Chapter 3 - Physics of Quantum Information

Introduction

It was a great realization that information is physical and that a (classical) Turing machine is not the end of the story of computation. The physical system in which the information is stored and manipulated is important and qubits are quite different from bits.

In this chapter, some background in quantum mechanics is provided. Not all of this chapter will be directly relevant to our discussion, but it is included for the sake of completeness of our understanding of how quantum mechanics from a textbook is related to quantum computing. The connection is, as of yet, clear but the story seems incomplete from a physicists perspective and for the subject of error prevention methods, some of this chapter will be vital. In particular, the section(s) concerning the density matrix. Not only is this vital, but not usually covered in most quantum mechanics classes, either undergraduate or graduate.

It is also worth emphasizing that this chapter is primarily aimed at physicists and for those others which are interested in the background physics. It is not necessary for much of what follows.

Schrodinger's Equation

A common starting point in quantum mechanics is Schrodinger's equation. This equation is not derived, or justified here, but is given in a general form:

	$H\left\vert \Psi \right\rangle =i\hbar {\frac {\partial }{\partial t}}\left\vert \Psi \right\rangle ,$	(3.1)

where $H\,\!$ is the Hamiltonian, $\hbar \,\!$ is Planck's constant (divided by $2\pi \,\!$), and $t\,\!$ is time. The Hamiltonian contains what is known about the system's evolution. Most of the time in these notes, we let $\hbar =1\,\!$.

This equation is (formally) solved by taking the time derivative to be an ordinary derivative (we assume no explicit time dependence for $H\,\!$), so

	$H\left\vert \Psi \right\rangle =i{\frac {d\left\vert \Psi \right\rangle }{dt}}.$	(3.2)

This means that

	$-iHdt={\frac {d\left\vert \Psi \right\rangle }{\left\vert \Psi \right\rangle }},$	(3.3)

so

	${\begin{aligned}\ln \left\vert \Psi \right\rangle &=-iHt+C,\\\Rightarrow \left\vert \Psi (t)\right\rangle &=e^{-iHt}\left\vert \Psi (0)\right\rangle .\end{aligned}}$	(3.4)

Now if $H\,\!$ is Hermitian, and it is, then the matrix

	$U=e^{-iHt}$	(3.5)

is unitary. (See Appendix C - Vectors and Linear Algebra, in particular the section entitled Unitary Matrices.) Any transformation on a closed system can be described by a unitary transformation and any unitary transformation can be obtained by the exponentiation of Hermitian matrix.

Aside, for inclusion later in an appendix about Group theory.

It may seem strange to exponentiate a matrix. However, you can define a function of a matrix according to its Taylor expansion. The details of this are primarily unimportant here, but just to show how it goes, it is written out.

The Taylor expansion of an exponential is the following:

	$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$	(3.6)

and this can be used to exponentiate a matrix by letting the matrix replace $x/,/!$ in the equation. This can also be used to prove that

	$e^{ix}=\cos x+i\sin x.$	(3.7)

End Aside

The end result and important point is that the evolution of a quantum state is, in general, given by a unitary matrix

	$\left\vert \Psi (t)\right\rangle =U\left\vert \Psi (0)\right\rangle .$	(3.8)

So our objective in quantum information processing is to create a unitary evolution, and eventual measurement, which will produce a particular outcome.

Density Matrix for Pure States

Now let us consider the object (a density matrix, or density operator, of rank one)

	$\rho =\left\vert \psi \right\rangle \left\langle \psi \right\vert ,$	(3.9)

which is just the outter product of two vectors. For example, if

	$\left\langle \psi \right\vert =(0,0,1),$	(3.10)

then

	$\left\langle \psi \mid \psi \right\rangle =\left\langle \psi \right\vert (\left\langle \psi \right\vert )^{\dagger }=1,\;\;\;{\mbox{but}}\;\;\;\left\vert \psi \right\rangle \left\langle \psi \right\vert =\left({\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}}\right).$	(3.11)

Again $\left\vert \psi \right\rangle =\left\vert \psi (t)\right\rangle \,\!$, so $\rho =\rho (t)\,\!$. If we differentiate this with respect to $t\,\!$,

	${\begin{aligned}{\frac {\partial \rho }{\partial t}}&=\left({\frac {\partial \left\vert \psi \right\rangle }{\partial t}}\right)\left\langle \psi \right\vert +\left\vert \psi \right\rangle \left({\frac {\partial \left\langle \psi \right\vert }{\partial t}}\right)\\&=(-iH)\rho +\rho (iH)=-i[H,\rho ],\end{aligned}}$	(3.12)

which is the Schrodinger equation for the density matrix, with solution,

	$\rho (t)=U\rho (0)U^{\dagger }.$	(3.13)

This follows from $\left\vert \psi (t)\right\rangle \left\langle \psi (t)\right\vert =U\left\vert \psi (0)\right\rangle \left\langle \psi (0)\right\vert U^{\dagger }\,\!$.

Consider our two-state system

	$\left\vert 0\right\rangle =\left({\begin{array}{c}1\\0\end{array}}\right),\;\;\;{\mbox{and}}\;\;\;\left\vert 1\right\rangle =\left({\begin{array}{c}0\\1\end{array}}\right).$	(3.14)

A superposition of these two states is

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle =\left({\begin{array}{c}\alpha _{0}\\\alpha _{1}\end{array}}\right),$	(3.15)

where $\alpha _{0}\,\!$ and $\alpha _{1}\,\!$ are complex numbers such that $|\alpha _{0}|^{2}+|\alpha _{1}|^{2}=1\,\!$. The corresponding pure state, (i.e. rank one) density matrix is given by

	$\rho _{p}=\left\vert \psi \right\rangle \left\langle \psi \right\vert =\left({\begin{array}{cc}|\alpha _{0}|^{2}&\alpha _{0}\alpha _{1}^{*}\\\alpha _{0}^{*}\alpha _{1}&|\alpha _{1}|^{2}\end{array}}\right).$	(3.16)

Note that the superposition in Eq.(3.15) can be obtained from any pure state by a unitary transformation. Here, the trace of the density matrix is an important quantity; it is

	${\mbox{Tr}}(\rho _{p})=|\alpha _{0}|^{2}+|\alpha _{1}|^{2}=1.\,\!$	(3.17)

Notice also that the determinant of this matrix is zero:

	$\det(\rho _{p})=|\alpha _{0}|^{2}|\alpha _{1}|^{2}-\alpha _{0}\alpha _{1}^{*}\alpha _{0}^{*}\alpha _{1}=0.$	(3.18)

To see this another way, note that the density operator of rank one can be written as $U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{\dagger }\,\!$, so that the determinant is

	${\begin{aligned}\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{\dagger })&=\det(U(\left\vert 0\right\rangle \left\langle 0\right\vert )U^{-1})\\&=\det(U)\det(\left\vert 0\right\rangle \left\langle 0\right\vert ){\frac {1}{\det(U)}}\\&=\det(\left\vert 0\right\rangle \left\langle 0\right\vert )=0.\end{aligned}}$	(3.19)

Measurements Revisited

If the state of a quantum system is described by

	$\left\vert \psi \right\rangle =\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle ,$	(3.20)

the probability of finding it in the state $\left\vert 0\right\rangle \,\!$ when measured in the computational basis is $|\alpha _{0}|^{2}\,\!$. However, this is a particular superposition which could be written as

	$\left\vert \psi \right\rangle =U\left\vert 0\right\rangle .$	(3.21)

In Sec.~\ref{sec:schreq} it was shown that this matrix $U\,\!$ results from the exponentiation of a Hermitian matrix and from Sec.~\ref{sec:Paulimatrices} any $2\times 2\,\!$ Hermitian matrix can be written in terms of the Pauli matrices. To make this explicit using standard conventions,

	Failed to parse (unknown function "\mahtbb"): {\displaystyle \begin{align} \left\vert \psi\right\rangle &= U\left\vert 0\right\rangle \\ &= \exp(-i\vec{n}\cdot\vec{\sigma} \theta) \left\vert 0\right\rangle \\ &= (\mahtbb{I}\cos(\theta) -i\vec{n}\cdot\vec{\sigma} \sin(\theta))\left\vert 0\right\rangle, \end{align} }	(3.22)

where ${\vec {n}}\,\!$ is a unit vector, $|{\vec {n}}|=1\,\!$ and ${\vec {n}}\cdot {\vec {\sigma }}=n_{1}\sigma _{1}+n_{2}\sigma _{2}+n_{3}\sigma _{3}\,\!$. One can write this matrix out explicitly

	${\begin{aligned}\exp(-i{\vec {n}}\cdot {\vec {\sigma }}\theta )&=\left({\begin{array}{cc}1&0\\0&1\end{array}}\right)\cos(\theta )\\&+(-i)\left[n_{1}\left({\begin{array}{cc}0&1\\1&0\end{array}}\right)+n_{2}\left({\begin{array}{cc}0&-i\\i&0\end{array}}\right)+n_{3}\left({\begin{array}{cc}1&0\\0&-1\end{array}}\right)\right]\sin(\theta )\\&=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.23)

Notice this is a special unitary matrix. (Sec. {sec:linalg}.)

To see that any state $\left\vert \psi \right\rangle \,\!$ for arbitrary coefficients $\alpha _{0}\,\!$, $\alpha _{1}\,\!$ can be obtained by choosing ${\vec {n}}\,\!$ and $\theta \,\!$ appropriately, the state $\left\vert 0\right\rangle \,\!$ can be chosen as a starting point. Then

	${\begin{aligned}U\left\vert 0\right\rangle &=\left({\begin{array}{cc}\cos(\theta )-in_{3}\sin(\theta )&(-in_{1}-n_{2})\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )&\cos(\theta )+in_{3}\sin(\theta )\end{array}}\right)\left({\begin{array}{c}1\\0\end{array}}\right)\\&=\left({\begin{array}{c}\cos(\theta )-in_{3}\sin(\theta )\\(-in_{1}+n_{2})\sin(\theta )\end{array}}\right).\end{aligned}}$	(3.24)

For example, choosing $\theta =0\,\!$ gives the original state; choosing ${\vec {n}}=(0,1,0)\,\!$ and $\theta =\pi /2\,\!$ gives $\left\vert 1\right\rangle \,\!$; and choosing ${\vec {n}}=(0,1,0)\,\!$ and $\theta =\pi /4\,\!$ gives an equal superposition. In general, when the system is in the state $\left\vert \psi \right\rangle =U\left\vert 0\right\rangle \,\!$, the probability of finding the state $\left\vert 0\right\rangle \,\!$ when a measurement is made in the computational basis is given by

	${\begin{aligned}|\ll 0{\vec {r}}U\left\vert 0\right\rangle |^{2}&=|\cos(\theta )-in_{3}\sin(\theta )|^{2}\\&=\cos ^{2}(\theta )+n_{3}^{2}\sin ^{2}(\theta ),\end{aligned}}$	(3.25)

and the probability of finding $\left\vert 1\right\rangle \,\!$ is

	${\begin{aligned}|\left\langle 1\right\vert U\left\vert 1\right\rangle |^{2}&=|(-in_{1}+n_{2})\sin(\theta )|^{2}\\&=(n_{1}^{2}+n_{2}^{2})\sin ^{2}(\theta ).\end{aligned}}$	(3.26)

Notice the probabilities add up to one if ${\vec {n}}\,\!$ is a unit vector.

What this shows is that there is a transformation that takes the state $\left\vert 0\right\rangle \,\!$, which has probability $1\,\!$ of being in the state $\left\vert 0\right\rangle \,\!$ and probability 0 of being in the state $\left\vert 1\right\rangle \,\!$ and transform it (using a "rotation'' into a state with a different (and generic) probability of each. This means that the density matrix corresponding to this system always has determinant zero, meaning (for a two-state system) it has one eigenvalue 1 and another eigenvalue 0. (The determinant is the product of the eigenvalues.)