Perturbations - Quantum Mechanics

Quantum Perturbation

Assume the Hamiltonian \(H\) cannot be solved analytically. However, if it is close to a solvable Hamiltonian, say \(H_0\), there are ways to approximate it. This technique is called “Perturbation” and the theory is called “Perturbation Theory”. i.e., we have:

\[\hat{H}=\hat{H_0}+\hat{H'}\]

For simplicity, we assume the system does not have “degeneracy”, i.e., each eigenvalue only corresponds to an eigenvector, or, if you are mathematically minded, the geometric multiplicity is smaller than the algebraic multiplicity.

By doing a taylor series, we obtain: (the eigenstates are no longer normalized!)

\[E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+...\]

and

\[|n\rangle=\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^2\left|n^{(2)}\right\rangle+\ldots\]

where the (0), (1), (2) implies taking to the 0th, 1st, 2nd order respectively.

If we expand everything, we will get:

\[\begin{aligned} & \left(\hat{H}_0+\lambda \hat{H}^{\prime}\right)\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^2\left|n^{(2)}\right\rangle+\ldots\right)= \\ & \quad\left(E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\ldots\right)\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^2\left|n^{(2)}\right\rangle+\ldots\right) \end{aligned}\]

By comparing terms, we get:

\[\lambda^1: \left(\hat{H}_0-E_n^{(0)}\right)\left|n^{(1)}\right\rangle =\left(E_n^{(1)}-\hat{H}^{\prime}\right)\left|n^{(0)}\right\rangle\]

Consider Left hand side. From the term:

\[\hat{H}_0 \left|n^{(1)}\right\rangle\]

We can operate \(\langle k^{(0)} \,\lvert\) and obtain:

\[\left\langle k^{(0)}\left|\hat{H}_0\right| n^{(1)}\right\rangle = E_k^{(0)}\left\langle k^{(0)} \mid n^{(1)}\right\rangle\]

Hence we have:

\[\begin{aligned} &\langle k^{(0)} \lvert \left(\hat{H}_0 - E_n^{(0)}\right) \lvert n^{(1)} \rangle \\ &\implies (E_k^{(0)}-E_n^{(0)})\left\langle k^{(0)} \mid n^{(1)}\right\rangle=E_n^{(1)} \delta_{k, n}-\left\langle k^{(0)}\left|\hat{H}^{\prime}\right| n^{(0)}\right\rangle \end{aligned}\]

Choosing \(k=n\) will give us:

\[\begin{aligned} E_n^{(1)} \delta_{n, n}-\left\langle n^{(0)}\left|\hat{H}^{\prime}\right| n^{(0)}\right\rangle=0 \\ \implies E_n^{(1)}=\left\langle n^{(0)}\left|\hat{H}^{\prime}\right| n^{(0)}\right\rangle \end{aligned}\]

The astute reader will find out that this is an expectation value! So the expected “change in energy” is just the expectation of the perturbation!

Classic Example

Now consider a standard problem in perturbation:

\[V(x)=V_0 \cos\left(\frac{\pi x}{2a}\right)\]

between \(-a\) to \(a\). What is the ground state perturbation?

From above, we have:

\[\begin{aligned} \Delta E_{n} & =\left\langle\psi^{(n)}\mid V \mid \psi^{(n)}\right\rangle \\ & =\int_{-a}^{+a} \left[\psi^{(n)}(x)^*\right] V(x) \left[\psi^{(n)}(x)\right] dx \end{aligned}\]

Considering :

\[\psi^1(x)=\frac{1}{\sqrt{a}} \cos{\left(\frac{\pi x}{2a}\right)}\]

We can substitute and get:

\[\begin{aligned} \Delta E_n&=\frac{V_0}{a} \int_{-a}^{+a} \cos \left(\frac{\pi x}{2 a}\right) \cos \left(\frac{\pi x}{2 a}\right) \cos \left(\frac{\pi x}{2 a}\right) dx \\ &=\frac{2 V_0}{\pi} \int_{-\pi / 2}^{+\pi / 2} (\cos ^3 \theta) d\theta\\ &=\frac{8 V_0}{3 \pi} \end{aligned}\]

The details are left for an exercise for the reader.