Central Forces: Spring-2025
HW 06 (SOLUTION): Due W3 D5

1. Scattering S1 5242S

   Consider a very light particle of mass \(\mu\) scattering from a very heavy, stationary particle of mass \(M\). The force between the two particles is a repulsive Coulomb force \(\frac{k}{r^2}\) (neglect the gravitational force). The impact parameter \(b\) in a scattering problem is defined to be the distance which would be the closest approach if there were no interaction (See Figure). The initial velocity (far from the scattering event) of the mass \(\mu\) is \(\vec v_0\).

   Answer the following questions about this situation in terms of \(k\), \(M\), \(\mu\), \(\vec v_0\), and \(b\). (It is not necessarily wise to answer these questions in order.)

   

   1. What is the initial angular momentum of the system?

      \begin{align} \vec L &= \vec r \times \vec p\\ &= (-x \hat{x} + b \hat{y})\times \mu v_0 \hat{x}\\ &= -\mu v_0 b \hat{z} \qquad \textrm{i.e. into the page}\\ \Rightarrow& \ell=\mu v_0 b \end{align}

   2. What is the initial total energy of the system?

      \begin{equation} E_{i}=\frac{1}{2} \mu v_0^2 + \frac{k}{\sqrt{x^2 + b^2}} \approx \frac{1}{2} \mu v_0^2 \end{equation} The potential energy term in this equation can be made arbitrarily small by taking the initial position arbitrarily far from \(M\).

   3. What is the distance of closest approach \(r_{\rm{min}}\) with the interaction?

      At \(r_{\rm{min}}\), the effective potential is equal to the total energy. So, we write down this condition and solve for \(r_{\rm{min}}\). \begin{align} E_i=V_{eff}(r_{\textrm{min}})\\ E_i=\frac{\ell^2}{2\mu r_{\textrm{min}}^2} + \frac{k}{r_{\textrm{min}}}\\ E r_{\textrm{min}}^2 - k r_{\textrm{min}} - \frac{\ell^2}{2\mu}=0 \end{align} Using the quadratic formula, we obtain: \begin{align} r_{\textrm{min}} &= \frac{1}{2E}\left[ k \pm \sqrt{k^2 + \frac{4E\ell^2}{2\mu}}\right] \\ &= \frac{k}{2E}\left[ 1 \pm \sqrt{1+\frac{4E\ell^2}{2\mu k^2}}\right]\\ &= \frac{k}{\mu v_0^2}\left[ 1 \pm \sqrt{1+\frac{\ell^2\, v_0^2}{k^2}}\right]\\ &= \frac{k}{\mu v_0^2}\left[ 1 \pm \sqrt{1+\frac{(\mu v_0 b)^2\, v_0^2}{k^2}}\right]\\ &= \frac{k}{\mu v_0^2}\pm\sqrt{\frac{k^2}{\mu^2 v_0^4}+b^2} \end{align} It is a little ambiguous here to know when to stop simplifying. The expression for \(r_{\rm{min}}\) is messy no matter what you do. As long as your answer is written in terms of the required variables, you are fine.

   4. Sketch the effective potential.

      See Figure (2)

      

   5. What is the angular momentum at \(r_{\rm{min}}\)?

      Angular momentum is conserved. Therefore, the angular momentum at \(r_{\rm{min}}\) is the same as the answer to part (a), i.e. \begin{equation} -\mu v_0 b \hat{z} \; \hbox{(into the page)}. \end{equation}

   6. What is the total energy of the system at \(r_{\rm{min}}\)?

      Energy is conserved. Therefore, the energy at \(r_{\rm{min}}\) is the same as the initial total energy, \begin{equation} E_i=\frac{1}{2} \mu v_0^2 . \end{equation}

   7. What is the radial component of the velocity at \(r_{\rm{min}}\)?

      The radial component of the velocity at \(r_{\rm{min}}\) is zero.
      That is the definition of \(r_{\rm{min}}\).

   8. What is the tangential component of the velocity at \(r_{\rm{min}}\)?

      From the definition of the magnitude of the angular momentum, \(\ell=\mu r^2\dot\phi\), we can solve for the tangential component of the velocity, \(r\dot\phi\): \begin{align} r\dot\phi&=\frac{\ell}{\mu r}\\ &=\frac{\mu v_0\, b}{\mu r}\\ &=\frac{v_0\, b}{r} \end{align} Then we can evaluate this last expression at \(r_{\hbox{min}}\).

   9. What is the value of the effective potential at \(r_{\rm{min}}\)?

      The value of the effective potential at \(r_{\rm{min}}\) has to be the same as the constant total energy, \(V_{eff}=E_i=\frac{1}{2} \mu v_0^2\).

   10. For what values of the initial total energy are there bound orbits?

       Since there is no local minimum in the effective potential, there are no bound orbits for this system. I know it may seem like this is a "trick" question. But remember, in the job world, your boss may well ask you to do something that turns out to be impossible. Being able to show that something is impossible can save yourself and your company lots of time and money.

   11. Using your results above, write a short essay describing this type of scattering problem, at a level appropriate to share with another Paradigm student.

       A few of the key features:

       • Energy is conserved
       • Angular momentum is conserved
       • The distance of closest approach with the interaction is larger than without the interaction (\(r_{min}>b\))
       • There is no initial energy that would produce a bound orbit

2. Effective Potential Diagrams, version 2 S1 5242S

   Consider a mass \(\mu\) in the potential shown in the graph below.

   

   1. Sketch the effective potential if the angular momentum is not zero.

      The effective potential is the pointwise sum of these two graphs (plotted on the same axes).

      Here is the plot of the effective potential:

   2. Describe qualitatively, the shapes of all possible types of orbits, indicating the energy for each in your diagram.

      If the energy is below the minimum of the effective potential (\(r\approx 4.2\), \(E\approx -1.8\)), no orbit exists.

      If the energy is exactly at the minimum of the effective potential (\(r\approx 4.2\), \(E\approx -1.8\)),, the orbit will be a circle.

      If the energy is above the minimum of the effective potential (\(r\approx 4.2\), \(E\approx -1.8\)), there are always two classical turning points and the orbit will be bounded. Because this potential is not exactly \(1/r\) the orbits will NOT be ellipses.

3. Working with Representations on the Ring S1 5242S

   The following are 3 different representations for the \(\textbf{same}\) state on a quantum ring for \(r_0=1\) \begin{equation} \left|{\Phi_a}\right\rangle = i\sqrt{\frac{ 2}{12}}\left|{3}\right\rangle - \sqrt{\frac{ 1}{12}}\left|{1}\right\rangle +\sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i\sqrt{\frac{ 2}{ 12}}\left|{-1}\right\rangle +\sqrt{\frac{ 4}{12}}\left|{-3}\right\rangle \end{equation} \begin{equation} \left| \Phi_b\right\rangle \doteq \left( \begin{matrix} \vdots \\ i\sqrt{\frac{ 2}{12}}\\ 0 \\ -\sqrt{\frac{ 1}{12}} \\ \sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}} \\ -i\sqrt{\frac{ 2}{12}}\\ 0 \\ \sqrt{\frac{4}{12} }\\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0 \end{matrix} \end{equation} \begin{equation} \Phi_c(\phi) \doteq \sqrt{\frac{1}{24 \pi}} \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) \end{equation}

   1. With each representation of the state given above, explicitly calculate the probability that \(L_z=-1\hbar\). Then, calculate all other non-zero probabilities for values of \(L_z\) with a method/representation of your choice.

      We always calculate the probabilties as the norm squared of the inner product of our given state on the basis state corresponding to the desired measurement. For \(L_z=-1\hbar\), the desired state is \(\left|{-1}\right\rangle \): \[{\cal{P}}(L_z=-1\hbar)=\left|\left\langle {-1}\middle|{\Phi_a}\right\rangle \right|^2\] \[=\left|\left\langle {-1}\right|\left(i\sqrt\frac{ 2}{12}\left|{3}\right\rangle - \sqrt\frac{ 1}{12}\left|{1}\right\rangle +\sqrt\frac{ 3}{12}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i\sqrt\frac{ 2}{ 12}\left|{-1}\right\rangle +\sqrt\frac{ 4}{12}\left|{-3}\right\rangle \right) \right|^2\] Foiling this out, only the inner product of the ket \(\left|{-1}\right\rangle \) isn't orthogonal to \(\left\langle {-1}\right|\) and so only that inner product yields a non-zero result: \[{\cal{P}}(L_z=-1\hbar)=\left|-i\sqrt\frac{ 2}{ 12}\left\langle {-1}\middle|{-1}\right\rangle \right|^2\] \[{\cal{P}}(L_z=-1\hbar)=\left|-i\sqrt\frac{ 2}{ 12} \right|^2=\frac{2}{12}=\frac{1}{6}\]

      For the matrix notation: \begin{equation} \left|{-1}\right\rangle \doteq \left( \begin{matrix} \vdots \\ 0 \\0 \\0 \\0\\1 \\0 \\0 \\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0\end{matrix} \end{equation} We need the bra version of this vector which is the Hermitian Adjoint (the complex conjugate and transpose) \begin{equation} \left\langle {-1}\right|\doteq \stackrel{\stackrel{m=0}{\downarrow}} {\left( \begin{matrix} \cdots &0 &0 &0 &0 &1 &0 &0 &\cdots \end{matrix}\right)} \end{equation} We can now perform the needed matrix multiplication \begin{align} \left\langle {-1}\middle|{\Phi_b}\right\rangle &= \left( \begin{matrix} \cdots &0 &0 &0 &0 &1 &0 &0 &\cdots \end{matrix}\right) \left( \begin{matrix} \vdots \\ i\sqrt\frac{ 2}{12}\\ 0 \\ -\sqrt\frac{ 1}{12} \\ \sqrt\frac{ 3}{12}e^{i\frac{\pi}{4}} \\ -i\sqrt\frac{ 2}{12}\\ 0 \\ \sqrt\frac{4}{12} \\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0\end{matrix}\\ &= -i\sqrt{\frac{2}{12}} \end{align} And take the norm squared to get the probability \begin{equation} {\cal{P}}(L_z=-1\hbar)= \frac{2}{12}=\frac{1}{6} \end{equation}

      Finally for the wavefunction representation of the eigenstates, we can find the inner product: \begin{align} \left\langle {-1}\middle|{\Phi_c}\right\rangle &=\int_0^{2\pi}\Phi_{-1}^*(\phi)\Phi_c(\phi)~d\phi \\&=\int_0^{2\pi}\left(\frac{1}{\sqrt{2\pi}}e^{-i\phi}\right)^* \sqrt {\frac{1}{24 \pi}} \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) ~d\phi \\&=\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi}\left(e^{i\phi}\right) \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) ~d\phi \\&=\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi} \left( i\sqrt{2}e^{i 4 \phi} -e^{i2\phi} +\sqrt{3}e^{i\frac{\pi}{4}}e^{i\phi} -i \sqrt{2} + \sqrt{4}e^{-i 2 \phi} \right) ~d\phi \end{align} This looks like a sum of Kronecker deltas. \begin{equation} \delta_{nm}=\frac{1}{2\pi}\int_0^{2\pi}e^{i(n-m)\phi}d\phi \end{equation} This means that all terms of this integration will be zero, except the term with the \(0\) in the exponent. \begin{equation} \left\langle {-1}\middle|{\Phi_c}\right\rangle =\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi}-i \sqrt{2}d\phi = \frac{-i}{2\pi}\sqrt{\frac{1}{6}}\int_0^{2\pi}d\phi=-i\sqrt{\frac{1}{6}} \end{equation} We take the norm squared of this inner product to get the probability, so: \[\cal{P}(L_z=-1\hbar)=\frac{1}{6}\] Through any of the representation, we can find the remaining non-zero probabilities, with the easiest way being to read off the probabilities from from the ket representatio n and norm square them: \[{\cal{P}}(L_z=3\hbar)=\frac{1}{6}\] \[{\cal{P}}(L_z=1\hbar)=\frac{1}{12}\] \[{\cal{P}}(L_z=0\hbar)=\frac{1}{4}\] \[{\cal{P}}(L_z=-1\hbar)=\frac{1}{6}\] \[{\cal{P}}(L_z=-3\hbar)=\frac{1}{3}\]

   2. Explain how you could be sure you calculated all of the non-zero probabilities.

      The easiest way would be to check if the sum of all the probabilities add into 1, and they do. Or you can argue that only kets represented in the state can be thought of as having non-zero probabilities of measurement.

   3. If you measured the \(z\)-component of angular momentum to be \(3\hbar\), what would the state of the particle be immediately after the measurement is made?

      It will be in the \(\vert 3\rangle\) state, since the function is unambigiously in that state based on our measurment (this simple case will always happen when non-degenerate values are measured).

   4. With each representation of the state given above, explicitly calculate the probability that \(E=\frac{9}{2}\frac{\hbar^2}{I}\). Then, calculate all other non-zero probabilities for values of \(E\) with a method of your choice.

      The big thing to remember here is that energies on the ring can be degenerate, which means we need to take the probabilities of being in the \(\left|{+m}\right\rangle \) and \(\left|{-m}\right\rangle \) and add them together in order to get the total probability. Like so: \[{\cal{P}}\left(E=\frac{m^2\hbar^2}{2I}\right)=\left|\left\langle {+m}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-m}\middle|{\Phi}\right\rangle \right|^2\] The individual inner products and probabilities will be exactly the same as for \(L_z\) we just need to add them together properly to take account for the degeneracy in the energies: \[{\cal{P}}\left(E=\frac{9}{2}\frac{\hbar^2}{I}\right)=\left|\left\langle {3}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-3}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\] \[{\cal{P}}\left(E=\frac{1}{2}\frac{\hbar^2}{I}\right)=\left|\left\langle {1}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-1}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{12}+\frac{1}{6}=\frac{1}{4}\] \[{\cal{P}}\left(E=0\frac{\hbar^2}{I}\right)=\left|\left\langle {0}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{4}\]

   5. If you measured the energy of the state to be \(\frac{9}{2}\frac{\hbar^2}{I}\), what would the state of the particle be immediately after the measurement is made?

      Because the state corresponding to this energy is degenerate, we need to consider the a superposition of states coming out, and use the projection postulate! \[\left|{\Psi}\right\rangle = \frac{P_n\left|{\Phi}\right\rangle }{\sqrt{\langle{\Phi}|P_n|{\Phi}\rangle}},\] where \(P_n\) is the projection operator for the subset of eigenstates associated with \(\frac{9}{2}\frac{\hbar^2}{I}\) and \(\left|{\Psi}\right\rangle \) is the new state after the measurement. Therefore, our projection operator is:

      \[P_n = \left|{3}\right\rangle \left\langle {3}\right| + \left|{-3}\right\rangle \left\langle {-3}\right|.\]

      I'll do the calculation in pieces so that it fits on the page: \begin{align} P_n\left|{\Phi}\right\rangle &=\left(\left|{3}\right\rangle \left\langle {3}\right| + \left|{-3}\right\rangle \left\langle {-3}\right|\right) \left(i{\scriptstyle\sqrt{\frac{ 2}{12}}}\left|{3}\right\rangle - {\scriptstyle\sqrt{\frac{ 1}{12}}}\left|{1}\right\rangle +{\scriptstyle\sqrt{\frac{ 3}{12}}}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i{\scriptstyle\sqrt{\frac{ 2}{ 12}}}\left|{-1}\right\rangle +{\scriptstyle\sqrt{\frac{ 4}{12}}}\left|{-3}\right\rangle \right)\\ &=\left(\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}} \right) \end{align} and \begin{align} \left\langle {\Phi}\right|P_n\left|{\Phi}\right\rangle &=\left(-i{\scriptstyle\sqrt{\frac{ 2}{12}}}\left\langle {3}\right| - {\scriptstyle\sqrt{\frac{ 1}{12}}}\left\langle {1}\right| +{\scriptstyle\sqrt{\frac{ 3}{12}}}e^{-i\frac{\pi}{4}}\left\langle {0}\right| +i{\scriptstyle\sqrt{\frac{ 2}{ 12}}}\left\langle {-1}\right| +{\scriptstyle\sqrt{\frac{ 4}{12}}}\left\langle {-3}\right| \right) \left(\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}} \right)\\ &= (-i{\scriptstyle\sqrt{\frac{ 2}{12}}})(i{\scriptstyle\sqrt{\frac{ 2}{12}}}) +({\scriptstyle\sqrt{\frac{ 4}{12}}})({\scriptstyle\sqrt{\frac{ 4}{12}}})\\ &=\frac{1}{2} \end{align}

      Thus, the state of the particle immediately after the measurement of \(\displaystyle E = \frac{9}{2}\frac{\hbar^2}{I}\) is: \begin{align} \left|{\Psi}\right\rangle &= \frac{P_n\left|{\Phi}\right\rangle }{\sqrt{\langle{\Phi}|P_n|{\Phi}\rangle}}\\ &=\frac{\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}}} {\sqrt{\frac{1}{2}}}\\ &=i\sqrt{\frac{1}{3}}\left|{3}\right\rangle +\sqrt{\frac{2}{3}}\left|{-3}\right\rangle \end{align}

      Sensemaking: Notice that this state is normalized, as it must be, i.e. the sum of the squares of the norms of the coefficients is equal to one.

4. Confidence Rating S1 5242S 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?