How to calculate the 4th order perturbation energy of harmonic oscillator with diagram correctly?

Question

I'm learning Feynman diagram. It is said that:

$$\Delta E=\frac {i} {T_{tot}}\sum connected\ vacuum\ diagram$$

in which $T_{tot}$ is the total time for perturbation acting. I want to test the theorem in simple isolated harmonic oscillator whose Hamiltonian reads:

$$H=p^2/2m+\frac 1 2 m \omega_0 ^2 x^2=\omega_0(a^\dagger a+1/2)$$

The ground state $|0\rangle$ has energy of $\omega_0 /2$.

Now I perturb it with squared term:

$$H=p^2/2m+\frac 1 2 m \omega_0 ^2 x^2 + \frac 1 2 m \omega _1 ^2 x^2=\sqrt{\omega_0^2+\omega_1^2}(a^\dagger a+1/2).$$

The exact ground state energy now is:

$$\frac 1 2\sqrt{\omega_0^2+\omega_1^2}=\frac {\omega_0} 2 \sqrt{1+\frac{\omega_1^2}{\omega_0^2}}=\frac {\omega_0} 2(1+\frac 1 2\frac{\omega_1^2}{\omega_0^2}-\frac 1 8(\frac{\omega_1^2}{\omega_0^2})^2+\frac{1}{16}(\frac{\omega_1^2}{\omega_0^2})^3-\frac{5}{128}(\frac{\omega_1^2}{\omega_0^2})^4+\ldots)$$

in which the square root is expanded to polynomials.

Next, I want to calculate the perturbed energy order by order using Feynman diagram. Since the perturbation term is quadratic $\frac 1 2 m \omega _1 ^2 x^2$, I think the possible connected vacuum diagrams are single circles, with several vertexes on arcs. I can calculate the 1st, 2nd, 3ird order $\Delta E$ consistent with former exact solution. However, I countered inconsistency in 4th order. The exact 4th order energy $$\Delta E^{(4)}=-\frac{5}{256}\omega_0(\frac{\omega_1^2}{\omega_0^2})^4$$ has a factor of 5 in numerator. The corresponding diagram is circle with 4 vertexes, and I really have no idea how the 4 vertexes circle diagram generate combination numbers involving factor 5. Actually, I come up with $$\Delta E_{diagram}^{(4)}=-\frac{6}{256}\omega_0(\frac{\omega_1^2}{\omega_0^2})^4$$ via calculate diagram. Am I wrong? Here is my calculation details (only real factors without units are concerned):

1. Every vertex provide factor of 1/2;
2. Green's function of oscillator is: $$G(t_1-t_2)=-i\langle 0|Tx(t_1)x(t_2)|0 \rangle =-i\frac 1 {2m\omega}\exp(-i\omega |t_1-t_2|).$$ Then every Green's function provide factor 1/2;
3. Every time-integral also leads to 1/2;
4. Multiplicity of four-node circle is $3\times 2^4$
5. I have 4 vertexes, 4 Green's function, 3 time-integrals, then come to the factor: $\frac 3 {128}=\frac 6 {256}$.

Answer 1

Hints:

1. Let us for simplicity put mass $m=1$ and go to Euclidean signature as e.g. my Phys.SE answer here.

2. The free propagator is $$\Delta(t,t^{\prime})~=~\frac{1}{2\omega_0}e^{-\omega_0 |t-t^{\prime}|},\tag{A} $$ while the Feynman rule for the 2-vertex insertion is $$-\frac{\omega_1^2}{\hbar}\int_{[0,T]}\!dt. \tag{B} $$

3. The connected Feynman vacuum bubble diagrams are 1-loop diagrams of the form

    x--x--x--x
    |        |
    x--x--x--x
   

   $\uparrow$ Fig. 1. A vacuum bubble with $n=8$ 2-vertex insertions.

   The Feynman 1-loop bubble diagram with $n$ 2-vertex insertions has a symmetry factor $S=2n$.

4. The time integrals in the Feynman 1-loop diagram with $n$ 2-vertex insertions $$ \int_{[0,T]} \!dt_1 \ldots \int_{[0,T]} \!dt_n $$ $$\exp\left\{-\omega_0\left[ |t_1-t_n|+|t_n-t_{n-1}|+\ldots+|t_2-t_1|\right]\right\} $$ $$~=~ \frac{(\#)}{\omega_0^{n-1}}T+{\cal O}(T^0) \tag{C}$$ are a bit more complicated to evaluate than OP suggests. Here $(\#)\in\mathbb{Q}$ is some rational number. One may check that the first few numbers for small $n$ are $$(\#)~=~1,1,\frac{3}{2},\frac{5}{2},\ldots,\tag{D}$$ cf. OP's main question.

5. Altogether, the $n$th energy contribution therefore takes the form $$E_n~=~ \frac{-\hbar}{S} \left(\frac{-\omega_1^2}{2\omega_0}\right)^n \frac{(\#)}{\omega_0^{n-1}}. \tag{E}$$