Physics 7440 Spring 2003
Solutions 2
1. Introduction to variational wavefunctions
This problem is a tune-up to get you familiar with the idea of variational wavefunctions. Throughout the course, we will require approximate ground state wavefunctions for solids. The variational technique is a fine way to start. In this problem, you were supposed to construct three different guesses for a possible ground state wavefunction for the hydrogen atom. This problem is nice because we actually know the exact answer for the ground state. Therefore, its easy to check how good a job the approximations actually do.
In each case below, we follow the same sequence of steps:
- We guess a functional form for the ground state wavefunction. In this particular case, all the wavefunctions are chosen to be spherically symmetric (they only depend upon the radius).
- We adjust the prefactor to assure that the wavefunctions are properly normalized.
- We calculate the expectation value of the system energy. This calculated energy will be a function of the remaining parameters in the wavefunction.
- We minimize the energy function to find the values of the wavefunction parameters which result in the lowest possible energy. This lowest value of energy is our guess for the ground state energy. Plugging the minimizing parameters back into the wavefunction yields our best guess (for that function) for the shape of the ground state wavefunction.
All variational calculations follow these steps. For many particle systems, the wavefucntion will have many parameters, but the idea is identical.
a) A really good guess.
We started by guessing a variational wavefunction that happens to be of exactly the same functional form as the known exact ground state. My point here was to show you that if you happen to make a great guess, the variation technique will give you the exact ground state. In this case, we guessed the function:
Now let’s go through the rest of the steps:
Make sure that it is normalized:
Normalization means that the probability of finding the electron somewhere in all of space is unity. Functionally, it means that we require the integrated square of the wavefucntion to yield a value of 1 i.e.,
In this problem set, all the wavefunctions are real-valued functions, so the complex conjugate of the wavefunction is just the wavefunction. For our exponential guess, we have the requirement that:
Convert the integral to spherical coordinates and rewrite it in terms of a new variable, :
The definite integral is just some number. Look it up in any integral table, or use Mathematica to determine that the number is just the Gamma function of 3:
G functions are often referred to as the Generalized Factorials because they satisfy the relation:
Iteration of this relation shows you that . Thus, the normalization relation shows that the A and B parameters must be related by:
Of course, this result should immediately look sensible to you (at least upon reflection): The pre-factor, A, assures the normalization. Therefore, it should have a value, which is the square root of the typical volume occupied by the electron. For the exponential case, the electron is most probably going to be found inside a radius of r=B. Therefore, the typical volume within which the electron will be found is just , so we might well have guessed that A would be something like .
In any case, the wavefunction now depends only upon one parameter. The other is specified by the normalization condition.
Energy expectation value:
To get the expected energy value, we take the expectation value of the Hamiltonian:
I find it easier to break this apart into two integrals, one for the average kinetic energy and a second one for the average potential energy:
Again, this problem works out best if done in spherical coordinates. Again, since there is no angular dependence in the wavefunction, we use:
Thus, in the kinetic energy integral, the double gradient results in two pieces. The potential energy piece remains a single term. We find:
Again, we rewrite in terms of the dimensionless variable, , and identify the G-functions. We get:
Plug in the relationship between A and B to find our final result for the expectation value:
The first term is the kinetic energy due to the characteristic derivative of the wavefunction. The second term says that the electron is typically at a distance of B from the proton, so there is a typical Coulombic energy.
Minimization of the energy:
Now look for the value of B that makes the energy as small as possible:
Here, I have identified the Bohr radius, a0. Plugging this value for B back into the energy gives us:
This result is the exact ground state energy of the hydrogen problem (where Z=1 for actual hydrogen and other Z values are used for a single electron around an arb. nucleus). Plugging this value for B into the wavefunction yields:
We find the exact hydrogenic ground state.
b) A different guess.
Now try a variational wavefunction of the form:
Make sure that it is normalized:
We require that:
Again, convert the integral to spherical coordinates and rewrite it in terms of a new variable, :
The definite integral is again just some number. Mathematica evaluates it as , so that we find:
This result is very similar to the relation we found for the previous wavefunction guess. As you can imagine, all the normalization relations will be similar, but for numerical factors.
Energy expectation value:
This time, we have a different guess, but the expectation value problem is similar. Again, it pays to do the problem in spherical coordinates. Again, the derivatives in the kinetic energy piece result in that term breaking into two pieces. Again, we have a single integration for the potential energy piece. We find that:
With the substitution, , the integrals become dimensionless:
Each of these integrals is some definite number. Respectively, Mathematica evaluates them as . Substitute in and use the relation between A and B from normalization to get:
The first term is the kinetic energy due to the characteristic derivative of the wavefunction; it is three times larger than what we found for the first wavefunction due to the larger curvature. The second term is again a Coulombic energy, but with a somewhat different length scale.
Minimization of the energy:
Now look for the value of B that makes the energy as small as possible:
Where again I have identified the Bohr radius, a0. Plugging this value for B back into the energy gives us:
This result is almost as low as the exact ground state energy of the hydrogen problem. It misses being as low by the factor of , i.e., it is about 15% higher than the true ground state energy. The approximate wavefunction is then:
c) A third guess.
Finally, try a variational wavefunction of the form:
i.e., a Lorentzian form.
Make sure that it is normalized:
We require that:
With the substitution that, , we find that:
Energy expectation value:
This time we get:
With the substitution, , the integrals become dimensionless:
Respectively, Mathematica evaluates them as . Substitute in and use the relation between A and B from normalization to get:
The kinetic energy is smaller than what we found for the first wavefunction due to the smaller curvature. The second term is again a Coulombic energy.
Minimization of the energy:
Again, we minimize to find:
Where again I have identified the Bohr radius, a0. Plugging this value for B back into the energy gives us:
This result is again, almost as low as the exact ground state energy of the hydrogen problem. It misses being as low by the factor of , i.e., it is about 19% higher than the true ground state energy. The approximate wavefunction is then:
d) Comparisons.
The first interesting thing about the variational method is that it does a surprisingly good job of nailing down the ground state energy. Notice that we used three rather different functions, two types of exponential and an algebraic, and all of them gave the ground state energy to within 20%. The general fact is that the ground state energy estimate from the variational technique is far better than the quality of the wavefunction might suggest. It also tends to be better than estimates of other quantities.
As an example of one such quantity, the characteristic length scale, B, which determines roughly at what distance the electron will be found from the nucleus, varies by more than a factor of 2 among the three wavefunctions.
Most of you had no trouble plotting things. If you think about the integrand of the normalization integral, you can see that it represents the probability per unit radial distance that the electron will be found between r and r+dr. Thus, it is the probability density of finding the electron near a given radius. In each case, this probability per unit radius is of the form:
For the three cases above, we therefore find:
The natural unit of length for putting these on the same plot would be the Bohr radius divided by the nuclear charge number. Also, since the probablity is per unit length, the natural dimensionless probability is just . Then, the dimensionless probabilities vs. dimensionless radius are given by:
Here is a plot with the three probability distributions together, which shows the variation in most probable radius I discussed above:
2. Variational wavefunctions and the Hartree Hamiltonian.
As I said in the original assignment, this problem is shamelessly stolen in part from Ashcroft and Mermin, Prob 17.1. The goal of this problem is to show how more general guesses for wavefunctions can be handled with the variational calculations. In the simple hydrogen case above, we had simple enough functions that direct minimization was an argument. In this case, we don't know anything about the explicit single particle functions, we only know that the many-body wavefunction is a product of functions of the individual coordinates. However, the variational minimization of the energy expectation value eventually tells us how to find this single-particle functions.
a) Expectation value of the hamiltonian.
Suppose that you guess a variational wavefunction of the form of a product of single particle wavefunctions, like Marder's (6.79). Take the expectation value of Eq. (1) above with the product wavefunction (6.79) and show that you get:
Assume that the single particle wavefunctions are normalized.
We are told to assume that:
The individual single-particle wavefunctions are normalized. Then, the expectation value of the hamiltonian is:
The first two pieces are single particle pieces and are fairly simple to consider. Therefore, concentrating on the first two terms only, we get:
or, since each of the single particle functions is normalized, we have:
Dropping back out of Dirac notation, this result is just:
We are half way to the desired result. The third term is more complicated, because the potential energy between pairs of electrons causes mixing of the associated pairs of wavefunctions. Let's just write things out explicitly:
Again, noticing that the single particle functions are normalized we find:
Thus, we have the total desired result:
b) Variational calculation.
Now, let's see what the variational calculation tells us about the single-particle wavefunctions. As for the case of hydrogen, we want to look for an extremum in the expectation value of the energy. In other words, given an assumed product wavefunction:
and some evaluated expectation value:
In this approach, I am explicitly normalizing by the squared magnitude of the wavefunction, so that the expectation value is independent of the normalization of the wavefunction. Thus, we don't need the Lagrange multipliers (thanks to Rob Chiaramonte!). We want to find the conditions on the single particle wavefunctions such that the expectation value is stationary with respect to variation in the single particle functions:
where
Continuing with this notation, we can rewrite the variation in expectation value as:
In the second line, we have only retained terms to first order in the variation. Next, expand the denominator, again to first order in the variation to get:
Thus, the total variation is:
We insist that this variation be zero for a good estimate of the ground state energy and wavefunction. Since the variation of the wavefunction and its complex conjugate are independent, each of the two terms above must be independently zero. Let's consider the second one:
Now, we need to directly insert the assumed product wavefunction and an explicit variation. I suggested that the most obvious variation is to simply replace one of the single particle functions by some other single particle function. In other words, change one function out of a mole of functions. That certainly sounds like a small change. For example, let's change the first single particle function to some other one:
Then, we have:
At this point, you can continue with exactly the same math and manipulations as in part a) to whittle the various pieces down into distinct terms. Again, let's look at the single particle pieces:
You get two terms. The first is the expectation value of the single particle parts of the hamiltonian taken between the varied state and the original state. The second part has the sum of all the other single particle energies over the entire system excluding the electron of interest and all multiplied by some overlap integral between the old single particle state and the new state. Similarly, if you look at the two particle operator, you find two terms: