Notes - Quantum Information HT24, Bloch sphere


Flashcards

Quickly derive the fact that the pure states of a qubit can be represented as points on the unit sphere.


Suppose $ \vert \psi\rangle = \alpha \vert 0 \rangle + \beta \vert 1\rangle$ is a unit vector. Then

\[\alpha = |\alpha|e^{i\gamma} , \beta = |\beta|e^{i\delta}\]

Where $\gamma$ and $\delta$ are suitable angles. Furthermore, $ \vert \alpha \vert ^2 + \vert \beta \vert ^2 = 1$ implies that

\[|\alpha| = \cos\frac\theta 2 , |\beta| = \sin \frac\theta 2\]

for some $\theta \in [0, \pi]$ (since non-negative). Then

\[|\psi\rangle = e^{i\gamma} \left[\cos\frac \theta 2 |0\rangle + \sin\frac\theta 2 e^{i\phi} |1\rangle\right]\]

where $\phi = \delta - \gamma$. Since $e^{i\gamma}$ is a global phase, this vector respresents the same state as

\[\cos\frac \theta 2 |0\rangle + \sin\frac\theta 2 e^{i\phi} |1\rangle\]

Hence we can use $\theta$ and $\phi$ to represent the spherical coordinates of a point in $\mathbb R^3$.

If $ \vert \psi \rangle$ is a quantum state represented by a Bloch vector $\pmb r$, where is the vector $\pmb r’$ that is orthogonal to $ \vert \psi\rangle$, and what does this tell you about the orthonormal bases and the Bloch sphere?


It is antipodal to $\pmb r$. This tells you that the pairs of antipodal points form the orthnormal bases for qubits.

What transformation do reversible processes (i.e. unitary) represent on the Bloch sphere?


Rotations.

Suppose we have a state

\[\cos\frac\theta 2 |0\rangle + \sin \frac\theta 2 e^{i\phi} |1\rangle\]

Where $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi)$. In spherical coordinates, this would be represented as $(\theta, \phi)$. How does the correspond to a vector $\pmb r \in \mathbb R^3$?


\[\pmb r = \begin{pmatrix}\sin \theta \cos \phi \\\\ \sin\theta\sin\phi \\\\ \cos \theta\end{pmatrix}\]

Suppose we have two states $ \vert \psi\rangle$ and $ \vert \psi’\rangle$, corresponding to Bloch vectors $\pmb r$ and $\pmb r’$. When using Born’s rule, we need to calculate

\[|\langle\psi | \psi'\rangle|^2\]

How can we do this using just the Bloch vectors?


\[|\langle\psi | \psi'\rangle|^2 = \frac{1 + \pmb r \cdot \pmb r'}{2}\]

Can you define the states $ \vert {+y}\rangle$ and $ \vert {-y}\rangle$, and describe their position on the Bloch sphere?


\[|{+y}\rangle = \frac{1}{\sqrt 2} (|0\rangle + i|1\rangle)\] \[|{-y}\rangle = \frac{1}{\sqrt 2} (|0\rangle -i|1\rangle)\]

They lie at the $\pm$ positions of the $y$-axis. This is why they are also called the $y$-basis.

What is meant by the $X$, $Y$ or $Z$ basis, where do they lie on the Bloch sphere, and how do these link to the $X$, $Y$ and $Z$ Pauli matrices?


  • $X$ basis:
    • Fourier basis, $ \vert +\rangle$ and $ \vert -\rangle$.
    • Lie on the intersection of the $x$-axis with the Bloch sphere.
    • The Pauli-$X$ matrix rotates by $\pi$ radians about the $x$-axis anticlockwise.
  • $Y$ basis:
    • $ \vert {+y}\rangle = \frac{1}{\sqrt 2} ( \vert 0\rangle + i \vert 1\rangle)$
    • $ \vert {-y}\rangle = \frac{1}{\sqrt 2} ( \vert 0\rangle -i \vert 1\rangle)$
    • Lie on the intersection of the $y$-axis with the Bloch sphere.
    • The Pauli-$Y$ matrix rotates by $\pi$ radians about the $y$-axis anticlockwise.
  • $Z$ basis:
    • Computational basis, $ \vert 0\rangle$, $ \vert 1\rangle$.
    • Lie on the intersection of the $z$-axis with the Bloch sphere.
    • The Pauli-$Z$ matrix rotates by $\pi$ radians about the $z$-axis anticlockwise.

Sketch the Bloch sphere with the correct axes.


One possible correct orientation is

  • $z$ pointing up
  • $y$ going left to right
  • $x$ going back to front

(The diagram above is slightly wrong, the arrow for the rotation about the $Y$-axis (i.e. the application of the $Y$ gate) should be going in the opposite direction, so that it fits the general rule where if you were looking directly at the point of the axis, the rotation goes anticlockwise).

You can remember by using the right-hand rule, where the thumb is $x$, the first finger is $y$ and the middle finger is $z$.

Give a decomposition of a $2\times 2$ unitary gate that demonstrates the single-qubit gates correspond to rotations of the Bloch sphere.


\[U = e^{i\gamma} \left[ \cos \frac \alpha 2 I - i \sin \frac \alpha 2 \pmb n \cdot \pmb \sigma \right]\]

where $\alpha \in [0, 2\pi)$ is a rotation angle, $n = (n _ x, n _ y, n _ z)^\top$ is the rotation axis, and $\phi = (X, Y, Z)^\top$ are the Pauli matrices.

How can you parameterise any state $ \vert \psi\rangle$, and what is the state orthogonal to $ \vert \psi^\perp\rangle$?


\[\begin{aligned} &|\psi\rangle = \cos\frac{\theta}{2} |0\rangle + e^{i\phi}\sin\frac{\theta}{2} |1\rangle \\\\ &|\psi^\perp\rangle = \sin\frac{\theta}{2} |0\rangle - e^{-i\phi}\cos\frac{\theta}{2} |1\rangle \end{aligned}\]

(in fact, this parameterisation is the natural one to correspond to Bloch vectors $\begin{pmatrix}\sin \theta \cos \phi \\ \sin\theta\sin\phi \\ \cos \theta\end{pmatrix}$)

Suppose we have two states $ \vert \psi\rangle$ and $ \vert \psi’\rangle$, corresponding to Bloch vectors $\pmb r$ and $\pmb r’$. Quickly prove that

\[|\langle\psi | \psi'\rangle|^2 = \frac{1 + \pmb r \cdot \pmb r'}{2}\]

Suppose

\[\begin{aligned} &|\psi\rangle := \cos\frac\theta 2 |0\rangle +e^{i\phi} \sin \frac \theta 2 |1\rangle \\\\ &|\psi'\rangle := \cos\frac{\theta'} 2 |0\rangle +e^{i\phi'} \sin \frac {\theta'} 2 |1\rangle \end{aligned}\]

Then

\[\begin{aligned} |\langle \psi | \psi' \rangle|^2 &= \left| \cos\frac{\theta}{2} \cos\frac{\theta'}{2} + \sin \frac{\theta}{2} \sin\frac{\theta'}{2} e^{i(\phi' - \phi)}\right|^2 \\\\ &= \left( \cos \frac\theta 2 \cos \frac{\theta'}{2} \right)^2 + \left( \sin \frac{\theta}{2} \sin\frac{\theta'}{2} \right)^2 + 2 \cos(\phi' - \phi) \sin \frac \theta 2 \cos \frac \theta 2 \sin \frac{\theta'}{2} \cos\frac{\theta'}{2} && (1\star) \\\\ &= \left(\frac{1 + \cos\theta}{2}\right) \left(\frac{1 + \cos\theta'}{2}\right) + \left(\frac{1 - \cos\theta}{2}\right) \left(\frac{1 - \cos\theta'}{2}\right) + \frac 1 2 \cos(\phi' - \phi)\sin\theta \sin\theta' \\\\ &= \frac{1 + \cos \theta \cos \theta' + \sin \theta \sin \theta' \cos(\phi' - \phi)}{2} \end{aligned}\]

$(1\star)$ comes from the relation $ \vert a + b \vert ^2 = \vert a \vert ^2 + \vert b \vert ^2 + 2\text{Re}(\overline a b)$.

But then

\[\begin{aligned} \pmb r \cdot \pmb r' &= \sin \theta \sin \theta' (\cos \phi \cos \phi' + \sin \phi \sin \phi') + \cos \theta \cos \theta' \\\\ &= \sin \theta \sin \theta' \cos(\phi' - \phi) + \cos \theta \cos \theta' \end{aligned}\]

so we see the result.

How can you write

\[\pmb n \cdot \pmb \sigma\]

where

\[\pmb n =\begin{pmatrix}\sin \theta \cos \phi \\\\ \sin\theta\sin\phi \\\\ \cos \theta\end{pmatrix}\]

and

\[\pmb \sigma = (X, Y, Z)^\top\]

?


\[\begin{pmatrix} \cos \theta & e^{-i\phi} \sin \theta \\\\ e^{i\phi}\sin \theta & {-\cos} \theta \end{pmatrix}\]

How can you write

\[\begin{pmatrix} \cos \theta & e^{-i\phi} \sin \theta \\\\ e^{i\phi}\sin \theta & {-\cos} \theta \end{pmatrix}\]

in terms of the Pauli matrices?


\[\pmb n \cdot \pmb \sigma\]

where

\[\pmb n =\begin{pmatrix}\sin \theta \cos \phi \\\\ \sin\theta\sin\phi \\\\ \cos \theta\end{pmatrix}\]

and

\[\pmb \sigma = (X, Y, Z)^\top\]

Quickly prove the decomposition of a $2\times 2$ unitary gate that demonstrates the single-qubit gates correspond to rotations of the Bloch sphere:

\[U = e^{i\gamma} \left[ \cos \frac \alpha 2 I - i \sin \frac \alpha 2 \pmb n \cdot \pmb \sigma \right]\]

where $\alpha \in [0, 2\pi)$ is a rotation angle, $n = (n _ x, n _ y, n _ z)^\top$ is the rotation axis, and $\phi = (X, Y, Z)^\top$ are the Pauli matrices.


Diagonalise $U$ to obtain

\[U = e^{i\lambda_0} |\psi_0\rangle\langle \psi_0| + e^{i\lambda_1} |\psi_1\rangle\langle\psi_1|\]

for two real numbers $\lambda _ 0$ and $\lambda _ 1$ and two orthonormal vectors $ \vert \psi _ 0\rangle$ and $ \vert \psi _ 1\rangle$. Defining

\[\gamma := \frac{\lambda_0 + \lambda_1}{2}\]

and

\[\alpha := \lambda_1 - \lambda_0\]

we can write

\[\begin{aligned} U &= e^{i\gamma} \left( e^{-i\alpha/2} |\psi_0\rangle\langle\psi_0| + e^{ia/2} |\psi_1\rangle\langle \psi_1| \right) \\\\ &= e^{i\gamma} \left((\cos \frac \alpha 2 (|\psi_0\rangle\langle\psi_0| + |\psi_1\rangle\langle\psi_1|) - i \sin \frac \alpha 2(|\psi_0\rangle\langle\psi_0| - |\psi_1\rangle\langle\psi_1|)\right) \\\\ &= e^{i\gamma} \left( \cos \frac \alpha 2 I - i\sin \frac \alpha 2 A \right) \end{aligned}\]

where

\[A = |\psi_0\rangle\langle\psi_0| - |\psi_1\rangle\langle\psi_1|\]

So it suffices to show that $A$ can be written as $\pmb n \cdot \pmb \sigma$ for some $\pmb n \in \mathbb R^3$. Since orthonormal vectors correspond to opposite points on the Bloch spehere, write

\[\begin{aligned} &|\psi_0\rangle = \cos\frac{\theta}{2} |0\rangle + e^{i\phi}\sin\frac{\theta}{2} |1\rangle \\\\ &|\psi_1\rangle = \sin\frac{\theta}{2} |0\rangle - e^{i\phi}\cos\frac{\theta}{2} |1\rangle \end{aligned}\]

Then

\[\begin{align*} A &= \left| \psi_0 \right\rangle \left\langle \psi_0 \right| - \left| \psi_1 \right\rangle \left\langle \psi_1 \right| \\\\ &= \left( \begin{array}{cc} \cos^2 \frac{\theta}{2} & e^{-i \varphi} \sin \frac{\theta}{2} \cos \frac{\theta}{2} \\\\ e^{i \varphi} \sin \frac{\theta}{2} \cos \frac{\theta}{2} & \sin^2 \frac{\theta}{2} \end{array} \right) - \left( \begin{array}{cc} \sin^2 \frac{\theta}{2} & -e^{-i \varphi} \sin \frac{\theta}{2} \cos \frac{\theta}{2} \\\\ -e^{i \varphi} \sin \frac{\theta}{2} \cos \frac{\theta}{2} & \cos^2 \frac{\theta}{2} \end{array} \right) \\\\ &= \left( \begin{array}{cc} \cos \theta & e^{-i \varphi} \sin \theta \\\\ e^{i \varphi} \sin \theta & -\cos \theta \end{array} \right) \\\\ &= \pmb n \cdot \pmb \sigma \end{align*}\]

Suppose a pure state $ \vert \psi\rangle$ has Bloch vector

\[\begin{pmatrix}\sin \theta \cos \phi \\\\ \sin\theta\sin\phi \\\\ \cos \theta\end{pmatrix}\]

Up to global phase, how can you write $ \vert \psi\rangle$ in terms of $ \vert 0\rangle$ and $ \vert 1\rangle$?


\[|\psi\rangle := \cos\frac\theta 2 |0\rangle +e^{i\phi} \sin \frac \theta 2 |1\rangle\]

Proofs




Related posts