Second-order energies of a quartic pertubation of a harmonic oscillator A homework exercise was to calculate the second-order perturbation of a quantum anharmonic oscillator with the potential
$$
V(x) = \frac{1}{2}x^2 + \lambda x^4.
$$
We set $\hbar = 1$, $m=1$, etc. Using the harmonic oscillator $H = \frac{1}{2}p^2 + \frac{1}{2}x^2$ as my basis hamiltonian, I calculated the perturbed ground state energy multiplication factors as
$$
E_0(\lambda) = 1 + \frac{3}{4} \lambda - \frac{21}{\color{red}{8}} \lambda^2 + \mathcal{O}(\lambda^3)
$$
while our lecture notes state
$$
E_0(\lambda) = 1 + \frac{3}{4} \lambda - \frac{21}{\color{red}{16}} \lambda^2 + \mathcal{O}(\lambda^3).
$$
I did not find any sources in literature, neither did I find a mistake in my calculations yet. Which one is correct?
 A: It seems that your result is not correct (with your convention for the hamiltonian):
Starting from your hamiltonian $H = \frac{1}{2}p^2 + \frac{1}{2}x^2 + \lambda x^4$, and with :
$$E_0^{(1)} = V_{00}, \quad E_0^{(2)} = \sum\limits_{m \neq 0} \frac{|V_{0m}|^2}{E_0-E_m} \tag{1}$$
Here : $E_n^{(0)} = n + \frac{1}{2}$, with $V_{00} = \lambda \langle 0|X^4|0\rangle$ and  $|V_{0m}|^2 = \lambda^2 |\langle 0|X^4|m\rangle|^2$, and with $X = \frac{1}{\sqrt{2}} (a+a^+), P = \frac{i}{\sqrt{2}}(a^+-a)$
By applying successively ($4$ times) the operator $X = \frac{1}{\sqrt{2}} (a+a^+)$, on the state $|0\rangle$ (with the rules $a|n\rangle = \sqrt{n}|n-1\rangle$ and $a^+|n\rangle = \sqrt{n+1}|n+1\rangle$)  you find :
$X^4|0\rangle = \dfrac{1}{\sqrt{2}^4}(\sqrt{24}|4\rangle + 2\sqrt{18}|2\rangle + 3|0\rangle) \tag{2}$
So, finally : 
$$E_0^{(1)} = \frac{3}{4} \lambda\tag{3}$$
$$E_0^{(2)} = -\frac{1}{2^4}(\frac{24}{4} + \frac{4*18}{2})\lambda^2= - \frac{42}{16} {\lambda^2} = - \frac{21}{8}\tag{4} {\lambda^2}$$
So, finally, the (absolute) modifyed energy for the ground state is  : 
$$E_0(\lambda) = \frac{1}{2} + \frac{3}{4} \lambda - \frac{21}{8} {\lambda^2} \tag{5}$$
This is compatible with the other convention for the hamiltonian (in my reference) which is :
$H' = p^2 + x^2 + \lambda x^4 = 2(\frac{1}{2}p^2 + \frac{1}{2}x^2 + \frac{\lambda}{2} x^4)$, the (absolute) modifyed energy for the ground state is then : 
$E'_0(\lambda) = 2 E_0(\frac{\lambda}{2}) = 2(\frac{1}{2} + \frac{3}{4} \frac{\lambda}{2} - \frac{21}{8} \frac{{\lambda^2}}{4})=1 + \frac{3}{4} \lambda - \frac{21}{16} {\lambda^2} \tag{6}$
Now, if you want relative factors, you have to consider $\frac{E_0(\lambda)}{E_0(0)}$ or $\frac{E'_0(\lambda)}{E'_0(0)}$, depending on the hamiltonian you are considering, so with your hamiltonian $H$, you have : 
$$\frac{E_0(\lambda)}{E_0(0)} = 1+ \frac{3}{2} \lambda - \frac{21}{4} {\lambda^2} \tag{7}$$
while, with the hamiltonian $H'$, we get : 
$$\frac{E'_0(\lambda)}{E'_0(0)}=1 + \frac{3}{4} \lambda - \frac{21}{16} {\lambda^2}\tag{8}$$
A: I did this both by hand, and using Mathematica, and obtained the same result you obtained for the second order correction term.  Here's my work.
Write $\hat x$ in terms of creation and annihilation operators;
\begin{align}
  \hat x = \frac{1}{\sqrt{2}}(\hat a+\hat a^\dagger).
\end{align}
Recall what they do to energy eigenstates;
\begin{align}
  a^\dagger|n\rangle =\sqrt{n+1}|n+1\rangle, \qquad a|n\rangle = \sqrt{n}|n-1\rangle. 
\end{align}
use these relations to show that
\begin{align}
  \hat x^4|0\rangle &= \frac{1}{4}(3|0\rangle + 6\sqrt{2}|2\rangle + 2\sqrt{6}|4\rangle)
\end{align}
and thus
\begin{align}
  E_0^{(2)} 
&= \sum_{m=1}^\infty \frac{|\langle m|\lambda \hat x^4|0\rangle|^2}{(0+\frac{1}{2})-(m+\frac{1}{2})}
=-\frac{\lambda^2}{16}\left(\frac{(6\sqrt{2})^2}{2}+\frac{(2\sqrt{6})^2}{4}\right)
= -\frac{21}{8}\lambda^2
\end{align}
However, as Trimok points out, your zeroth order term should be a $1/2$, which changes the overall normalization of the answer.
