### Lei Mao

Machine Learning, Artificial Intelligence, Computer Science.

# Quantum System States

### Introduction

In quantum theory, a system state could be represented using a complex vector $| \psi \rangle$. Usually, we prefer to use a normalized complex vector $| \psi \rangle$ whose norm $|| \psi \rangle| = 1$, because it is mathematically convenience to derive some important properties of the system state. Any complex vector could be normalized by being divided by its norm, i.e., $\frac{| \psi \rangle}{|| \psi \rangle|}$.

The unormalized and normalized complex vector only differ by a constant real valued factor $|| \psi \rangle|$. In quantum theory, actually, for any complex vector $| \psi \rangle$ and $c | \psi \rangle$ where $c \in \mathbb{C}$ and $c \neq 0$, they represent the same system state. In this blog post, we would like to discuss why this is the case mathematically.

### Quantum System State

In quantum theory, a system state $| \psi \rangle \in \mathbb{C}^n$could be described as the a linear combination of basic state vectors $| x_i \rangle \in \mathbb{C}^n$, by suitable complex amplitude $c_i$, which is typically called superposition.

\begin{align} | \psi \rangle &= c_0 | x_0 \rangle + c_1 | x_1 \rangle + \cdots + c_{n-1} | x_{n-1} \rangle \\ &= \sum_{i=0}^{n-1} c_i | x_i \rangle \end{align}

The basic state vectors, $| x_i \rangle$, are orthonormal to each other, and they form a basis for $\mathbb{C}^n$. When system state $| \psi \rangle$ is observed, we would find it in one of the basic states $| x_i \rangle$. The probability of finding the system being in the state $| x_i \rangle$ after observation is

\begin{align} p_{| \psi \rangle}( x_i ) &= \frac{|c_i|^2}{| | \psi \rangle |^2} \end{align}

The square norm of a complex vector $v$ could be computed using the inner product of itself, i.e.,

$|v|^2 = \langle v, v \rangle$

If you are not familiar with the concept of inner product and how it was computed, please check my previous blog post on the topic of inner product.

Computing $|c_i|^2$ looks trivial. Let’s check what $| | \psi \rangle |^2$ is.

\begin{align} | | \psi \rangle |^2 &= \langle | \psi \rangle, | \psi \rangle \rangle \\ &= \langle \sum_{i=0}^{n-1} c_i | x_i \rangle, \sum_{i=0}^{n-1} c_i | x_i \rangle \rangle \\ &= \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} \langle c_i | x_i \rangle, c_j | x_j \rangle \rangle \\ &= \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} c_i \overline{c_j} \langle | x_i \rangle, | x_j \rangle \rangle \\ &= \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} \langle c_i, c_j \rangle \langle | x_i \rangle, | x_j \rangle \rangle \\ &= \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} \langle c_i, c_j \rangle \langle | x_i \rangle, | x_j \rangle \rangle \\ &= \sum_{i=0}^{n-1} \langle c_i, c_i \rangle \langle | x_i \rangle, | x_i \rangle \rangle \\ &= \sum_{i=0}^{n-1} |c_i|^2 \times 1 \\ &= \sum_{i=0}^{n-1} |c_i|^2 \\ \end{align}

Therefore,

\begin{align} p_{| \psi \rangle}( x_i ) &= \frac{|c_i|^2}{| | \psi \rangle |^2} \\ &= \frac{|c_i|^2}{\sum_{i=0}^{n-1} |c_i|^2} \end{align}

Let’s further check what $c | \psi \rangle$ is where $c \in \mathbb{C}$ and $c \neq 0$.

\begin{align} | c \psi \rangle &= c c_0 | x_0 \rangle + c c_1 | x_1 \rangle + \cdots + c c_{n-1} | x_{n-1} \rangle \\ &= \sum_{i=0}^{n-1} c c_i | x_i \rangle \end{align}

Similarly,

\begin{align} p_{c | \psi \rangle}( x_i ) &= \frac{|c c_i|^2}{| | c \psi \rangle |^2} \\ \end{align}

For $|c c_i|^2$, we have

\begin{align} |c c_i|^2 &= \langle c c_i, c c_i \rangle \\ &= c \overline{c} \langle c_i, c_i \rangle \\ &= \langle c, c \rangle \langle c_i, c_i \rangle \\ &= |c|^2 |c_i|^2 \\ \end{align}

For $| c | \psi \rangle |^2$, we have

\begin{align} | c | \psi \rangle |^2 &= \langle c | \psi \rangle, c | \psi \rangle \rangle \\ &= c \overline{c} \langle | \psi \rangle, | \psi \rangle \rangle \\ &= |c|^2 \langle | \psi \rangle, | \psi \rangle \rangle \\ &= |c|^2 | | \psi \rangle |^2 \\ &= |c|^2 \sum_{i=0}^{n-1} |c_i|^2 \\ \end{align}

Therefore,

\begin{align} p_{c | \psi \rangle}( x_i ) &= \frac{|c c_i|^2}{| | c \psi \rangle |^2} \\ &= \frac{|c|^2 |c_i|^2}{|c|^2 \sum_{i=0}^{n-1} |c_i|^2} \\ &= \frac{|c_i|^2}{\sum_{i=0}^{n-1} |c_i|^2} \\ &= p_{| \psi \rangle}( x_i ) \end{align}

### Conclusions

Because it turns out that $p_{c | \psi \rangle}( x_i ) = p_{| \psi \rangle}( x_i )$ for any $i \in [0,n-1]$, this means $| \psi \rangle$ and $c | \psi \rangle$, where $c \in \mathbb{C}$ and $c \neq 0$, represent the same system state.