The bra vector may be seem as one of the most unintuitive concepts in introductory quantum mechanics. One may understand that it provides a close analogy to inner products from regular linear algebra. Some may see it as a duo vector, but then the mapping $\phi: \ket{\psi}\rightsquigarrow \bra{\psi}$ is certainly not uniquely given and must be induced by an arbitrary basis $e$. I hope to demonstrate in this discussion that by choosing the mapping $\phi: \ket{\psi}\rightsquigarrow \bra{\psi}$ for its inner product space, nature uniquely selects a basis $e$ that prescribes its definition of orthogonality.
Let $V$ be a vector space. Let $F$ be the underlying field of the vector space.
- The inner product $(\cdot,\cdot)$ is an operator $V\times V\rightarrow F$ satisfying
- Linearity: $\forall u,v \in V, \forall c\in F$: $$(u,cv) = c(u,v).$$
- Conjugate symmetry: $$(u,v) = \overline{(v,u)}.$$
- Positive-semi definite on $(u,v)$: $$\forall v\in V, (v,v) \geq 0,$$ $$(v,v) = 0 \iff v=0.$$
The orthogonal representation via $\R^n$
- According to Gramschmidt, for any inner product space $V$, $\exists$ an orthogonal basis $\beta = \{\beta_i\}$, s.t. $(\beta_i, \beta_j) = \delta_{ij}.$ The choice of $\beta$ is not unique.
- We call a representation \begin{align} v =\sum_{i=1}^n v_i \beta_i &= \begin{bmatrix} v_1\\ \vdots \\ v_n\end{bmatrix}_\beta \text{the orthogonal representation. } \end{align}
- In particular, for any two vectors $u = \sum_{i=1}^n u_i\beta_i, v = \sum_{i=1}^n v_i\beta_i$, their inner products satisfy: \begin{align} (u,v) &= \left(\sum_{i=1}^n u_i\beta_i, \sum_{i=1}^n v_i\beta_i\right)\\&= \sum_{ij}^n u_i^*v_i(\beta_i, \beta_j)\\&= \sum_{ij}^n u_i^*v_i\delta_{ij}\\ &= \sum_{i=1}^n u_i^* v_i,\end{align} which is compliant with the matrix multiplication between the adjoint of the orthogonal representation of $u$ and the orthogonal representation of $v$: $$(u,v) = [u]^\dagger_\beta[v]_\beta.$$
The adjoint of a linear operator
Definition. The adjoint of a matrix $M\in Mn_2(\R)$ is given by: $$(M^\dagger)_{ij} = (M)_{ji}^*.$$
Definition. The adjoint of a linear operator $T:V\rightarrow V$ is denoted by $T^\dagger$, which satisfy the property of $\forall u,v\in V$, $$(u,Tv) = (T^\dagger u, v). $$
- It can be proven that such a matrix exists and is unique for any given inner product.
- It is clear from the definition of $[M]_\beta^\gamma$ that $$[Tv]_\beta = [T]_\beta[v]_\beta.$$
Then it follows that $$(u,Tv) = [u]^\dagger_\beta[Tv]_\beta = [u]_\beta^\dagger[T]_\beta[v]_\beta. (1)$$
- Now noting that matrix multiplication is associative, and that $([T][u])^\dagger = [u]^\dagger[T]^\dagger:$
- \begin{align}
([A][B])_{ij} &= \sum_k^n ([A])_{ik}([B])_{kj}\\
\Rightarrow ([A][B])_{ij}^\dagger &= \sum_k^n ([B])_{ki}^*([A])_{jk}^*\\
&= \sum_k^n ([B]^\dagger)_{ik}([A]^\dagger)_{jk}\\
&= \left([B]^\dagger [A]^\dagger\right)_{ij}.
\end{align}
$$\text{Hence } ([A][B])^\dagger = [B]^\dagger[A]^\dagger.$$ - Going back to (1), we now have:
\begin{align}
(u,Tv) &= [u]_\beta^\dagger[T]_\beta[v]_\beta\\
&=\left([T]_\beta^\dagger [u]_\beta\right)^\dagger [v]_\beta\\
&=(T^\dagger u, v).
\end{align}
Note that $T^\dagger = [T]^\dagger_\beta$ here immediately advocates itself as the matrix of the operator adjoint of $T$. Its uniqueness is given by the consideration of $[T^\dagger]$ on orthogonal set $\beta$ of $(\cdot ,\cdot )$: $\forall u,v \in \beta, (u, Tv) = [u]_\beta^\dagger[T]_\beta[v]_\beta$ gives the explicit value of each entry of $[T^\dagger]_\beta$. - Hence for any $T:V\rightarrow V$ linear, $\exists ! T^\dagger: V\rightarrow V$, s.t. $\forall u,v \in V, (u,Tv) =(T^\dagger u,v).$ Additionally it is given by $[T^\dagger]_\beta = [T]^\dagger_\beta$.
The bra-ket notations
Definition. Let H be the vector space of concern, we use notation $\ket{\psi} \in H$ to denote a vector representation from H.
Definition. The duo space $H^*$ of $H$ is given by the set of linear forms mapping from $H$ to F $$H^* = \{T \big| T:H\rightarrow F \text{ linear}\}.$$
- Note 1: $dim(H^*) = dim(H).$
- Note 2: Under a basis $\beta$, we can create bijection between $H$ and $H^*$ given by: $$[v]_\beta \leftrightsquigarrow [v]_\beta^\dagger.$$
Definition. We now define the duo of a vector $\ket{\psi}$ as $\bra{\psi} = \phi(\ket{\psi})$, where $\phi$ is a linear mapping $\phi$ from $H$ to $H^*$ satisfying the properties (Dirac) that:
- $\forall \ket{u}, \ket{v}, \ket{w} \in H,$
$$\bra{w}(\ket{u}+\ket{v}) = \braket{w|u}+\braket{w|v}.$$ - $\forall \ket{u}, \ket{v} \in H,$
$$\braket{u|v} = \braket{v|u}^* \Rightarrow (\bra{u}+\bra{v})\ket{w}=\braket{u|w}+\braket{v|w}.$$ - $\forall \ket{v}\in H$, $\braket{v|v} \geq 0$.
The duo $\bra{\psi} = \phi(\ket{\psi})$ is often called the bra vector and $\ket{\psi}$ is often called the ket vector. Since the choice of $\phi$ can be arbitrary as long as it satisfies the aforementioned 3 criterion, it is not yet valid to claim that $\bra{u} = \ket{u}^\dagger$ (conjugate transposition). We will show later that there exists a unique basis $e$ under which this relation becomes true.
From 1. and 2. it follows that the operation $\phi(\ket{u})(\ket{v}) = \braket{u|v}$ is bilinear and conjugate symmetric, so the function $\phi(\ket{u}, \ket{v}) = \braket{u|v}$ is a quadratic form and must be represented by $[\ket{u}]_\beta^\dagger M [\ket{v}]_\beta$, where $M\in Mn(\R)$, and $\beta$ is an arbitrary basis.
Further by 3, we deduce that M is positive definite and hence Hermittian, that is $M^\dagger = M$.
This implies $\exists B\in Mn(\R)$ s.t. $M = B^\dagger B$.
Then $\forall u,v\in H,$ $$\braket{u|v} = [\ket{u}]_\beta^\dagger M [\ket{v}]_\beta=\left(B[\ket{u}]_\beta\right)^\dagger B [\ket{v}]_\beta.$$
If we consider basis $e = \{e_i\}$ s.t. $B$ (written in $\beta$) $=[I]_{\beta}^e$ is the change of basis matrix. That is, $[e_i]_\beta=\left([I]^\beta_e\right)_{\cdot,i}=\left(B^{-1}\right)_{\cdot,i}$
then $[\ket{u}]_e = B[\ket{u}]_\beta$ staisfy the natural property that $$[\ket{u}]^\dagger_e[\ket{v}]_e = \braket{u|v}. (2)$$
So definition of $\phi$ is now explicitly rewritten as:
$$[\bra{u}]_e = [\phi(\ket{u})]_e = [\ket{u}]_e^\dagger.$$
- The basis $\{e\}$ here is orthogonal when we consider $\braket{u|v}$ as defining a unique innerproduct space $(u,v)$. All subsequent operations on $H$ must thus be compatible with the notion of orthogonality set forth by $\{e\}$.
We have now concluded that $[\bra{u}]_e = [\ket{u}]_e^\dagger$. Since $e$ computed from $\phi$ and $\beta$ will be from now on the default basis of representation, we may ommit the $[\cdot]_e$ denoting representation and simply write, $\bra{u} = \ket{u}^\dagger$.
Properties of adjoint on operators in basis $e$
When we have a ket vector $\ket{\psi}=T\ket{u}$, where $T$ is a linear operator, and want to take its duo: $$\ket{\psi} \xrightarrow[]{\phi}\bra{\psi},$$
we will first note that $\phi$ is now well-defined as $$[\phi(T\ket{u})]_e = [T\ket{u}]^\dagger_e.$$ Then we obtain
\begin{align}
[\phi(T\ket{u})]_e &= [\ket{u}]_e^\dagger[T]_e^\dagger\\
&=[\phi(\ket{u})]_e[T]^\dagger_e\\
\bra{Tu} = \phi(T\ket{u}) &= \bra{u} T^\dagger.
\end{align}
Some times when basis ${e}$ is implied, we may simply write \begin{align} \ket{\psi} &= \begin{bmatrix} \psi_1\\ \vdots \\ \psi_n\end{bmatrix},\end{align} since all the notations are compatible.
It’s also easily recognized that for all $c\in \C$, $$\phi(c\ket{\psi}) = c^* \bra{\psi} \text{ due to the conjugate property.}$$