) =\pm \sqrt{\dfrac{1}{a_o}} e^{-|x|/a_o} \)

\( |\psi |^2=\psi^*\psi = A^* (e^{-|x|/a_o})^* Ae^{-|x|/a_o}=A^2 e^{-2|x|/a_o} =(\pm \sqrt{\dfrac{1}{a_o}})^2 e^{-2|x|/a_o}=\dfrac{1}{a_o} e^{-2|x|/a_o} \)

b. \( \psi(x) = A \cos (\pi x/2a) \) restricted to the region \(-a < x < a \)

\(\int_{-a}^{a}\psi^*\psi dx= \int_{-a}^{a} A^* (\cos (\pi x/2a))^* A \cos (\pi x/2a) dx=A^2 \int_{-a}^{a} \cos^2 (\pi x/2a) dx= A^2 \int_{-a}^{a} \dfrac{1+\cos(\pi x/a)}{2}dx= A^2 \int_{-a}^{a} \dfrac{1}{2}dx+ A^2 \int_{-a}^{a} \dfrac{\cos(\pi x/a)}{2}dx=(\dfrac{A^2}{2} x + \dfrac{A^2}{2} \dfrac{a}{\pi} \sin(\pi x/a)) \Biggr\rvert_{-a}^{a} = \\ \dfrac{A^2}{2} a - \dfrac{A^2}{2} (-a) + \dfrac{A^2 a}{2\pi} \sin(\pi ) - \dfrac{A^2 a}{2\pi} \sin(-\pi ) = A^2+ \dfrac{A^2 a}{\pi}\sin(\pi) = A^2=1\)

\(A=1\)

Normalized wavefunction is \( \psi(x) = \cos (\pi x/2a) \)

\( |\psi |^2=\psi^*\psi =A^* (\cos (\pi x/2a))^* A \cos (\pi x/2a)=A^2 \cos^2 (\pi x/2a) = \cos^2 (\pi x/2a)=\dfrac{1+\cos(\pi x/a)}{2}=\dfrac{1}{2} + \dfrac{1}{2} \cos(\pi x/a)\)

c. \( \psi(x,t) = Ae^{i(\omega t - kx)} \) with \(A,k,\omega\) all real

\(\int_{-\infty}^{\infty}\psi^*\psi dx= \int_{-\infty}^{\infty} A^* (e^{i(\omega t - kx)})^*Ae^{i(\omega t - kx)} dx= A^2 \int_{-\infty}^{\infty} e^{-i(\omega t - kx)} e^{i(\omega t - kx)} dx= A^2 \int_{-\infty}^{\infty} 1 dx = 2A^2 \int_{0}^{\infty} 1 dx=2A^2 \infty (?) \)

\(A= (?)\)

Normalized wavefunction is \( \psi(x) = (?) \)

For unnormalized wavefunction \( |\psi |^2=\psi^*\psi =A^* (e^{i(\omega t - kx)})^*Ae^{i(\omega t - kx)} =A^2 e^{-i(\omega t - kx)} e^{i(\omega t - kx)}=A^2 \)

d. \( \psi(x,t) = Ae^{-\lambda |x|} e^{i(\omega t)} \) with \(A,\lambda,\omega \) all real

\(\int_{-\infty}^{\infty}\psi^*\psi dx= \int_{-\infty}^{\infty} A^* (e^{-\lambda |x|} e^{i(\omega t)})^* Ae^{-\lambda |x|} e^{i(\omega t)} dx = A^2 \int_{-\infty}^{\infty} e^{-\lambda |x|} e^{-i(\omega t)} e^{-\lambda |x|} e^{i(\omega t)} dx=A^2 \int_{-\infty}^{\infty} e^{-2\lambda |x|} dx =2A^2 \int_{0}^{\infty} e^{-2\lambda x} dx=2A^2 (-\dfrac{1}{2 \lambda}) e^{-2\lambda x} \Biggr\rvert_{0}^{\infty}= \dfrac{A^2}{\lambda} (1-0)= \dfrac{A^2}{\lambda}=1\)

\(A=\pm \sqrt{\lambda}\)

Normalized wavefunction is \( \psi(x) =\pm \sqrt{\lambda} e^{-\lambda |x|} e^{i(\omega t)} \)

\( |\psi |^2=\psi^*\psi = A^* (e^{-\lambda |x|} e^{i(\omega t)})^* Ae^{-\lambda |x|} e^{i(\omega t)}=A^2 e^{-\lambda |x|} e^{-i(\omega t)} e^{-\lambda |x|} e^{i(\omega t)}=A^2 e^{-2\lambda |x|}= (\pm \sqrt{\lambda})^2 e^{-2\lambda |x|}=\lambda e^{-2\lambda |x|} \)