3220 Physics Quantum Mechanics 1 Fall 2008 Problem Set #5
Due Wednesday, September 24 at 2pm Problem 5.1: Properties of the simple harmonic oscillator. (20 points) Some of these results are discussed in class or in the book, but it's quite useful to work through them. a) Given the definitions 1 (-ip + mx) a+ 2 m h 1 a- (ip + mx) , 2 m h (1)
^ demonstrate that the simple harmonic oscillator Hamiltonian H = -( 2 /2m)( 2 /x2 ) + h 2 2 (1/2)m x can be written in two ways, 1 1 ^ H = h a+ a- + = h a- a+ - 2 2 . (2)
b) Use the energy formula En = h(n + 1/2) along with part a) to demonstrate that a- a+ un = (n + 1)un , a+ a- un = nun , (3)
where un (x) are the normalized stationary state wavefunctions. c) Use (??) and integration by parts to demonstrate that
-
f (a g) dx =
-
(a f ) g dx ,
(4)
for any functions f (x), g(x) that go to zero at infinity, f () = g() = 0; for example, they could be normalizable wavefunctions. (As we will discuss more later, this implies that a+ and a- are adjoints or Hermitian conjugates of each other.) d) The raising and lowering operators must take one stationary state to the next, times an overall constant: a+ un = cn un+1 , a- un = dn un-1 , (5)
where cn and dn are constants to be determined. Consider the expression - (a+ un ) (a+ un )dx. Evaluate it two ways, one of them using the results from parts b) and c), to solve for cn . Now consider - (a- un ) (a- un )dx and do something similar to solve for dn . You should find: a+ un = n + 1un+1 , a- un = nun-1 . (6)
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By using the formulas demonstrated in this problem, you will be able to evaluate a lot of things about the SHO without ever having to "get your hands dirty" with the explicit forms of the wavefunctions. You will see this in the next couple problems. Problem 5.2: Expectation values in the SHO. (20 points) a) Find an expression for the operators x and p in terms of the raising and lowering operators ^ ^ a+ and a- as well as constants. b) Calculate x , p , x2 and p2 in the nth stationary state using the expressions from part a). Hint: You don't ever need to write out the functional form of the un if you use results from the previous problem. c) How must H be related to the expectation values you calculated in the previous part? Check that this relationship works given what you know H must be for a stationary state. How much do the kinetic and potential energies each contribute to the total expectation value H ? d) Calculate the product of uncertainties x p for the nth stationary state and verify the Heisenberg Uncertainty Principle x p h/2. For which values of n is the minimum possible uncertainty achieved? Besides learning some basic properties of the stationary states, here you get to practice with the algebraic method of using operators. These sorts of techniques will come up again when we study angular momentum in three dimensions. Problem 5.3: Coherent states. (20 points) In this problem, consider a wavefunction (x) in the simple harmonic oscillator defined as (x) = A n un (x) , n! n=0
(7)
where is a complex number, and we assume A is chosen so as to normalize the wavefunction; don't worry about the value of A. This coherent state wavefunction has a lot of applications in optics and atomic physics, and we will explore some of its properties. a) Demonstrate that a coherent state is an eigenvector of the lowering operator with eigenvalue : a- (x) = (x) . (8)
b) Calculate the expectation values x and p . As in the last problem, you should be able to do this without writing down the exact form of the wavefunction! Equation (??) will help.
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c) Let (x, t) be the wavefunction at all times, so (x, t = 0) = (x). Demonstrate that (x, t) is also an eigenvector of a- , with eigenvalue (t); what is (t) in terms of and other quantities? d) In this part, for simplicity assume is real (but (t) might not be real). Take the results for x and p from part b) and put the value of (t) into them to find x (t) and p (t) for (x, t). How does the result compare to the classical motion of a particle in a simple harmonic oscillator? Lasers and Bose-Einstein condensates are examples of coherent states. Another property of the coherent state is that it minimzes the uncertainty, for all values of . This can be seen by calculating x2 and p2 -- if you want extra practice, take a look! Problem 5.4: Analytic solution of simple harmonic oscillator. (20 points) There are two ways to solve the harmonic oscillator. In class we pursued the method using raising and lowering operators, which is more powerful but difficult to guess if you didn't already know about it. In this problem we go through the method of solving the differential equation directly. Recall that the TISE for the simple harmonic oscillator is h2 d2 u 1 - + m 2 x2 u = Eu . 2 2m dx 2 (9)
a) It is convenient to simplify the problem by introducing a dimensionless variable. To define x/ with dimensionless, we must construct x with units of length. What are the units x of the constants h, m and ? What is the unique combination (without any extra pure numbers) that gives a unit of length? Define this as x and switch variables to x/. x Show that the equation can be written d2 u = ( 2 - K)u , d 2 where K 2E/ contains the energy E. What are the units of h/2? h b) Consider the limit of the equation far from the center, |x| . Explain why 2 this limit, leading to the approximate equation d2 u 2u . 2 d Show that in this limit, an approximate solution is u Ae-
2 /2
(10)
K in
(11)
+ Be
3
2 /2
.
(12)
Be sure to explain any terms you neglect. We have to contstrain one of A and B to make sure that u can be normalized; what is this constraint? Explain why. (Don't actually try to normalize function.) the c) The original equation turns out to simplify if we "extract" the asymptotic (large |x|) behavior of u and solve for what's left. Accordingly, define h() by means of u() h()e-
2 /2
.
(13)
Substitute this back into the full TISE (not just the asymptotic version) and show that it is equivalent to h () - 2h () + (K - 1)h() = 0 . d) To solve this differential equation, postulate a series solution for h:
(14)
h() =
j=0
aj j ,
(15)
where the aj are constants. Show that the result of part c) implies the recursion relation for the constants: aj+2 = (2j + 1 - K) aj . (j + 1)(j + 2) (16)
If we had tried to do this for u directly, we would have obtained a more complicated recursion relation. e) Normalizability of the wavefunction implies that the series (??) can't go on forever. Thus to ensure we can normalize our wavefunction, the series must stop at some point. Assume that there exists a value n such that an = 0 but an+2 = 0, and find the resulting constraint on the energy En associated with this wavefunction. The SHO wavefunctions are thus of the form un = (nth order polynomial in ) e- = (nth order polynomial in x) e
2 /2
, ,
(17) (18)
-mx2 /2 h
where the polynomials are determined by the recursion relation (??) and are called (up to an overall constant) Hermite polynomials. This agrees with the results dervied using raising and lowering operators. These sorts of methods are common in solving differential equations, and also show up in solving the radial part of the hydrogen wavefunction.
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Problem 5.5: Finding quantized energies numerically. (20 points) Only certain values of the energy E lead to normalizable solutions of the TISE for the simple harmonic oscillator. If a different E is picked, solutions exist, but they cannot be normalized they blow up at large |x|. They can blow up either to positive or negative infinity; the allowed values of E "thread the needle" in between blowing up to plus infinity or to minus infinity by going to zero instead. One can thus numerically determine the allowed energies for the simple harmonic oscillator (or a different potential) by guessing values of E and trying to find solutions that don't blow up at large |x|, tweaking your guess of E each time to get closer. Even if you don't guess the exact value of E, if you can find the cross-over region where the wavefunction switches from blowing up to positive infinity to blowing up to negative infinity, the allowed value of E must be in between. a) Consider the version of the simple harmonic oscillator TISE derived in the previous problem, 2u = ( 2 - K)u , 2 (19)
where K = 2E/ contains the energy and x/ is the rescaled position variable. h x Numerically solve and graph this equation for various values of K to find the ground state energy and plot the results. Use the boundary conditions: u(0) = 1, u (0) = 0. What property of the ground state are we assuming in order to set the first derivative equal to zero at the origin? (The choice u(0) = 1 is for convenience; we won't worry about actually normalizing the answer.) If you use Mathematica, some useful code might be K = 1.1; a = 0; b = 10; c = -10; d= 10; Plot[ Evaluate[u[z] /. NDSolve[{u''[z] - (z2 - K)*u[z] == 0, u[0] == 1, u'[0] == 0}, u[z], {z, .000000001, 10}, MaxSteps -> 10000]], {z, a, b}, PlotRange -> {c, d}, PlotStyle -> Thick] Here we used z instead of for convenience. (a, b) and (c, d) are the ranges on the two axes of the plot; feel free to tweak these to get a better view. We already know the answer: K = 1. Try values like K = 1.1 and K = 0.9 and see how the asymptotic (large-|x|) behavior flips between these values; by tuning the guesses for K one can get closer and closer to the correct value. In your write-up, give plots for the result for one choice just above K = 1 and one choice just below K = 1. (Note that by numerically solving, you are only working to a finite precision, and so even K = 1 may seem to blow up; this is because not enough significant digits are being kept. This is okay because we can still see that the asymptotic behavior flips right around K = 1, and hence the allowed energy is there to within the precision of the solution.) b) Now consider the second-lowest energy state of the SHO, the so-called "first-excited state". How must you change the boundary conditions in the numerical evaluation to find this state?
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What value of K corresponds with this state? Check that you've chosen the right boundary conditions by doing the numerical solution and seeing that the asymptotic behavior flips at this value of K. (You don't need to include the plots in your write-up.) c) Now let's do something new: consider a different system, with potential energy 1 V (x) = x4 . 2 (20)
What are the units of ? We would like to write the TISE for this potential with a dimensionless variable; show that the unique combination of , m and h that gives a unit of length is of the form x= h2 m
,
(21)
and determine the power . Now define x/ as before and rewrite the TISE as x 2u = ( 4 - K)u , 2 where K will be different from the SHO case, of the form K = E/E; what is E? d) Now find to three significant digits the values of K and the corresponding energies of both the ground state and first excited state of this system using the numerical method, assuming that = 1 in the appropriate MKS units. Don't forget to think about boundary conditions as in the SHO case. This method helps make explicit the mechanics of why only certain energies are allowed in bound state problems in quantum systems. The method gets very useful for cases where the potential is too complicated to solve analytically, as V = x4 /2 begins to show. (22)
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