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Notation: If A is an operator then I'll denote the Hermitian conjugate of A by C(A)
The annihilation operator in terms of p and x is:
a = 1/sqrt(2m hbar omega) [m omega x + i p]
And the creation operator is its Hermitian conjugate:
C(a) = 1/sqrt(2m hbar omega) [m omega x - i p]
It is convenient to have the expressions for x and p in terms of a and C(a). Solving the above equations for x and p gives:
x = Squareroot[h-bar/(2 m omega)] [a + C(a)]
p = i Squareroot[m omega h-bar/2] [C(a) - a]
The Hamiltonian is:
H = [C(a) a + 1/2] hbar omega
a and C(a) satisfy the commutation relation:
[a, C(a)] = 1 (1)
We also have:
[H, a] = [[C(a) a + 1/2] hbar omega, a] = hbar omega [C(a) a, a] = hbar omega [C(a)a a - a C(a) a ]
From (1) it follows that a C(a) = 1 + C(a) a. The second term in the above formula can thus be written as:
a C(a) a = [1 + C(a) a] a = a + C(a) a a
The expression for the commutator thus simplifies to:
[H, a] = -hbar omega a (2)
We can compute the eigenvectors and eigenvalues of H as follows. We put C(a) a = N. Then since H = hbar omega(N+1/2)
we can just as well compute the eigenstates of N. These are then also the eigenstates of H. If |n> is a normalized eigenstate of N with eigenvalue n, then N|n> = n|n> and thus H|n> = (n+ 1/2)hbar omega|n>, so the corresponding eigenvalue of H is
(n + 1/2)hbar omega. Note that we are not assuming that n is an integer.
Consider the action of N on the state a|n>:
N a|n> = C(a) a a|n>
Eq. (1) says that
a C(a) - C(a) a = 1 ------>
C(a) a = a C(a) - 1
and we see that
N a|n> = C(a) a a|n> = [a C(a) - 1]a |n> = a C(a) a|n> - a|n> = a N|n> - a|n> = n a |n> - a |n> = (n-1)|n> (3)
So, it looks like a|n> is an eigenstate with eigenvalue n-1. Let's compute the norm of this state. The absolute value squared of the norm is:
<n|C(a) a|n> = <n|N|n> = <n|n|n> = n <n|n> = n. (4)
The absolute value squared of any ...
A detailed solution is derived from first principles.
Expectation Values For Various States on a Harmonic Oscillator
See the attached file.
3. (a) Calculate the expectation values < x >, < p >, < x^2 > and < p^2 > for the ground state, | 0 >, and the first excited state, | 1 >, of the harmonic oscillator.
(b) Now compute delta(x)delta(p), does this satisfy the uncertainty principle?
4. Using the results from (3), find the expectation values of kinetic, <K> and potential energies, < V >, for the above states, and then that of the total energy, < E >. Are they what you would have expected?
My biggest problem with this homework problem is that I don't know how to do it with the information provided. I have calculated the expectation values in many problems before, but there was more information provided describing the states. Calculating expectation values doesn't usually take me more than a couple minutes, but I just don't know how to approach the problem with the information provided. Thank you very much for your help with this problem!
Note: I have modified this problem to include number 4, because it follows from problem 3. I increased the number of credits for helping with this problem because of that.View Full Posting Details