Bras and kets
Until now we have been using a specific notation (as in $ |\psi ⟩$) to refer to quantum states. This is not just its name, but it indicates a complex column vector $ |\psi ⟩$ that describes the quantum state. It is an element of a vector space $\mathbb{C}^d$, where $d $ is the dimension of the vector space. $d $ depends on the physical system under consideration.
Such a column vector is called a ket vector in quantum theory.
Corresponding to $|\psi⟩$ (a column vector!), there exists a row vector $⟨ \psi|$. A row vector is called a bra vector in quantum theory.
Let $\{|e_1 ⟩, \ldots, |e_d ⟩\} $ be a set of basis vectors. We assume $|e_k⟩ $has unit length $⟨ e_k|e_k ⟩=1 $ and orthogonal$⟨ e_j |e_k ⟩=0 \ (j \neq k)$. Such a basis is called "orthonormal". We will employ an orthonormal basis in the following unless mentioned otherwise. The standard assignment is$|e_k ⟩=(0,0,\ldots, 0,1,0, \ldots,0)^t$, where 1 is in the $k$-th position. Note that $x_k =⟨ e_k|\psi⟩ $ if$|\psi ⟩ = \sum_{k=1}^{d} x_k| e_k ⟩$. The complex number $x_k=⟨ e_k|\psi ⟩ $ is called the $k$-th component of $|\psi ⟩ $and $|\psi⟩ $ is expressed as$|\psi ⟩ =(x_1, x_2, \ldots, x_d)^t$.
Suppose $|\psi ⟩ = (x_1, x_2, \ldots, x_d)^t$. Then $⟨ \psi| $ is defined as $ (\bar{x}_1, \bar{x}_2, \ldots \bar{x}_d)$.
What is the origin of complex conjugation in the definition of the bra vector?
For this, let us consider the product
$ ⟨ \psi|\cdot |\psi ⟩ = (\bar{x}_1, \bar{x}_2, \ldots \bar{x}_d) \begin{pmatrix} x_1\\ x_2\\ \vdots \\ x_d \end{pmatrix} = \sum_{k=1}^d |x_k|^2. $
$⟨ \psi|\cdot |\psi ⟩ $is often written as $⟨ \psi|\psi ⟩$.
It is clear that $⟨ \psi|\psi ⟩ $ is a non-negative real number because of complex conjugation in the definition of a bra vector. It vanishes if and only if $|\psi ⟩ $is a zero-vector $(0,0,\ldots, 0)^t$. As in the case of real vectors, we define the length (norm) of $|\psi⟩ $by $\||\psi⟩\|=\sqrt{⟨ \psi|\psi ⟩}$.Now the nomenclatures "bra" and "ket&34; should be clear. They are combined to produce a "bracket" $⟨ \psi|\psi ⟩$.
A norm satisfies $\| |\psi ⟩\| \geq 0$, $\|c|\psi ⟩\|=|c|\||\psi ⟩\| $and $\||\psi ⟩ + |\phi⟩\|\leq \||\psi ⟩ \|+\||\phi⟩\| $(triangular inequality), where $c $ is a complex number.
Due to the reason mentioned later, we often require $|\psi ⟩ $be a unit vector, $\| |\psi ⟩\|=1$. Multiplying a constant to make a non-zero vector a unit vector is called normalization.
Deepening your understanding of bras and kets
To get yourself familiar with bras and kets, we have compiled three exercises for you. Feel free to use them to deepen your understanding. There will be less math after this page ;-)
!
Let $|\psi ⟩ = (1, 1+i, 2-i)^t$. Find $⟨ \psi|\psi ⟩$ and normalize $|\psi ⟩ $.
A
$⟨ \psi|\psi ⟩ = 8$, and the normalized vector is $\frac{1}{\sqrt{8}}|\psi⟩$
B
$⟨ \psi|\psi ⟩ = 4$, and the normalized vector is $\frac{1}{4}|\psi⟩$
C
$⟨ \psi|\psi ⟩ = 2$, and the normalized vector is $\frac{1}{\sqrt{2}}|\psi⟩$
!
An inner product of two vectors $|\psi ⟩ $and $|\phi⟩ $is given by $⟨ \phi|\psi ⟩$or $⟨ \psi|\phi⟩$ with:$
⟨ \phi|\psi ⟩ = (\bar{y}_1, \bar{y}_2, \ldots \bar{y}_d)
\begin{pmatrix}
x_1\\
x_2\\
\vdots \\
x_d
\end{pmatrix} = \sum_{k=1}^d \bar{y}_i x_i
$Note that these two inner products are complex conjugate of each other; $⟨ \phi|\psi ⟩= \overline{⟨ \psi|\phi ⟩}$.
Let $|\psi ⟩ = (1, 1+i, 2-i)^t $and $|\phi ⟩= (2i, 1-i, 0)^t$. Find $⟨ \phi|\psi ⟩ $ and $⟨ \psi|\phi⟩$.
A
$⟨ \phi|\psi ⟩ = 1 $ and $⟨ \psi|\phi⟩=1$
B
$⟨ \phi|\psi ⟩ = 1+i $ and $⟨ \psi|\phi⟩=1-i$
C
$⟨ \phi|\psi ⟩ = 0 $ and $⟨ \psi|\phi⟩=0$
Vectors are added and multiplied by a complex number. Let $|\psi ⟩ = (1, 1+i, 2-i)^t $and $|\phi ⟩= (2i, 1-i, 0)^t$. Then $|\psi ⟩ + |\phi⟩= (1+2i, 2, 2-i)^t$ and $(1-i)|\psi ⟩= (1-i, 2, 1-3i)^t $ for example. Addition of two or more vectors is called superposition of vectors in the context of quantum theory. Note that one cannot add a ket vector and a bra vector or two ket vectors whose dimensions are different.
!
Let $|\psi ⟩ = (1, 1+i, 2-i)^t $and $|\phi ⟩= (2i, 1-i, 0)^t$. Find $|\chi⟩= (1-i)|\psi ⟩ + (1+i)|\phi⟩$ and $\| |\chi⟩\|$.
A
$|\chi⟩=(-1+i,4, 1-3i)^t) $ and $\| |\chi⟩||=\sqrt{28}$
B
$|\chi⟩=(0,1, 2i)^t) $ and $\||\chi⟩||=\sqrt{6}$
C
$|\chi⟩=(1+i,2, 1-3i)^t) $ and $\| |\chi⟩||=\sqrt{11}$