Notes › PHYS 3571: Quantum Computing Lecture 1

Scalar Product and Braket Notation

calendar_today January 22, 2026 article 430 words access_time 2 min replay Modified February 5, 2026

Table of Contents

Scalar Product

Qubit states are $\{\ket{0}, \ket{1}\}$
- Each state corresponds to a basis for the $\hat{i}$ and $\hat{j}$ unit vectors
Scalar product notation:
- $\bra{0}\ket{0} = \hat{i} \cdot \hat{i} = 1$
- $\bra{1}\ket{0}$ = 0
- $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$ ($\alpha, \beta$ complex-valued)
Conjugate property:
- $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}\implies \bra{\psi} = \alpha^* \bra{0} + \beta^* \bra{1}$
- $\bra{\psi}\ket{\psi} = \alpha^*\alpha + \beta^* \beta$
Normalization property:
- $\bra{\psi}\ket{\psi} = 1$ means $\ket{\psi}$ is normalized
- $\bra{\psi}\ket{\phi} = 0$ means $\ket{\psi}$ and $\ket{\phi}$ are orthogonal
- If both properties hold for a set of bases, the set is orthonormal

Properties of the Scalar Product

For any 2 vectors $\ket{\psi}$ and $\ket{\phi}$:

$$\bra{\psi}\ket{\phi} = \bra{\phi}\ket{\psi}^*$$

For any vector $\ket{\psi}$:
- Only 0 iff $\ket{\psi} = \mathbf{0}$

$$\bra{\psi}\ket{\psi} \ge 0$$

For any two vectors $\ket{\psi}$ and $\ket{\phi}$ (Schwarz/Triangle Inequality):
- Basically only used to prove the Uncertainty Principle

$$|\bra{\psi}\ket{\phi}|^2 \le \bra{\psi}\ket{\psi} \bra{\phi}\ket{\phi}$$

BraKet Notation

How to turn $\ket{\psi}$ into $\bra{\psi}$?
- $\hat{A}, \hat{B}$ are operators
- $\hat{A} ^{\dagger}$ is the Hermitian conjugate of $\hat{A}$ defined as: $\ket{\phi} = \hat{A} \ket{\psi} \implies \bra{\phi} = \bra{\psi}\hat{A} ^{\dagger}$
  - If $\hat{A} = \hat{A}^{\dagger}$, then $\hat{A}$ is Hermitian

$$\ket{\psi} = \alpha \hat{A} \ket{\phi_{1}} + \beta \hat{B} \ket{\phi_{2}}$$$$\bra{\psi} = \alpha^* \bra{\phi} \hat{A}^{\dagger} + \beta^* \bra{\phi_{2}} \hat{B ^{\dagger}}$$

Bloch Sphere

Recall $z = re^{i \theta}$ where $|z| = r$ and $\theta = \tan^{-1}\left( \frac{z_{y}}{z_{x}} \right)$
Therefore:

$$\begin{align} \ket{\psi} = \alpha \ket{0} + \beta \ket{1} &= |\alpha| e^{i \theta_{\alpha}} \ket{0} + |\beta| e^{i \theta_{\beta}} \ket{1} \\ &= e^{i \theta_{\alpha}} \left( |\alpha| \ket{0} + |\beta| e^{i (\theta_{\beta} - \theta_{\alpha})} \ket{1}\right) \\ &= |\alpha| \ket{0} + |\beta| e^{i \phi} \ket{1} \end{align}$$

If we denoted $\theta_\alpha$ as the overall phase, then no prediction of QM depends on the overall phase, since it’s a scalar multiple
If $\ket{\psi}$ is normalized (which it must be to represent a quantum system), then:

$$\bra{\psi} \ket{\psi} = 1 = |\alpha|^2 + |\beta|^2$$

We can parameterize the previous result with $|\alpha| = \cos \left( \frac{\theta}{2} \right), |\beta| = \sin\left( \frac{\theta}{2} \right)$ such that $\ket{\psi}$ only depends on $\theta, \phi$:
- $\theta$ is the angle formed from the $z$-axis to the point’s displacement vector
- $\phi$ is the angle formed from the $x$-axis to the projection of the point on the $xy$-plane

$$\ket{\psi} = \cos \left( \frac{\phi}{2} \right) \ket{0} + e^{i \psi} \sin \left( \frac{\theta}{2} \right) \ket{1}; \theta \in [0, \pi], \phi \in [0, 2\pi)$$

Which point $(\theta, \phi)$ is $\ket{0}$?
- $(0, \_)$
Which point is $\ket{1}$?
- $(\pi, \_)$
This is why $\{\ket{0}, \ket{1}\}$ are called $z$-basis, since they both intersect the sphere at the $z$-axis
- $x$-basis is $\{\ket{+}, \ket{-}\} = \left\{ \left( \frac{\pi}{2}, 0 \right), \left( \frac{\pi}{2}, \pi \right) \right\}$

Graph

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