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Grading notes.
The pdf version above has been adjusted with some grading commentary. [Click here for the PDF for the original submission, as found below.
Problem 1.
Statement
Is it possible to derive the eigenvalues and eigenvectors presented in Section 8.2 from those in Section 8.1.2? What does this say about the potential energy operator in these two situations?
For reference 8.1.2 was a finite potential barrier, , and zero in the interior of the well. This had trigonometric solutions in the interior, and died off exponentially past the boundary of the well.
On the other hand, 8.2 was a delta function potential , which had the solution , where .
Solution
The pair of figures in the text [1] for these potentials doesn’t make it clear that there are possibly any similarities. The attractive delta function potential isn’t illustrated (although the delta function is, but with opposite sign), and the scaling and the reference energy levels are different. Let’s illustrate these using the same reference energy level and sign conventions to make the similarities more obvious.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{FiniteWellPotential}
\caption{8.1.2 Finite Well potential (with energy shifted downwards by )}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{deltaFunctionPotential}
\caption{8.2 Delta function potential.}
\end{figure}
The physics isn’t changed by picking a different point for the reference energy level, so let’s compare the two potentials, and their solutions using outside of the well for both cases. The method used to solve the finite well problem in the text is hard to follow, so re-doing this from scratch in a slightly tidier way doesn’t hurt.
Schr\”{o}dinger’s equation for the finite well, in the region is
where a positive bound state energy has been introduced.
Writing
the wave functions outside of the well are
latex x a$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.3)$
Within the well Schr\”{o}dinger’s equation is
or
Noting that the bound state energies are the values, let , so that the solutions are of the form
As was done for the wave functions outside of the well, the normalization constants can be expressed in terms of the values of the wave functions on the boundary. That provides a pair of equations to solve
Inverting this and substitution back into 2.6 yields
Expanding the last of these matrix products the wave function is close to completely specified.
latex x < -a$} \\ u(a) \frac{\sin(\alpha (a + x))}{\sin(2 \alpha a)} +u(-a) \frac{\sin(\alpha (a – x))}{\sin(2 \alpha a)} & \quad \mbox{$latex {\left\lvert{x}\right\rvert} a$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.8)$
There are still two unspecified constants and the constraints on have not been determined (both and are functions of that energy level). It should be possible to eliminate at least one of the by computing the wavefunction normalization, and since the well is being narrowed the term will not be relevant. Since only the vanishingly narrow case where is of interest, the wave function in that interval approaches
Since no discontinuity is expected this is just . Let’s write for short, and the limited width well wave function becomes
latex x 0$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.10)$
This is now the same form as the delta function potential, and normalization also gives .
One task remains before the attractive delta function potential can be considered a limiting case for the finite well, since the relation between , and has not been established. To do so integrate the Schr\”{o}dinger equation over the infinitesimal range . This was done in the text for the delta function potential, and that provided the relation
For the finite well this is
In the limit as this is
Some care is required with the term since as , but the term is unambiguously killed, leaving
The exponential vanishes in the limit and leaves
Comparing to 2.11 from the attractive delta function completes the problem. The conclusion is that when the finite well is narrowed with , also letting such that the absolute area of the well is maintained, the finite potential well produces exactly the attractive delta function wave function and associated bound state energy.
Problem 2.
Statement
For the hydrogen atom, determine and such that and is the radial position operator . What do these quantities represent physically and are they the same?
Solution
Both of the computation tasks for the hydrogen like atom require expansion of a braket of the following form
where or .
The spherical representation of the identity resolution is required to convert this braket into integral form
where the spherical wave function is given by the braket .
Additionally, the radial form of the delta function will be required, which is
Two applications of the identity operator to the braket yield
To continue an assumption about the matrix element is required. It seems reasonable that this would be
The braket can now be written completely in integral form as
Application of the delta functions then reduces the integral, since the only , and dependence is in the (orthonormal) terms they are found to drop out
This leaves just the radial wave functions in the integral
As a consistency check, observe that with , this integral evaluates to 1 according to equation (8.274) in the text, so we can think of as the radial probability density for functions of .
The problem asks specifically for these expectation values for the state. For that state the radial wavefunction is found in (8.277) as
The braket can now be written explicitly
Now, let’s consider the two functions separately. First for we have
The last integral evaluates to , leaving
The expectation value associated with this state for the radial position is found to be proportional to the Bohr radius. For the hydrogen atom where this average value for repeated measurements of the physical quantity associated with the operator is found to be 5 times the Bohr radius for states.
Our problem actually asks for the inverse of this expectation value, and for reference this is
Performing the same task for
This last integral has value , and we have the second part of the computational task complete
The question of whether or not 3.24, and 3.25 are equal is answered. They are not.
Still remaining for this problem is the question of the what these quantities represent physically.
The quantity is the expectation value for the radial position of the particle measured from the center of mass of the system. This is the average outcome for many measurements of this radial distance when the system is prepared in the state prior to each measurement.
Interestingly, the physical quantity that we associate with the operator has a different measurable value than the inverse of the expectation value for the inverted operator . Regardless, we have a physical (observable) quantity associated with the operator , and when the system is prepared in state prior to each measurement, the average outcome of many measurements of this physical quantity produces this value , a quantity inversely proportional to the Bohr radius.
ASIDE: Comparing to the general case.
As a confirmation of the results obtained, we can check 3.24, and 3.25 against the general form of the expectation values for various powers of the radial position operator. These can be found in locations such as farside.ph.utexas.edu which gives for (without proof), and in [2] (where these and harder looking ones expectation values are left as an exercise for the reader to prove). Both of those give:
It is curious to me that the general expectation values noted in 3.26 we have a quantum number dependence for , but only the quantum number dependence for . It is not obvious to me why this would be the case.
References
[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.
[2] R. Liboff. Introductory quantum mechanics. Cambridge: Addison-Wesley Press, Inc, 2003.