PHYS 424 Notes

Our next problem is the harmonic oscillator, solving Hy = Ey where H=p²/(2m) + (1/2)mw²x² or

where w²=k/m

There are two ways to solve this equation: directly, by power-series substitution, and indirectly by algebraic manipulations. We will do the second of these first. The technique is to find a way to write H = a₊a_- + (1/2) hbarw. We then find that there is an eigenfunction of H, call it |0>, whose eigenvalue is E₀ = (1/2) hbar w. Moreover, given an eigenfunction y with eigenvalue E, a₊ y is an eigenfunction of H with eigenvalue E + hbar w, so we can get as many states as we want by applying a₊ repeatedly to |0>.

• Method 1 -- Algebraic Solution

  First, some new notation. We define the commutator of A and B as

  [A,B] = AB - BA

  which is most often used in the form AB = BA + [A, B]. For ordinary numbers, [A, B] = 0, but for operators this result is not necessarily true. For instance,

  [x,p]f(x)= x(-i hbar d/dx)f(x) - [-i hbar d/dx][xf(x)]
  = -i hbar df/dx (x-x) + i hbar f(x)

  so that

  [x,p] = i hbar

  A pair of useful identities are

  [A+B,C] = [A,C] + [B,C]

  [AB,C] = A[B,C] + [A,C]B

  Now define [pulling part of a rabbit out of a hat; Griffiths gives the motivation]

  a_± = [1/Ö(2m)] [p_op ± imwx]

  Then

  a₊ a_- = [1/(2m)][p + imwx] [p - imwx]
  = {1/(2m)} {p² + m²w²x² - [p,imwx]}
  = [1/(2m)] [p² + m²w²x²] - (1/2) hbar w]
  = H - (1/2) hbar w]
  

  Thus the Schroedinger Equation may be written

  [a₊ a_- + (1/2) hbar w] y = E y

  To remove the rest of the rabbit from the hat, calculate [H,a_±], which is mostly easily done if you know [a₊, a_-]. Calculating the latter

  [a₊, a_-] = [1/(2m)][p + imwx, p - imwx]
  = [1/(2m)][-im][p,wx] +[1/(2m)][+im][wx, p]
  = [1/(2m)[im+im][i hbar]

  [a₊, a_-] = - hbar w

  Now we can calculate

  [H,a₊] = [a₊a_- + (1/2)hbarw, a₊]
  = a₊[a_-, a₊] + 0
  

  [H, a₊] = hbar w a₊

  Similarly,

  [H, a_-] = - hbar w a_-

  These relations are too simple not to be useful. Remember that operator equations are always applied to some wave function, which we have been omitting during the derivation for algebraic simplicity. If we use a wave function y which is an eigenfunction (solution) of the Hamiltonian with energy E, so that Hy = Ey,

  [H, a_±] y = ±hbar w a_± y

  Ha_±y-a_±Hy = hbar w a_± y

  Using the Schroedinger equation and rearranging,

  Ha_±y = (E±hbar w) a_± y

  Hence a_±y is an eigenfunction of the Hamiltonian with eigenvalue E±hbar w.

  Unless, of course, a_±y=0 or a_±y is not normalizable.

  Griffiths problem 2.11 eliminates the second problem for a_-, and an analogous argument eliminates it for a₊. However, a_±y=0 might happen. In fact, since we know that E>0, there must be some state for which a_-y_n=0. Otherwise, repeated applications of a_- to y must eventually yield an energy below zero. Classically, we know that the energy of a harmonic oscillator may be as large as we wish, and we should expect the same of the quantum oscillator. Hence we expect that a₊y_n=0 does not occur. We need to prove both of these.

  In a slightly gimpy notation, label as y₀ or |0> the state for which a_±|0> = 0. There should really be two labels, since these are two different states, but one of the states doesn't actually exist, so we will be lazy and use only one label. Explicitly,

  [1/Ö(2m)] [-i hbar d/dx ± imwx] y₀ = 0

  dy₀/dx = (± m w x / hbar) y₀

  The solution of this equation is

  y₀ = A₀exp[(± 1/2)(m w / hbar)x²]

  With the + sign, the function is not normalizable, so there is no state for which a₊ y=0. However, if we define a_-y₀ = 0, then there is a unique state y₀

  y₀ = A₀exp[-(1/2)(m w / hbar)x²]

  For this state

  H y₀ = hbar w [a₊a_- + 1/2] y₀
  = (1/2) hbar w y₀
  

  so that E₀ = (1/2) hbar w

  We can build an infinite number of states by

  y_n = A_na₊ⁿ exp[-(1/2)(m w / hbar)x²]

  with

  E_n = (n + 1/2) hbar w

  Since y₀ is unique, we get all the states this way.

  We could get all the properties of the wave functions that we would ever need this way, but there is a second way to get the functions which we will need for other potentials. We will go through this method next.