Central Forces: Spring-2025
HW 03 2025 (SOLUTION): Due W2 D2

1. Recurrence Relations S1 5248S For each of the problems below, suppose you have been solving a differential equation using power series methods around the indicated point and you have derived the indicated recurrence relation. Write out the first five nonzero terms in the power series expansion. If the recurrence relation allows two solutions, write out the first four nonzero terms in each such solution.

   1. In an expansion around the point \(z=1\), the recurrence relation is: \[a_{n+1}=\frac{1}{n+1}\, a_n\]

      Sensemaking: The solution will look like \[y(z)= A\left[a_0+a_1 (z-1)+a_2 (z-1)^2+a_3 (z-1)^3+a_4 (z-1)^4+\dots\right]\] where my job is to find and plug in the \(a_n\)'s. The given recurrence relation involves only the indices \(n\) and \(n+1\) as would be typical for a first order ODE, so I expect one unknown constant, call it \(A\).

      Using a recurrence relation, you will never find a value for the first one or two coefficents. These early coefficients will become the unknown constants in the solution of the homogeous differential equation. So let's set \(a_0=A\).

      Then, the recurrence relation for \(n=0\) tells me that \[a_1=\frac{1}{0+1}\, a_0=a_0=A.\]

      The recurrence relation for \(n=1\) tells me that \[a_2=\frac{1}{1+1}\, a_1= \frac{1}{2}\, a_1.\] But I can't stop there. I need every coefficient in terms of \(a_0=A\), so I plug in the result from \(n=0\): \[a_2=\frac{1}{2}\, a_1=\frac{1}{2}\, a_0=\frac{1}{2}\, A.\]

      Continuing in this way, I find that: \begin{align} a_3 &= \frac{1}{3}\, a_2 =\frac{1}{3}\frac{1}{2}\, A =\frac{1}{3!}\, A\\ a_4 &= \frac{1}{4}\, a_3 =\frac{1}{4}\frac{1}{3!}\, A =\frac{1}{4!}\, A \end{align} Notice that I am leaving the coefficients with factorials in them so that I can see the pattern that is developing.

      I now have 5 nonzero coefficients, so I can stop calculating, but I'm not done. I still need to write out all the terms in the power series. I'll call the unknown function that I am trying to find \(f\). I'm told that the recurrence relation is for an expansion around \(z=1\), so I expect to have powers of \(z-1\). Altogether: \[y(z)= A\left[1+(z-1)+\frac{1}{2!}\, (z-1)^2+\frac{1}{3!}\, (z-1)^3+\frac{1}{4!}\, (z-1)^4+\dots\right]\] Don't forget the \(a_0\) term which looks like \((z-1)^0=1\)! Note that in this case, because there is an \(n=0\) term, five nonzero terms means that this solution is correct to fourth order.

   2. In an expansion around the point \(z=0\), the recurrence relation is: \[a_{n+2}=-\frac{(5-n)(6+n)}{(n+2)(n+1)}\, a_n\]

      Sensemaking: The solution will look like \[y(z)= A\left[a_0+a_2 z^2+a_4 z^4+a_6 z^6+\dots\right] +B\left[a_1 z+a_3 z^3+a_5 z^5 +a_7 z^7+\dots\right]\] where my job is to find and plug in the \(a_n\)'s. The given recurrence relation involves only the indices \(n\) and \(n+2\) as would be typical for a second order ODE, so I expect two unknown constants, call them \(A\) and \(B\). One of the solutions will be even and one of the solutions will be odd.

      Letting \(a_0=A\), where \(A\) is an unknown constant and using the same strategy as above, for \(n=0\), \(2\) and \(4\), I get: \begin{align} a_2 &= -\frac{5\cdot 6}{2\cdot 1}\, a_0= -\frac{5\cdot 6}{2\cdot 1}\, A \\ a_4 &= -\frac{3\cdot 8}{4\cdot 3}\, a_2 =\frac{5\cdot 3\cdot 6\cdot 8}{2\cdot 4\cdot 1\cdot 3}\, A \\ a_6 &= -\frac{1\cdot 10}{6\cdot 5}\, a_4 =-\frac{5\cdot 3\cdot 1\cdot 6\cdot 8\cdot 10}{2\cdot 4\cdot 6\cdot 1\cdot 3\cdot 5}\, A \\ \end{align}

      I will never get the odd terms from the sequence above. Instead, letting \(a_1=B\), where \(B\) is an unknown constant and using the same strategy as above, for \(n=1\), \(3\) and \(5\), I get: \begin{align} a_3 &= -\frac{4\cdot 7}{3\cdot 2}\, a_1= -\frac{4\cdot 7}{3\cdot 2}\, B \\ a_5 &= -\frac{2\cdot 9}{5\cdot 4}\, a_3 =\frac{4\cdot 2\cdot 7\cdot 9}{3\cdot 5\cdot 2\cdot 4}\, B \\ a_7 &= -\frac{0\cdot 11}{7\cdot 6}\, a_5 =0\, B \\ \end{align} Notice that the \(B\) series terminates. All the coefficients beyond \(a_5\) will be zero.

      Putting it all together, I get: \begin{align} y(z)&= A\left[1-\frac{5\cdot 6}{2\cdot 1} z^2+\frac{5\cdot 3\cdot 6\cdot 8}{2\cdot 4\cdot 1\cdot 3} z^4 -\frac{5\cdot 3\cdot 1\cdot 6\cdot 8\cdot 10}{2\cdot 4\cdot 6\cdot 1\cdot 3\cdot 5} z^6+\dots\right]\\ &+B\left[1 z-\frac{4\cdot 7}{3\cdot 2} z^3+\frac{4\cdot 2\cdot 7\cdot 9}{3\cdot 5\cdot 2\cdot 4} z^5 \right] \end{align} Notice that if you are asked for a certain number of nonzero terms and the series terminates, you should just give as many nonzero terms as there are.

2. ODE Power Series Solutions One S1 5248S

   Consider the differential equation \(y^{\prime\prime} - 2y^{\prime} + y = 0\).

   1. Use the power series method to find the first six terms in each of two independent solutions to this differential equation.

      Our answer at the end should have the form \(y(x) \approx A(c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5) + B(d_0 + d_1x + d_2x^2 + d_3x^3 + d_4x^4 + d_5x^5)\). There are two solutions because it is a second-order differential equation. Our task is to find the \(c\) and \(d\) coefficients.
      We start by assuming \begin{equation} y(x) = \sum_{n=0}^\infty c_nx^n, \end{equation} giving \begin{equation} y^\prime(x) = \sum_{n=1}^\infty nc_nx^{n-1} \end{equation} and \begin{equation} y^{\prime\prime}(x) = \sum_{n=2}^\infty n(n-1)c_nx^{n-2} \end{equation} Rearranging the terms in the sums gives us:
      \begin{equation} \sum_{n=2}^\infty n(n-1)c_nx^{n-2} - 2\sum_{n=1}^\infty nc_nx^{n-1} + \sum_{n=0}^\infty c_nx^n = 0 \end{equation} \begin{equation} \sum_{n=0}^\infty ((n+2)(n+1)c_{n+2} - 2(n+1)c_{n+1} + c_n)(x^n) = 0 \end{equation} The parentheses must be equal to 0 for all \(n\), so:
      \begin{equation} c_{n+2} = \frac{2(n+1)c_{n+1} - c_n}{(n+1)(n+2)} \end{equation} We have the freedom to choose both \(c_0\) and \(c_1\). A first guess might be to try two solutions where these coefficients are each zero (with the other nonzero). This gives:
      \begin{align} &n=0: c_2 = -c_0/2 \\ &n=1: c_3 = 2c_2/3 = -c_0/3 \\ &n=2: c_4 = c_3/2 - c_2/12 = -c_0/8 \\ &n=3: c_5 = 2c_4/5 - c_3/20 = -c_0/30 \\ &n=4: c_6 = c_5/3 - c_4/30 = -c_0/144 \end{align} \(y_0(x) \approx c_0(1 - x^2/2 - x^3/3 - x^4/8 - x^5/30)\) does not immediately look like a familiar power series. If you want a cleverer choice (see next part), choose \(c_0 = c_1\). The other solution we try \(c_0 = 0\):
      \begin{align} &n=0: d_2 = d_1 \\ &n=1: d_3 = d_1/2 \\ &n=2: d_4 = d_3/2 - d_2/12 = d_1/6 \\ &n=3: d_5 = 2d_4/5 - d_3/20 = d_1/24 \\ &n=4: d_6 = d_5/3 - d_4/30 = d_1/120 \\ \end{align} This looks very familiar! \(y_1(x) \approx d_1(x + x^2 + x^3/2 + x^4/6 + x^5/24) = d_1xe^x\).
      The final answer puts these two solutions together as: \(y(x) \approx c_0(1 - x^2/2 - x^3/3 - x^4/8 - x^5/30) + d_1(x + x^2 + x^3/2 + x^4/6 + x^5/24)\)

   2. Solve this differential equation using a different method and show that your answers are the same as part a.

      This equation can also be solved using the method of constant coefficients. The characteristic polynomial is: \(r^2 - 2r + 1 = (r - 1)^2\). Since this equation has one repeated solution, an additional independent solution must be added in the form of \(y(x) = Ae^x + Bxe^x\). The second term matches one of our solutions from before. The other term can be found from the series solution either by an appropriate superposition of the two solutions we found, or by going back to the series coefficients and choosing the initial condition of \(c_1 = c_0\).

3. Quantum Particle in a 2D Box with Time Dependence S1 5248S The eigenstates for a quantum mechanical particle inside a 2-dimensional infinite potential well with sides of length \(L_x\), \(L_y\) are \[\left|{n,m}\right\rangle \doteq \sqrt{\frac{2}{L_x}}\sin{\frac{n_x\pi x}{L_x}} \sqrt{\frac{2}{L_y}}\sin{\frac{n_y\pi y}{L_y}}\]
   1. Find an exact expression for the initial wave function given by: \[\psi(x,y,0)= \frac{30}{\sqrt{L_x^5 L_y^5}}(L_x x-x^2)(L_y y-y^2)\] I have chosen coordinates so that one of the corners of the box is at the origin and all of the box is in the first quadrant (i.e. all positive values of the spatial coordinates).
   2. Find an exact expression for the wave function as a function of time.
   3. Bonus points: Plot an approximation for the probability density at \(t=0\) and at an interesting later time. Explain why you chose the later time that you did. Explain how you chose your approximation scheme and why.

   In the previous problem, we found the normalized energy eigenstates \begin{align} \left|{n_x n_y}\right\rangle &\doteq \sqrt{\frac{2}{L_x}}\sqrt{\frac{2}{L_y}}\sin\left( \frac{n_x\pi x}{L_x} \right)\sin\left( \frac{n_yy\pi y}{L_y} \right)e^{-iE_{n_xn_y}t/\hbar}\\ E_{n_xn_y}&= \frac{\pi^2\hbar^2}{2m^2}\left( \frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2} \right) \end{align}

   At this point we are finally prepared to write the full time-dependent wave function for the initial state \(\psi(x,y,0)\). Recall that we can do this by expanding the function in the energy basis, that is \begin{equation} \psi(x,y,t) = \sum_{n_x}\sum_{n_y}c_{n_xn_y}\sqrt{\frac{2}{L_x}}\sin\left( \frac{n_x\pi x}{L_x} \right) \sqrt{\frac{2}{L_y}} \sin\left( \frac{n_yy\pi y}{L_y} \right)e^{-iE_{n_xn_y}t/\hbar} \end{equation} which we would write in “ket-language” as \begin{equation} |\psi\rangle = \sum_{n_x}\sum_{n_y}c_{n_xn_y}e^{-iE_{n_xn_y}t/\hbar}|n_x,n_y\rangle \end{equation} All that we need to determine are the coefficients \(c_{n_xn_y}\). By now, this process should be familiar to you. The coefficients are given by \begin{align} c_{n_xn_y} &= \langle n_x,n_y|\psi \rangle\\ &= \int_0^{L_y}\int_0^{L_x}\sqrt{\frac{2}{L_x}}\sin\left( \frac{n_x\pi x}{L_x} \right) \sqrt{\frac{2}{L_y}} \sin\left( \frac{n_yy\pi y}{L_y} \right)\nonumber \\ &\qquad\times\frac{30}{\sqrt{L_xL_y}}(L_xx-x^2)(L_yy-y^2)\;dxdy \\ &= \frac{240\Big(-1+(-1)^{n_x}\Big)\Big(-1+(-1)^{n_y}\Big)}{n_x^3n_y^3\pi^6} \\ &= \begin{cases} \displaystyle{\frac{960}{n_x^3n_y^3\pi^6}}, & n_x,n_y\text{ are both odd} \\\ \;\;0, & \text{ otherwise} \end{cases} \end{align} It is perfectly reasonable to use Mathematica for this type of integral. If you do, take advantage of the Assumptions feature to tell Mathematica that \(n_x,n_y\) are integers .

   Our final, time-dependent wave function is given by \begin{equation} \psi(x,y,t) = \mathop{\sum\sum}_{n_x,n_y\text{odd}} \frac{960}{n_x^3n_y^3\pi^6}\sqrt{\frac{2}{L_x}}\sin\left( \frac{n_x\pi x}{L_x} \right) \sqrt{\frac{2}{L_y}} \sin\left( \frac{n_yy\pi y}{L_y} \right)e^{-iE_{n_xn_y}t/\hbar} \end{equation}

4. Confidence Rating S1 5248S After solving each problem on the assignment, indicate your answers to the following questions for each problem. Answer for the problem as a whole, even if the problem has multiple parts.
   1. Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?
      
      For the rest of the questions, assume you have interpreted the problem correctly
   2. Problem Confidence How confident are you that you could independentl come up with a correct solution process to a similar problem on a future problem set?
      
   3. Answer Confidence How confident are you that your final answer to this question is correct (not solution process)?
      
   4. Makes Sense To what degree do you understand how your answer fits (or does not fit) the physical or mathematical situation of the problem?