Appendix H: Topics in Quantum Mechanics

Introduction

In this section, some topics in quantum mechanics will be presented. These are outside the main topic of the book, which is focused on quantum computing and quantum error prevention. Here the topics are of quantum mechanics, but not directly having to do with those main topics of the book.

Time Independent Perturbation Theory

A fairly good introductory treatment of this topic can be found in Griffiths' book [4]. One may also consult many other good textbooks on quantum mechanics, such as Ballentine [47].

The Problem

The problem that we want to solve is the following. Suppose that we know how to solve a problem. Can we solve one that is "close" to this one? We have to be specific about what we mean by "close". But intuitively, if we know the solutions to a problem (we know the eigenfunctions and eigenvalues for a Hamiltonian), and the Hamiltonian is only changed slightly, we could expect that the solutions (e.g. the eigenvalues) for the new problem are not too different from the one we know how to solve. With this in mind, let us suppose that we know how to solve a problem:

	$H^{(0)}\left\vert {\Psi ^{(0)}}_{n}\right\rangle =E_{n}^{(0)}\left\vert {\Psi _{n}^{(0)}}\right\rangle .$	(H.1)

Now suppose there is a different problem that we would like to solve and it can be written in the form

	$(H^{(0)}+\lambda H^{(1)})\left\vert {\Psi }_{n}\right\rangle =E_{n}\left\vert {\Psi _{n}}\right\rangle ,$	(H.2)

where the term $\lambda H^{(1)}$ is small. The "smallness" of this term determines the "closeness" of this problem to the original one, and this will have to be made more precise later-when we can calculate the relative sizes. The parameter $\lambda$ is an expansion parameter that is used for convenience and it will be removed later by letting it go to one. But for now, we assume it to be "small". (This may sound a little funny, but it works just fine. The parameter can be an independent variable, in which case powers of it are independent and that is what we are going to use.)

Expanding and Solving

Let us expand the the unknown energies and wave functions in terms of this parameter by supposing that there is a sequence of energies and wave functions that are able to describe our system in a way that can be expressed as

	$E_{n}=E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda ^{2}E_{n}^{(2)}+\cdots$	(H.3)

and

	$\left\vert {\Psi }_{n}\right\rangle =\left\vert {\Psi _{n}^{(0)}}\right\rangle +\lambda \left\vert {\Psi _{n}^{(1)}}\right\rangle +\lambda ^{2}\left\vert {\Psi _{n}^{(2)}}\right\rangle +\cdots$	(H.4)

We can plug these into Eq.(H.2)

	${\begin{aligned}&(H^{(0)}+\lambda H^{(1)})\left(\left\vert {\Psi _{n}^{(0)}}\right\rangle +\lambda \left\vert {\Psi _{n}^{(1)}}\right\rangle +\lambda ^{2}\left\vert {\Psi _{n}^{(2)}}\right\rangle +\cdots \right)\\&=\left(E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda ^{2}E_{n}^{(2)}+\cdots \right)\left(\left\vert {\Psi _{n}^{(0)}}\right\rangle +\lambda \left\vert {\Psi _{n}^{(1)}}\right\rangle +\lambda ^{2}\left\vert {\Psi _{n}^{(2)}}\right\rangle +\cdots \right)\end{aligned}}\,\!$	(H.5)

Then we expand this and collect like powers of $\lambda$ to get

	${\begin{aligned}H^{(0)}\left\vert {\Psi _{n}^{(0)}}\right\rangle &+\lambda (H^{(1)}\left\vert {\Psi _{n}^{(0)}}\right\rangle +H^{(0)}\left\vert {\Psi _{n}^{(1)}}\right\rangle )+\lambda ^{2}(H^{(0)}\left\vert {\Psi _{n}^{(2)}}\right\rangle +H^{(1)}\left\vert {\Psi _{n}^{(1)}}\right\rangle )+\cdots \\=&E_{n}^{(0)}\left\vert {\Psi _{n}^{(0)}}\right\rangle +\lambda \left(E_{n}^{(0)}\left\vert {\Psi _{n}^{(1)}}\right\rangle +E_{n}^{(1)}\left\vert {\Psi _{n}^{(0)}}\right\rangle \right)\\&+\lambda ^{2}\left(E_{n}^{(2)}\left\vert {\Psi _{n}^{(0)}}\right\rangle +E_{n}^{(1)}\left\vert {\Psi _{n}^{(1)}}\right\rangle +E_{n}^{(0)}\left\vert {\Psi _{n}^{(2)}}\right\rangle \right)+\cdots \end{aligned}}\,\!$	(H.6)

Each power of $\lambda$ gives a separate equation by equating the powers of $\lambda$ on each side to each other. These equations are

	${\begin{aligned}\lambda ^{0}:&H^{(0)}\left\vert {\Psi _{n}^{(0)}}\right\rangle =E_{n}^{(0)}\left\vert {\Psi _{n}^{(0)}}\right\rangle \\\lambda ^{1}:&H^{(1)}\left\vert {\Psi _{n}^{(0)}}\right\rangle +H^{(0)}\left\vert {\Psi _{n}^{(1)}}\right\rangle =E_{n}^{(0)}\left\vert {\Psi _{n}^{(1)}}\right\rangle +E_{n}^{(1)}\left\vert {\Psi _{n}^{(0)}}\right\rangle \\\lambda ^{2}:&H^{(0)}\left\vert {\Psi _{n}^{(2)}}\right\rangle +H^{(1)}\left\vert {\Psi _{n}^{(1)}}\right\rangle =E_{n}^{(2)}\left\vert {\Psi _{n}^{(0)}}\right\rangle +E_{n}^{(1)}\left\vert {\Psi _{n}^{(1)}}\right\rangle +E_{n}^{(0)}\left\vert {\Psi _{n}^{(2)}}\right\rangle .\end{aligned}}\,\!$	(H.7)

The first equation is just the original problem for which we know the answer and is the zeroth order in $\lambda$. The second equation is the first order in $\lambda$, and so forth. Notice that the sum of the exponents is the same as the power of $\lambda$. This provides a way to check to see that we haven't made a mistake. Now, knowing the quantities in the original equation and the perturbing Hamiltonian and want to find the first order quantities in terms of those. This can be done by taking the inner product of the second equation with $\left\langle \Psi _{m}^{(0)}\right\vert$.

Time Independent Perturbation Theory with Degeneracies

Unitary transformations are represented in a circuit diagram with a box around the unitary transformation. Consider a unitary transformation $V$ on a single qubit state $\left\vert {\psi }\right\rangle$. If the result of the transformation is $\left\vert {\psi ^{\prime }}\right\rangle$, we can then write

	$\left\vert {\psi ^{\prime }}\right\rangle =V\left\vert {\psi }\right\rangle .$	(H.1)

The corresponding circuit diagram is shown in Fig. 2.1.

$\left\vert {\psi }\right\rangle$

$\left\vert {\psi ^{\prime }}\right\rangle$

Figure 2.1: Circuit diagram for a one-qubit gate that implements the unitary transformation $V\,\!$. The input state $\left\vert {\psi }\right\rangle$ is on the left and the output, $\left\vert {\psi ^{\prime }}\right\rangle$, is on the right.

Misc

We want to find the matrix form of the following, which is expressed in 3 different ways:

	${\begin{aligned}{\vec {\sigma }}^{(1)}\cdot {\vec {\sigma }}^{(2)}&=\sigma _{x}^{(1)}\sigma _{x}^{(2)}+\sigma _{y}^{(1)}\sigma _{y}^{(2)}+\sigma _{z}^{(1)}\sigma _{z}^{(2)}\\&=\sigma _{x}^{(1)}\otimes \sigma _{x}^{(2)}+\sigma _{y}^{(1)}\otimes \sigma _{y}^{(2)}+\sigma _{z}^{(1)}\otimes \sigma _{z}^{(2)}.\end{aligned}}$	(H.10)

Since

	$\sigma _{x}=\left({\begin{array}{cc}0&1\\1&0\end{array}}\right)$	(H.11)

Then

	${\begin{aligned}\sigma _{x}^{(1)}\otimes \sigma _{x}^{(2)}&=\left({\begin{array}{cc}0&1\\1&0\end{array}}\right)\otimes \left({\begin{array}{cc}0&1\\1&0\end{array}}\right)\\&=\left({\begin{array}{cccc}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{array}}\right)\end{aligned}}$	(H.11)

$[AB,C]=A[B,C]+[A,C]B$