0
$\begingroup$

I am currently reading the "The Mathematics of Quantum Coin-Flipping" by Carl A. Miller. On page 1909 (the second page in the pdf linked above) the author defines the state of a qubit as a matrix of the form

$$\phi = \begin{bmatrix} a &b \\ \overline{b} &c\end{bmatrix},$$

where $a$ and $c$ are real numbers such that $a + c = 1$ and $b$ is a complex number such that the determinant $ac-\lvert b \rvert^2 \ge 0$. In case of a measurement said qubit is $0$ with probability $a$ and $1$ with probability $c$.

However, from the lecture I only know the following (standard?) definition: A qubit is in the state $\alpha \vert 0 \rangle + \beta \vert 1 \rangle$, where $\alpha, \beta \in \mathbb{C}$, such that $\lvert\alpha\rvert^2 + \lvert \beta \rvert^2 = 1$. In case of measurement said qubit is $0$ with probability $\lvert\alpha\rvert^2$ and $1$ with probability $\lvert \beta \rvert^2$.

If I am not mistaken these definitions should be equivalent, but I do not see it. Could you please explain this to me?

$\endgroup$
1
  • 1
    $\begingroup$ The state which you are referring to (the vector) is what is known as the pure state. The state from the paper is what is known as the density matrix. Basically, the density matrix is a weighted combination of pure states, where the weight (probability) is a classical probability, our ignorance towards the system, and not some inherent indeterminism. Try going through section 2.4 of Nielsen and Chaung or section III of this. $\endgroup$
    – FDGod
    May 30, 2023 at 18:58

1 Answer 1

2
$\begingroup$

The matrix $\phi$ you give is a density matrix representation of a qubit which is different to the statevector form you may be used to.

As you say the state of a qubit can be represented by a statevector $|\psi\rangle$ with the following form.

\begin{equation} |\psi\rangle = \alpha |0\rangle + \beta|1\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \end{equation}

Where $\alpha$ and $\beta$ are complex numbers satisfying \begin{equation} |\alpha|^2 +|\beta|^2=1. \end{equation}

The density matrix (an equally valid form usually denoted by $\rho$) allows you to write statistical mixtures of states rather than just pure quantum states.

The density matrix of a pure state $|\psi\rangle$ is given by the outer product $|\psi\rangle\langle \psi|$.

\begin{equation} \rho = |\psi \rangle\langle\psi| = \begin{pmatrix}\alpha \\ \beta\ \end{pmatrix}\begin{pmatrix}\bar{\alpha}& \bar{\beta} \end{pmatrix} = \begin{pmatrix}|\alpha|^2& \alpha\bar{\beta} \\ \beta \bar{\alpha} & |\beta|^2\end{pmatrix} \end{equation}

Clearly $|\alpha|^2$ and $|\beta|^2$ are both real numbers so we can replace them with the new letters $a$ and $c$. This is where $a+c=1$ comes from. The diagonal entires of a density matrix correspond to probabilities so the trace must equal 1.

Also if we conjugate the product $(\alpha \bar{\beta})\,$, we get $\beta \bar{\alpha}\,$. This means we can set $b=\alpha \bar{\beta}$.

Therefore the matrix $\rho$ above is equivalent to the matrix $\phi$ that you give with \begin{align} a&=|\alpha|^2, \\ b&=\alpha\bar{\beta}\,, \\ c &= |\beta|^2\,. \end{align}

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.