# Quantum states

### Vector spaces over complex numbers

Quantum states are elements of an abstract vector space, $\mathbb{V}$, over the complex numbers, $\mathbb{C}$. This is in contrast to the space of classical states, which is a set.

Conventionally, elements of $\mathbb{V}$ are column vectors of complex numbers. Using Dirac's notation, we call them ket vectors. Thus, if $$\ket{a} \in \mathbb{V}$$ then $$\exists\; a_i \in \mathbb{C} \text{ for } i = 1,2,\cdots,N$$ such that $$\ket{a} \to \left [ \begin{matrix} a_1 \\ \vdots \\ a_N \end{matrix} \right ]$$

Scalar multiplication is given by $$c \ket{a} \to \left [ \begin{matrix} c a_1 \\ \vdots \\ c a_N \end{matrix} \right ]$$ where $c \in \mathbb{C}$.

$$\ket{a} + \ket{b} \to \left [ \begin{matrix} a_1 + b_1 \\ \vdots \\ a_N + b_N \end{matrix} \right ]$$ where $\ket{b} \in \mathbb{V}$ as well.

The dual space of the space of $\mathbb{V}$, denoted $\mathbb{V}^*$, is called the space of complex conjugate vectors. We use Dirac notation as well, but conjugates are labelled as bra vectors.

Conjugates are represented by row vectors, but their elements are the complex conjugates of the associated vector in the dual space.

For example, if $\ket{a} \in \mathbb{V}$, then $$\bra{a} \to [a_1^*, \cdots, a_N^*] \in \mathbb{V}^*$$ is called the complex conjugate of $\ket{a}$.

Nb. We don't include the conjugate-sign $*$ in the bra-label - this is implicit.

Vector multiplication, or the inner product, of two vectors, $\ket{a}, \ket{b} \in \mathbb{V}$, is given by \begin{align*}\braket{b}{a} &= [ b_1^* \cdots b_N^* ] \left [ \begin{matrix} a_1 \\ \vdots \\ a_N \end{matrix} \right ]\\&= b_1^* a_1+... +b_N^* a_n\end{align*}

The square of the vector is $$\braket{a}{a} = a_1^* a_1+... + a_N^* a_N \in \mathbb{R}$$ since the product of a complex number with its conjugate is always real.

We can consider the complex conjugate of the inner product, \begin{align*}\braket{b}{a}^*&= \left(b_1^* a_1 + \cdots + b_N^* a_N \right)^*\\&= \left(a_1^* b_1 + \cdots + a_N^* b_N \right)\\&= \braket{a}{b}\end{align*}

### Single bit quantum states

Recall, in the detection experiment, we prepared the electron at some angle to the vertical, and then tried to measure the electron's spin with respect to the vertical. We said there were only two distinct possibilities - either we don't detect a photon, corresponding to the up state, which we label $\ket{u}$, or we do detect a photon, corresponding to the down state, which we label $\ket{d}$.

We arbitrarily chose to set up to mean pointing in the $+z$ direction of normal space, and down to mean pointing in the $-z$ direction of normal space, but vectors in normal space do not define states in quantum mechanics. States in quantum mechanics are represented by ket vectors. The dimension of the space is determined by the number of bits that describes the system in question.

In our case, the number of bits is one, hence the dimension of the space, $\mathbb{V}$, is $2^1 = 2$. We want to choose vectors in this space to represent the two states (of the bit). By convention, we define the ket vectors, $\ket{u}, \ket{d}$ to have the components \begin{align*}\ket{u} & \to \begin{bmatrix} 1\\0 \end{bmatrix}\\\ket{d} & \to \begin{bmatrix} 0\\1 \end{bmatrix}\end{align*}

These form an orthonormal basis for $\mathbb{V}$. That is, they are both unit vectors $$\braket{u}{u} = \braket{d}{d} = 1$$ and they are mutually orthogonal $$\braket{u}{d} = \braket{d}{u} = 0$$

We now can define a general single-bit quantum state, $\ket{a} \in \mathbb{V}$ to be a unit linear combination of the two basis vectors $\ket{u}, \ket{d}$. That is, $\exists\; a_u, a_d \in \mathbb{C}$ such that $$\ket{a} \to a_u \ket{u} + a_d \ket{d} = \begin{bmatrix} a_u\\a_d\end{bmatrix}$$ and \begin{align*}\braket{a}{a} &= \begin{bmatrix} a_u^*,a_d^* \end{bmatrix}\begin{bmatrix} a_u\\a_d\end{bmatrix}\\&= a_u^*a_u + a_d^*a_d\\&= 1\end{align*}

Note that, instead of using columns and rows to calculate the square of a (ket) vector, we could use the fact that $\ket{u}, \ket{d}$ are basis vectors. Then, in detail \begin{align*}\braket{a}{a} &= \bra{a} \cdot \ket{a}\\&= \left(\bra{u} a_u^* + \bra{d} a_d^* \right) \cdot \left( a_u \ket{u} + a_d \ket{d} \right)\\&= \bra{u}{a_u^*a_u}\ket{u} + \bra{u}{a_u^*a_d}\ket{d} + \bra{d}{a_d^*a_u}\ket{u} + \bra{d}{a_d^*a_d}\ket{d}\\&= a_u^*a_u \braket{u}{u} + a_u^*a_d \braket{u}{d} + a_d^*a_u \braket{d}{u} + a_d^*a_d \braket{d}{d}\\&= a_u^*a_u (1) + a_u^*a_d (0) + a_d^*a_u (0) + a_d^*a_d (1)\\&= a_u^*a_u + a_d^*a_d\end{align*}

Notice that the ket vectors contain four real numbers in two complex components whereas the prepared directions of an electron can be represented by a unit-length spatial vector in $\mathbb{R}^3$, containing only two real numbers. This is resolved by the condition that quantum states are phase-independent. That is, we can multiply a state by the factor $$e^{\mathrm{i}\theta}$$ for any $\theta$, and not change the probability of being detected (see below) - meaning the physics doesn't change. This eliminates one real number from the abstract vector. A second real number is eliminated by insisting that the quantum state is also unit-length.

### Probability of being detected

Given the general state, $\ket{a}$, of an electron (prepared at some angle to the vertical), we define the probability of the electron being detected in the up-state as $$p_u=a_u^* a_u$$

Similarly, $$p_d=a_d^* a_d$$ is the probability that the electron is detected in the down-state.

For these values to be probabilities, we would require that they add up to one. But, $$p_u + p_d = a_u^* a_u + a_d^* a_d = 1$$ since this is just the square $\braket{a}{a}$ of a unit vector.

We refer to the components of a quantum state as probability amplitudes, which square to form probabilities.

Nb. Knowing the probabilities alone does not completely determine the coefficients, since we can multiply the co-efficients by a pure phase complex number, of the form $e^{\mathrm{i}\theta}$, and not change the probabilities.

### The three directions of space

We want to define the state of an electron that's been prepared in an arbitrary direction (in normal space), and then is detected to be in either the up or down state. We first consider preparation in the three standard directions.

Since we chose (arbitrarily) to associate the up state with the $+z$ direction, then we would expect that, if the electron was prepared in the $+z$ direction, then the probability of it being detected in the up state to be one. The probability of it being detected in the down state to be zero. That is, $$\ket{+z} = \begin{bmatrix}1 \\ 0 \end{bmatrix} = \ket{u}$$

Similarly, if the electron was prepared in the $-z$ direction, then the probability of it being detected in the up state to be zero and the probability of it being detected in the down state to be one, $$\ket{-z} = \begin{bmatrix}0 \\ 1 \end{bmatrix} = \ket{d}$$

The states corresponding to the electron being prepared in the other directions, $x, y$ have been determined by experiment. They are, \begin{align*}\ket{+x} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \end{bmatrix} & \ket{-x} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{-1}{\sqrt{2}} \end{bmatrix}\\\ket{+y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{\mathrm{i}}{\sqrt{2}} \end{bmatrix} & \ket{-y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{-\mathrm{i}}{\sqrt{2}} \end{bmatrix}\end{align*}

We can write explicitly in terms of the up and down states, \begin{align*}\ket{+x} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} + \ket{d}\right) &\quad \ket{-x} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} -\, \ket{d}\right)\\\ket{+y} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} + i\ket{d}\right) &\quad \ket{-y} &= \tfrac{1}{\sqrt{2}}\left(\ket{u} -\, \mathrm{i}\ket{d}\right)\end{align*}

We can read these as follows: the probability of the electron being detected in the up state, having been prepared in the $+x$ direction, is given by $$p_u(+x) = \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{2}$$

It is easily shown that all of these are unit vectors, for example, \begin{align*}\braket{+y}{+y} &= \begin{bmatrix}\tfrac{1}{\sqrt{2}} , \tfrac{-\mathrm{i}}{\sqrt{2}} \end{bmatrix} \begin{bmatrix}\tfrac{1}{\sqrt{2}} \\ \tfrac{\mathrm{i}}{\sqrt{2}} \end{bmatrix}\\&= \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} + \tfrac{-\mathrm{i}}{\sqrt{2}} \cdot \tfrac{\mathrm{i}}{\sqrt{2}}\\&= \tfrac{1}{2} + \tfrac{1}{2}\\&= 1\end{align*}

Also, note that the $\pm$ vectors representing each direction (of normal space) are mutually orthogonal, $$\braket{+x}{-x} = \braket{+y}{-y} = 0 \:( = \braket{+z}{-z} )$$ but a vector of one direction is not mutually orthogonal to a vector of another direction. For example, $$\braket{+x}{-y} \neq 0$$