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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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4 Answers 4

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
1  
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
    
Please, do not promulgate a silly typesetting mistake when symbols of standard sets are written in italic. Blackboard bold (ℂ, ℝ, ℚ, ℤ, ℕ…) vs bold (C, R, Q, Z, N…) is a question of preference. Either of two vs TeX default (i.e. italic) is a question of literacy – the latter should be corrected on sight and is no more acceptable from an educated guy than “l33t sp33ch”. –  Incnis Mrsi yesterday

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
    
Does the term “dual element” occur in literature? IMHO it should be called the “(formally) conjugate element”. –  Incnis Mrsi 22 hours ago

Really a good question (unfortunately Ī came too late to actually help). All this stuff perfectly makes mathematical sense, but its fundamentals lie in abstract algebra, and rarely are explained to students, and hence remain concealed from those who lack appropriate imagination.

To understand firmly bra and ket, not just multiply rows by columns and remember their conjugation rule, it is sufficient to learn two things: (certain pieces of) duality in linear algebra, including behaviour of linear operators, and the dual of a Hilbert space. A failure to understand either part of two may result in a mental havoc on any attempt to realize meaning of these symbols.

We see, original poster knows what is the dual vector space. In this paragraph we forget completely a Hilbert structure; there are only vector spaces (over ℂ) and linear maps. If A : V1 → V2 is a linear map and v : V2 → ℂ is a linear functional (element of V2*), then their composition v ∘ A maps V1 to ℂ, and hence belongs to V1*. This composition, for each given A , defines a linear map A⁠* : V2* → V1* that is called “transpose” of A; this thing is tautological and preserves linearity over ℂ. Whereas action of A on V1 is denoted by A ∣ u ⟩ in Dirac notation, action of A⁠* on V2* is denoted by ⟨ v ∣ A (note the order of v and  A in the composition). We do not need to distinguish A⁠* from A, because A always acts from the left and A⁠* always acts from the right.

Now the second part: the continuous dual ${\mathcal H}\text{*}$ to a Hilbert space $\mathcal H$ is canonically isomorphic (i.e. the same thing) to its complex conjugate $\overline{\mathcal H}$. It is a mathematical fact. Virtually that’s why Hilbert spaces are so convenient, and Ī have to explain better what means “complex conjugate” in this context. An element ∣ u ⟩ of $\mathcal H$, said a “ket vector”, and an element ⟨ u ∣ of $\overline{\mathcal H}$, said a “bra vector”, are identical. Two spaces are connected by bijection, i.e. don’t differ as sets. Moreover, they have identical laws of addition: ⟨ u ∣ + ⟨ v ∣ corresponds to ∣ u ⟩ + ∣ v ⟩, multiplication by real numbers r ∈ ℝ : r ⟨ u ∣ corresponds to r ∣ u ⟩, and also the same vector norm. The only difference is multiplication by complex scalars c ∈ ℂ : c ⟨ u ∣ corresponds to $\overline{c} ∣ u ⟩$, where overline means complex conjugation, not to “c ∣ u ⟩”. A pure state of a quantum system may be written as either ∣ ψ ⟩ or ⟨ ψ ∣; although as complex vectors they are different, they represent the same state (note that scalar multiplication of a state vector doesn’t change the state physically). We see: bras and kets are no different from the point of view of physics, set theory, metric geometry, topology, and even real linear algebra. All the difference between them are opposite structures of complex (scalar) multiplication.

We can see how two parts combine together. If $A: {\mathcal H}_1 → {\mathcal H}_2$, then $A\text{*}$ maps linearly ${\mathcal H}_2\text{*}$ to ${\mathcal H}_1\text{*}$ and hence $\overline{{\mathcal H}_2}$ to $\overline{{\mathcal H}_1}$, that means we have a mapping (not ℂ-linear a priori) from ${\mathcal H}_2$ to ${\mathcal H}_1$. It is denoted by $A^†$ and, in fact, is complex-linear because the complex structure is reversed twice: one time from ${\mathcal H}_2$ to ${\mathcal H}_2\text{*} ≅ \overline{{\mathcal H}_2}$, and another from ${\mathcal H}_1\text{*} ≅ \overline{{\mathcal H}_1}$ to ${\mathcal H}_1$. Or, in symbols, an identity

$$ \overline{⟨ u ∣ A ∣ v ⟩} = ⟨ v ∣ A^† ∣ u ⟩.$$

(Note that somewhere “*” is used in such contexts for complex conjugation, instead of overline, that certainly contributes to a thick atmosphere of confusion around the asterisk symbol.)

The operation “†” is called “Hermitian adjoint” for abstract linear operators and “conjugate transpose” for complex matrices. It is simply a consequence of transposition of operators (from linear algebra) and the ${\mathcal H}\text{*} ≅ \overline{\mathcal H}$ isomorphism (from the theory of Hilbert spaces), but, if ${\mathcal H}_2 = {\mathcal H}_1$, it defines a non-trivial involution on operators on Hilbert spaces. It is a bit tricky, because transpose operator is tautological and complex conjugate is nonsensical in isolation, but namely for Hilbert spaces they make an essential operation together. Two aforementioned parts of algebra must be combined to obtain another operator from the same operator space.

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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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