Diploma thesis - Fachbereich Physik

32 CHAPTER 2. EFFECTIVE ACTION
for arbitrary temperatures and until the third loop order for T = 0 by applying the results
from the previous section. To this end, one only needs to replace the derivatives of the
arbitrary potential from the last section by their actual values for the anharmonic oscillator.
The propagator G Ω (τ 1 , τ 2 ) of the anharmonic oscillator for arbitrary temperatures is given
by (2.141), and in the zero-temperature limit by (2.142), where the frequency Ω reads
Ω = ω
√
1 + 6gAX
Mω 2 + 12g2 BX 2
Mω 2 . (2.157)
Note that in order to ensure that Ω in (2.157) is real for all values of X, the condition
|A| ≤
√
4Mω2 B
3
(2.158)
must hold. Applying the result (2.151) to the anharmonic oscillator, one obtains its temperature
dependent effective potential up to the second order in ¯h:
V eff (X) = M 2 ω2 X 2 + gAX 3 + g 2 BX 4 + ¯h
¯hβ
ln (2 sinh ¯hβΩ/2) + ¯h2
3g
[
− ¯h2 (6gA + 24g 2 BX) 2 1
6Ω (2MΩ) 3 3 + 1
sinh 2 (¯hβΩ/2)
2 B
(2MΩ) 2 1
tanh 2 (¯hβΩ/2)
]
+ O(¯h 3 ) . (2.159)
According to (2.39), one obtains the free energy by evaluating (2.159) for that particular
average X that extremizes the effective potential. Since here the calculation is only up to
the second order in the reduced Planck constant ¯h, the background X can be expanded in
the form
X = X 0 + ¯hX 1 + ¯h 2 X 2 + O(¯h 3 ) . (2.160)
Inserting this identity into the first derivative of (2.159), expanding it in powers of ¯h, and
setting the resulting expression to zero leads to a system of three equations, each corresponding
to the respective order in ¯h. This system of equations for the extremizing average X e
can be solved consecutively. First, one obtains
X (1)
0 = 0 , X (2)
0 = − 3A + 9√ 9A 2 − 16BMω 2
8gB
, X (3)
0 = −3A + 9√ 9A 2 − 16BMω 2
.
8gB
(2.161)
Since the solution is required to be real for all allowed parameters A, B, only X (1)
0 = 0 is
a valid solution. The solutions X (2)
0 and X (3)
0 are complex due to condition (2.158). Using
this result, one further obtains
3gA
X 1 = −
2M 2 ω 3 tanh(¯hβω/2) . (2.162)
Since X 0 vanishes, one reads off from (2.159) that X 2 does not contribute to the free energy
in the second order of ¯h. This corresponds to the previous example of an ordinary integral in