Physics 487 Discussion 4 September 20, 2005
Solution
1. (I’ve deleted the preamble and footnotes.)
On the way to the complete BCS theory, Cooper proposed that
ˆ
U
is very symmetrical: all of
its matrix elements are equal. This means that
ˆ
U
is a very simple R×R matrix:
ˆ
U=!a
1!1!1
" # $ "
111
" $ # "
1!1!1
"
#
$
$
$
$
$
$
%
&
'
'
'
'
'
'
.
This problem can be solved exactly. Calculate the eigenvalues of
ˆ
U
. You will find that one
eigenvalue is special. What is its eigenvector?
Hint: Some properties of eigenvalue problems may be helpful:
• The sum of the diagonal elements (the trace) = the sum of the eigenvalues.
• The sum of the squares of all the matrix elements = the sum of the squares of the
eigenvalues. (I think this may only hold for Hermitian operators.)
• The determinant of the matrix = the product of the eigenvalues.
The equality of rows means the determinant = 0. Therefore, at least one eigenvalue = 0. In
fact, it’s easy to write down R-1 eigenvectors that all have ε = 0:
(1, -1, 0, 0, ... 0), (1, 0, -1, 0, ... 0), ... (1, 0, 0, 0, ... -1)
I haven’t chosen an orthonormal set of eigenvectors, but that’s not important here.
The trace of the matrix = -Ra, so Rth eigenvalue must equal -Ra. The eigenvector is
(1, 1, 1, 1 ... 1). Note that it is orthogonal to all of the others (because it has a different
eigenvalue).
So, we have an energy spectrum with a big gap:
In a macroscopic material, R can be very large, so the
gap can be big even if a is very small. Only the single
ground state will be affected by BEC.
R-1 degenerate
states
1 state
Ra
