Category Archives: Quantum Physics

The Eigenvectors of a Hermitian Operator Are Mutually Orthogonal

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In quantum mechanics, operators are required to be Hermitian. A reason for this is that the eigenvalues of a quantum mechanical operator are interpreted as physically measurable quantities such as positions, momenta, energies, etc. and therefore they are required to be real. As is well-known, Hermitian operators have all real eigenvalues. Hermitian operators also have another nice property. The eigenvectors of a Hermitian operator are mutually orthogonal. Here, we prove this only for the case of matrix operators. Let $A$ be a Hermitian operator and let $|\lambda_i\rangle$ be the eigenvectors of $A$ with distinct eigenvalues $\lambda_i$. Then we have $A|\lambda_i\rangle=\lambda_i|\lambda_i\rangle$. So, $\langle\lambda_j|A|\lambda_i\rangle=\lambda_i\langle\lambda_j|\lambda_i\rangle$. On the other hand,
\begin{align*} \langle\lambda_j|A&=(A^\dagger|\lambda_j)^\dagger\\ &=(A|\lambda_j\rangle)^\dagger\ (A^\dagger=A)\\ &=(\lambda_j|\lambda_i\rangle)^\dagger\\ &=\bar\lambda_j\langle\lambda_j|\\ &=\lambda_j\langle\lambda_j|\ (\mbox{the $\lambda_i$ are real}) \end{align*}
From this we also obtain $\langle\lambda_j|A|\lambda_i\rangle=\lambda_j\langle\lambda_j|\lambda_i\rangle$. This means that $\lambda_i\langle\lambda_j|\lambda_i\rangle=\lambda_j\langle\lambda_j|\lambda_i\rangle$, i.e. we have
$$(\lambda_i-\lambda_j)\langle\lambda_j|\lambda_i\rangle=0$$
If $i\ne j$ then $\lambda_i\ne\lambda_j$ and so $\langle\lambda_j|\lambda_i\rangle=0$.

Example. Let us consider the matrix $A=\begin{pmatrix}
3 & 1+i\\
-1+i & -3
\end{pmatrix}$. The adjoint (i.e. the conjugate transpose of this matrix) of $A$ is $A^\dagger=\begin{pmatrix}
3 & -1-i\\
1-i & -3
\end{pmatrix}$. Since $A\ne A^\dagger$, $A$ is not Hermitian. Although $A$ is not Hermitian, it has real eigenvalues $\pm\sqrt{7}$ and the eigenvectors corresponding to the eigenvectors $\sqrt{7}$ and $=-\sqrt{7}$ are, respectively,
$$v_1=\begin{pmatrix}
\frac{1+i}{3-\sqrt{7}}\\
1
\end{pmatrix},\ v_2=\begin{pmatrix}
\frac{1+i}{3+\sqrt{7}}\\
1
\end{pmatrix}$$
$\langle v_1|v_2\rangle=2$, so they are not orthogonal. Interestingly, $v_1$ and $v_2$ are orthogonal with respect to the inner product
\begin{equation}
\label{eq:jproduct}
\langle v_1|J|v_2\rangle
\end{equation}
where
$$J=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
The matrices of the form $$H=\begin{pmatrix}
a & b\\
-\bar b & d
\end{pmatrix},$$ where $a$ and $d$ are real and $(a-d)^2-4|b|^2>0$ (this condition is required to ensure that such a matrix has two distinct real eigenvalues), are self-adjoint with respect to the inner product \eqref{eq:jproduct} i.e. $H$ satisfies \begin{equation}\label{eq:jself-adjoint}\langle Hv_1|J|v_2\rangle=\langle v_1|J|Hv_2\rangle\end{equation} which is equivalent to $$H^\dagger J=JH$$

Exercise. Prove that the eigenvectors of a matrix which satisfies \eqref{eq:jself-adjoint} are mutually orthogonal withe respect to the inner product \eqref{eq:jproduct}.

Born Rule

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Let $A$ be a Hermitian (self-adjoint) operator. Due to Max Born, who proposed the statistical interpretation of quantum mechanics, the probability of measuring an eigenvalue $\lambda_i$ of $A$ in a state $\psi$ is $\langle\psi|P_i|\psi\rangle$, where $P_i$ is the projection onto the eigenspace of $A$ corresponding to $\lambda_i$ i.e. $P_i$ is a linear map $P_i: V_{\lambda_i}\longrightarrow V_{\lambda_i}$ such that $P^2=I$. If we assume no degeneracy of $\lambda_i$, then the eigenspace of $A$ corresponding to $\lambda_i$ is one-dimensional. In this case, $P_i=|\lambda_i\rangle\langle\lambda_i|$, where the $|\lambda_i\rangle$ are orthonormal. Now,
\begin{align*} \langle\psi|P_i|\psi\rangle&=\langle\psi|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\ &=\overline{\langle\lambda_i|\psi\rangle}\langle\lambda_i|\psi\rangle\\ &=|\langle\lambda_i|\psi\rangle|^2 \end{align*}
The complex number $\langle\lambda_i|\psi\rangle$ is called the probability amplitude, so the probability is the squared magnitude of the amplitude. This is called the Born rule. (It is named after Max Born.) Suppose $|\psi\rangle$ is a unit vector. Then since $\sum_i|\langle\lambda_i|\psi\rangle|^2=1$, we have \begin{equation}\label{eq:complete}\sum_iP_i=\sum_i|\lambda_i\rangle\langle\lambda_i|=I\end{equation} \eqref{eq:complete} is called the completeness of the $|\lambda_i\rangle$.

One may consider $A$ as a random variable with its eigenvalues as the values of random variable. (See Remark below.) Since for each $j$, $A|\lambda_j\rangle=\lambda_j|\lambda_j\rangle$,
$$\langle\lambda_i|A|\lambda_j\rangle=\lambda_j\langle\lambda_i|\lambda_j\rangle=\lambda_j\delta_{ij}$$
The expected value $\langle A\rangle$ of the self-adjoint operator $A$ in the state $|\psi\rangle$ is naturally defined as a weighted average
\begin{align*} \langle A\rangle&=\sum_i\lambda_i|\langle\lambda_i|\psi\rangle|^2\\ &=\sum_i\langle\lambda_i|\psi\rangle\langle\psi|\lambda_i|\lambda_i\rangle\\
&=\sum_i\langle\lambda_i|\psi\rangle\langle\psi|A|\lambda_i\rangle\ (A|\lambda_i\rangle=\lambda_i|\lambda_i\rangle)\\
&=\sum_i\langle\psi|A|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\&=\langle\psi|A|\sum_i|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\&=\langle\psi|A|I|\psi\rangle\ (\mbox{by the completeness of the}\ |\lambda_i\rangle)\\&=\langle\psi|A|\psi\rangle \end{align*}
Hence, we have
$$\langle A\rangle=\langle\psi|A|\psi\rangle$$

Remark. Let $\Omega=\bigcup_iV_{\lambda_i}$, i.e. $\Omega$ is the set of all eigenvectors of $A$. Let $\mathcal{U}$ consist of $\emptyset$, $\Omega$ and unions of subsequences of ${V_{\lambda_i}}$. Then $\mathcal{U}$ is a $\sigma$-algebra of subsets of $\Omega$. Define $X:\Omega\longrightarrow\mathbb{R}$ by
$$X|\lambda_i\rangle=\langle\lambda_i|A|\lambda_i\rangle=\lambda_i$$
Then $X$ is a random variable.

Quantum Degeneracy

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In quantum mechanics, an energy level is said to be degenerate if the energy corresponds to two or more states of a quantum system. Also, two or more states of a quantum system are said to be degenerate if they give the same energy upon measurement. The set of all degenerate states of a quantum system that correspond a particular energy $E$ forms a (Hilbert) subspace, called the eigenspace of $E$. To see this, let $H$ be a Hamiltonian and $|\psi_1\rangle$, $|\psi_2\rangle$ two linearly independent eigenstates corresponding to the same eigenvalue (energy) $E$. Then
\begin{align*} H|\psi_1\rangle&=E|\psi_1\rangle\\ H|\psi_2\rangle&=E|\psi_2\rangle \end{align*}
Let $|\psi\rangle=c_1|\psi_1\rangle+c_2|\psi_2\rangle$, a linear combination (superposition) of $|\psi_1\rangle$ and $|\psi_2\rangle$. Then
\begin{align*} H|\psi\rangle&=H(c_1|\psi_1\rangle+c_2|\psi_2\rangle)\\ &=c_1H|\psi_1\rangle+c_2H|\psi_1\rangle\\&=c_1E|\psi_1\rangle+c_2E|\psi_2\rangle\\ &=E(c_1|\psi_1\rangle+c_2|\psi_2)\\ &=E|\psi\rangle \end{align*}
The dimension of the eigenspace corresponding to an eigenvalue (energy) $E$ is called the degree of degeneracy of $E$.

Bell’s Theorem

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Suppose that Boris and Natasha are in different locations far away from each other. Their mutual friend Victor prepares a pair of particles and send one each to Boris and Natasha. Boris chooses to perform one of two possible measurements, say $A_0$ and $A_1$, associated with physical properties $P_{A_0}$ and $P_{A_1}$ of the particle he received. Each $A_0$ and $A_1$ has $+1$ or $-1$ for the outcomes of measurement. When Natasha receives one of the particles, she as well chooses to perform one of two possible measurements $B_0$, $B_1$, each of which has outcome $+1$ or $-1$. Let us consider the following quantity of the measurements $A_0$, $A_1$, $B_0$, $B_1$:
$$A_0B_0+A_0B_1+A_1B_0-A_1B_1=(A_0+A_1)B_0+(A_0-A_1)B_1$$
Since $A_0=\pm 1$ and $A_1=\pm 1$, either one of $A_0+A_1$ and $A_0-A_1$ is zero and the other is $\pm 2$. If the experiment is repeated over many trials with Victor preparing new pairs of particles, the expected value of all the outcomes satisfies the inequality
\begin{equation}
\label{eq:bell}
\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle\leq 2
\end{equation}
Proof of \eqref{eq:bell}:
\begin{align*}\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle&=\sum_{A_0,A_1,B_0,B_1}p(A_0,A_1,B_0,B_1)(A_0B_0+A_0B_1+A_1B_0-A_1B_1)\\&\leq \sum_{A_0,A_1,B_0,B_1}2p(A_0,A_1,B_0,B_1)\\&=2 \end{align*}
The inequality \eqref{eq:bell} is a variant of the Bell inequality called the CHSH inequality. CHSH stands for John Clauser, Michael Horne, Abner Simony and Richard Holt. The derivation of the CHSH inequality \eqref{eq:bell} depends on two assumptions:

  1. The physical properties $P_{A_0}$, $P_{A_1}$, $P_{B_0}$, $P_{B_1}$ have definite values $A_0$, $A_1$, $B_0$, $B_1$ which exist independently of observation or measurement. This is called the assumption of realism.
  2. Boris performing his measurement does not influence the result of Natasha’s measurement. This is called the assumption of locality.

These two assumptions together are known as the assumptions of local realism. Surprisingly this intuitively innocuous inequality can be violated in quantum mechanics. Here is an example. Let $|0\rangle=\begin{pmatrix}
1\\
0
\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}
0\\
1
\end{pmatrix}$. Then $|0\rangle$ and $|1\rangle$ are the eigenstates of
$$\sigma_z=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
Victor prepares a quantum system of two qubits in the state
\begin{align*}|\psi\rangle&=\frac{|0\rangle\otimes |1\rangle-|1\rangle\otimes |0\rangle}{\sqrt{2}}\\ &=\frac{|01\rangle – |10\rangle}{\sqrt{2}} \end{align*}
He passes the first qubit to Boris, and the second qubit to Natasha. Boris measures either of the observables
$$A_0=\sigma_z,\ A_1=\sigma_x=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$
and Natasha measures either of the observables
$$B_0=-\frac{\sigma_x+\sigma_z}{\sqrt{2}},\ B_1=\frac{\sigma_x-\sigma_z}{\sqrt{2}}$$
Since the system is in the state $|\psi\rangle$, the average value of $A_0\otimes B_0$ is
\begin{align*}\langle A_0\otimes B_0\rangle&=\langle\psi|A_0\otimes B_0|\psi\rangle\\ &=\frac{1}{\sqrt{2}} \end{align*}
Similarly, the average values of the other observables ar given by
\begin{align*}\langle A_0\otimes B_1\rangle&=\frac{1}{\sqrt{2}},\ \langle A_1\otimes B_0\rangle=\frac{1}{\sqrt{2}},\ \mbox{and}\\ \langle A_1\otimes B_1\rangle&=-\frac{1}{\sqrt{2}} \end{align*}
Since the expected value is linear, we have
\begin{equation}
\begin{aligned}
\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle&=\langle A_0B_0\rangle+\langle A_0B_1\rangle+\langle A_1B_0\rangle-\langle A_1B_1\rangle\\&=2\sqrt{2}\end{aligned}\label{eq:bell2}
\end{equation}
This means that the Bell inequality \eqref{eq:bell} is violated. Physicists have confirmed the prediction in \eqref{eq:bell2} by experiments using photons. It turns out that the Mother Nature does not obey the Bell inequality. What this means is that one or both of the two assumptions for the derivation of the Bell inequality \eqref{eq:bell} must be incorrect. There is no consensus among physicists which of the two assumptions needs to be dropped. An important lesson we learn from the Bell inequality is that the Mother Nature (Quantum Mechanics) defies our intuitive common sense. This also begs a troubling question. If we cannot rely on our intuition to understand how the universe works, what else can we rely on? One thing is certain. The world is not locally realistic.

References:

  1. Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2004

Quantum Gates: Single Qubit Gates

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A classical computer is built from electrical circuits that consist of logic gates and wires. The logic gates perform intended logical operations of information and wires pass information between logic gates. Similarly, a quantum computer would be built from quantum circuits that consist of quantum gates and wires. However, when it comes down to comparing how operations are performed at elementary gates level, there is no similarity between classical logic gates and quantum gates. Unlike classical logic gates, quantum gates operate not according to logical rules but to physical (quantum mechanical) laws. A single qubit gate acts on not only on qubits $|0\rangle=\begin{bmatrix}1\\0\end{bmatrix}$ and $|1\rangle=\begin{bmatrix}0\\1\end{bmatrix}$ but also on a superposition of $|0\rangle$ and $|1\rangle$, $\alpha|0\rangle+\beta|1\rangle$, where $\alpha$ and $\beta$ are complex numbers. For an obvious reason, single qubit gates are $2\times 2$ matrices. Furthermore, a quantum gate operation is performed through a unitary time evolution, and this means that they are required to be unitary matrices. Recall that a matrix $U$ of complex entries is unitary if $UU^\dagger=I$ where $\dagger$ denotes transpose conjugate i.e. $U^\dagger = \bar U^t$. If we write $U=\begin{bmatrix}a & b\\c & d\end{bmatrix}$, then $U$ is unitary if and only if $|a|^2+|b|^2=|c|^2+|d|^2=1$ and $a\bar c+b\bar d=0$. Any unitary matrix is qualified to be a quantum gate but not all unitary matrices would be interesting or useful gates. For practical purposes, it is important to see which gates are universal if there are any and that which gates are physically implementable. The following are some examples of well-known single cubit gates.

  1. Pauli-X, Pauli-Y, Pauli-Z gates. A matrix $H$ is Hermitian if $H=H^\dagger$. Hermitian matrices play a crucial role in quantum mechanics as Hamiltonians are required to be Hermitian. The eigenvalues of a Hamiltonian are the energies of the corresponding quantum system and of course they all have to be real. Hermitian matrices are guaranteed to have all real eigenvalues. A $2\times 2$ Hermitian matrix can be written in the form $H=\begin{bmatrix}x_0+x_3 & x_1-ix_2\\x_1+ix_2 & x_0-x_3\end{bmatrix}$ where the $x_i$ are all reals. The matrices \begin{equation}\label{eq:pauli}\sigma_0=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix},\ \sigma_1=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix},\ \sigma_2=\begin{bmatrix}0 & -i\\i & 0\end{bmatrix},\ \sigma_3=\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}\end{equation} form a basis for the space of all Hermitian matrices. The $\sigma_i$ in \eqref{eq:pauli} are called the Pauli spin matrices in physics. $\sigma_0$ is just the identity matrix but notice that $\sigma_1,\sigma_2,\sigma_3$ are also unitary matrices. They are called Pauli-X, Pauli-Y, Pauli-Z gates, respectively. Pauli-X gate is also called NOT gate as it acts like the classical NOT gate. \begin{align*}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}&=\begin{bmatrix}0\\1\end{bmatrix}\\\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}&=\begin{bmatrix}0\\1\end{bmatrix}\end{align*}
  2. Square root of NOT gate. The gate $$\sqrt{\mathrm{NOT}}=\frac{1}{2}\begin{bmatrix}1+i & 1-i\\1-i & 1+i\end{bmatrix}$$ is called square root of NOT gate because $$\sqrt{\mathrm{NOT}}\cdot\sqrt{\mathrm{NOT}}=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$ $\sqrt{\mathrm{NOT}}$ is one of the gates that are truly quantum because there is no classical counterpart.
  3. Hadamard gate. $$H=\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}$$ is called the Hadamard gate. It turns $|0\rangle$ to $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ and $|1\rangle$ to $\frac{|0\rangle-|1\rangle}{\sqrt{2}}$.

References:

  1. Michael A. Nielsen and Isaac L. Chung, Quantum Computation and Quantum Information, Cambridge University Press, 2000.
  2. Colin P. Williams and Scott H. Clearwater, Explorations in Quantum Computing, Spinger Telos, 1998.