I have started to study quantum mechanics. I know linear algebra,functional analysis, calculus, and so on, but at this moment I have a problem in Dirac bra-ket formalism. Namely, I have problem with "translation" from ordinary linear algebra "language" to this formalism. For better understanding of the problem, I'll give some definitions which I use:
1) Assume, that $\mid u\rangle$ is a vector in Hilbert vector space $V$.
Bra $\langle v \mid$ is a vector of dual vector space $V^{*}$ $\left(
\langle v \mid: V \longrightarrow C \right)$, defined by $\langle v \mid u \rangle=g\left(\mid v\rangle, \mid u\rangle \right) $, where $g: V \times V \longrightarrow C$ is metric on $V$.
2)$A$ is linear operator on $V$. Consider bilinear form $(\quad)$:
$\left(f,x\right )=f(x)$. In this notation we can define adjoint operator $A^{*}$ on $V^{*}$:$(f,Ax)=(A^*f,x)$.
I tried to understand two following equations:
- There is expression $\langle v \mid A \mid
u \rangle$. In my text book was written the following phrase:
"Operator $A$ acts on ket from the left and on bra on the right". But according to the difenitions that i use, adjoint operator $A^{*}$ acts on $V^{*}$.But in this case operator $A$ acts on $V^{*}$. I dont absolutely get it.
One possible solution which i see is that this is just a notation of the next thing:$\langle v \mid A \mid u \rangle=(v,Au)=(A^{*}v,u)= \langle A^{*}v \mid u \rangle; \quad A^*\langle v \mid :=\langle v \mid A$
The second way is to use isomorphism between $V$ and $V^{*}$, and then operator $A$ is able to act on $V$ and $V^{*}$ (Dual Correspondence).
The third way is that we use a matrix representation everywhere;and in expression $\langle v \mid A \mid
u \rangle$ we multiply a row $v$ on matrix of operator $A$ on column $u$. Then this expression absolutely clear because the multiplication of matrix's is associative.
- The same difficulties i have with expression $(A \mid v \rangle)^{*}=\langle v \mid A^ \dagger$. Could you explain it to?
I would be happy, if you will say which way is right and if all of my suggestions are wrong, please, tell me the right one.
Best Answer
The wording used in your textbook was sloppy.
$A$ acts as $A^*$ on a bra, as $\langle u\rvert A\lvert v\rangle:=\langle u\lvert Av\rangle~$ is the same as $\langle u\rvert A\lvert v\rangle=\langle A^*u\lvert v\rangle~$, by definition of the adjoint. The latter formula also shows that $\langle A^*u\rvert=\langle u\rvert A$.
Everything becomes very simple in linear algebra terms when interpreting a ket as a colum vector, the corresponding bra as the conjugate transposed row vector, an operator as a square matrix, and the adjoint as the conjugate transpose. This is indeed the special case when the Hilbert space is $C^n$.