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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.

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3 Answers 3

up vote 10 down vote accepted

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 $C^n$.

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Thank you, I mentioned about it in my post (see "the first possible solution"). But I have difficulties in $(A \mid u \rangle )^* = \langle u \mid A^{\dagger}$: On the one hand we have from bra-ket notation: $\langle u \mid A^{\dagger}$ . On the other hand we have from linear algebra way, that you described:$(A \mid u \rangle)^*=A^* \langle u \mid=\langle u \mid A$ .I'm very confused with it, because it seems to me, that this is different expressions. –  xxxxx Oct 31 '12 at 19:22
You can see your mistake when you translate the last chain of equations in your comment into matrix language. When transposing a product, you get the transpose of the product in the reverse order! –  Arnold Neumaier Nov 1 '12 at 13:43

If you think of $\mid u \rangle$ as column vectors and of $\langle u \mid$ as row vectors, then $A$ is just a $n \times n$ matrix (possibly with $n = \infty$).

You can then think of $A \mid u \rangle$ as the matrix $A$ acting on a vector $u$. However, since $\langle v \mid$ is a row and not a column vector, you cannot (for a sensible row vector) multiply $v$ with $A$ from the left but only from the right:

$$ v A = v' \in V^\star$$

One then usually defines $\langle v A \mid$ (or $\langle A v \mid$) as the result of acting on $v$ with $A$. If you then take the scalar product with $\mid u \rangle$, we can write:

$$\langle v A | u \rangle = \langle v | B | u \rangle = \langle v | B u \rangle = \langle v | u' \rangle $$

for some matrix $B$ such that $u' = Bu$ (from the left, since $\mid \rangle$ are column vectors). Furthermore, one finds that the relation between $A$ and $B$ is such that

$$ B = A^\dagger\quad,$$

that is, the hermitian adjoint: this sort of makes sense - if you let a real matrix act on a row rather than the usual column vector, you have to take it’s adjoint (i. e. $M_{ij}^T = M_{ji}$) and the magic of quantum mechanics simply adds the complex conjugate to this $A^\dagger_{ij} = \bar A_{ji}$

The second point is very much the same: $( A \mid v \rangle)^\star$ describes the dual element to $A \mid v \rangle$, which happens to be $\langle v A \mid = \langle v \mid A^\dagger$.

As a rule from a physicist’s point of view, you add a $^\dagger$ if you pull out an operator from a bra to then make it act on a ket. Of course (for sensile operators), $(A^\dagger)^\dagger = A$.

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Thank you. I think that i get it, i.e that we can work with matrixes, but it isn't absolutely clear about relation with operators. –  xxxxx Oct 31 '12 at 18:42

I just want to make a comment (requires 50 reputation...)

Acting from the "left" and from the "right" actually has a precise signification, cf group action or left/right modules. without writing any formulas, notice that you have two "multiplications", one is composition of operators, one is the "action". The distinction left/right action arises when you have two operators acting on a vector...

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