# Computation difficulty in variational method [closed]

This is essentially a computation question. I'm doing a question out of Griffith's Quantum text (7.16 in the old edition and 7.15 in the new), where I need to apply the variational method to a prescribed test wave functions ($cos(\phi)|\Psi_a> + sin(\phi)|\Psi_b>$ where $\phi$ is an unknown parameter and the $\Psi$s are the eigenfunctions of $H$.The Hamiltonian admits only 2 states and can be represented as $$\begin{pmatrix} E_a & 0 \\ 0 & E_b \end{pmatrix}$$ A perturbation is applied and can be represented thus:

$$\begin{pmatrix} 0 & h \\ h & 0 \end{pmatrix}$$

It's pretty straightforward to see that the eigenvalues o the pertubation are $h,~-h$. Now, using the provided test function I'm suppose to arrive back at this same answer, since obviously the given test function is a linear combination of the exact states.

So now if I understand correctly, I ought to take the expectation value of the Hamiltonian (total) for this linear combination of states. This is pretty straight forward, and using $$\begin{pmatrix} E_a & h \\ h & E_b \end{pmatrix}$$ as my total Hamiltonian I get $$<H> = E_acos^2(\phi) + 2hcos(\phi)sin(\phi) + E_bsin^2(\phi)$$

Then taking the derivative and fiddling around with it gives me $$\frac{\partial}{\partial \phi}<H> = (E_b-E_a)sin(2\phi) + 2hcos(2\phi) \rightarrow tan(2\phi) +\frac{2h}{E_b-E_a}$$

Now I'm supposed to find the zeros of this expression as a function of $\phi$ and then sub back in and somehow get out the exact values, but for the life of me I cannot figure out how. I plugged it into Wolfram Alpha and got an unwieldy mess.

• 1) That's not a mistake: $2sin(\phi)cos(\phi) = sin(2\phi)$ 2) There is another mistake, it should be $2hcos(2\phi)$.
– BenL
Commented Dec 4, 2016 at 19:59
• Rearrange $\langle H \rangle$ in terms of $\cos(2\phi)$ and $\tan(2\phi)$ only. From $\tan(2\phi) = \frac{2h}{E_a - E_b}$ obtain $\cos(2\phi)$, make sure you account for both possible signs. Sub back both, simplify, compare with the exact eigenvalues.
– udrv
Commented Dec 4, 2016 at 22:26
• Obtain $\cos(2\phi)$ in terms of sin? I don't see how I can find zeros that way.
– BenL
Commented Dec 4, 2016 at 22:36
• @BenL You need to address reply comments directly to me, otherwise I don't receive them, like this one. Here's how: $$E_a \cos^2(\phi) + E_b\sin^2(\phi) + h \sin(2\phi) = E_a[1 + \cos(2\phi)]/2 + E_b[1 - \cos(2\phi)]/2 + h \sin(2\phi) = \\=\frac {E_a + E_b}{2} + \frac{E_a - E_b}{2} \cos(2\phi) + h \sin(2\phi) = \frac {E_a + E_b}{2} + \frac{E_a - E_b}{2} \left[1 + \frac{2h}{E_a - E_b} \tan(2\phi) \right] \cos(2\phi)$$
– udrv
Commented Dec 6, 2016 at 20:27
• @BenL Now use $\tan(2\phi) = \frac{2h}{E_a - E_b}$ and notice that the expression in the square brackets is just $\cos^{-2}(2\phi)$, simplify, re-express $\cos^{-1}(2\phi)$ in terms of $\frac{2h}{E_a - E_b} \left(= \tan(2\phi) \right)$ and account for the two possible signs, substitute back. Done.
– udrv
Commented Dec 6, 2016 at 20:27