# Complex conjugation and vector space duality

Hi all. First of all I want to say that I am new here and I want to apologise if this is the wrong forum to post my question. :)
I've just finished a sort of a crash course on functional analysis that left much more questions in me than it answered but one thing bothers me a lot.
When we talk about dual Hilbert spaces it seems that the vector in the dual space has coordinates that are simply complex conjugates of the coordinates of the corresponding vector.
Or using the Dirac notation:
|x>*=<x|, where x is from H - a Hilbert space
|x>=(x1, x2,...); <x|=(x1*, x2*,...)
First I want to ask if I got this right. And second - is there a way to generalise this for dual spaces as a whole (not just Hilbert spaces where H=H*). In other words is there any connection between complex conjugation and dual vectors?

dextercioby
Homework Helper
There's a trick here. The Hilbert space must be complex and the scalar product must be a sequilinear form. Then, using Riesz' theorem and assuming separability of the HS, one shows that the action of a functional on a vector can be expressed as

$$F_{\phi} \psi = \langle \phi, \psi\rangle = \int_{\mathbb{R}^n} \phi^* (x) \psi (x) \, dx$$

Fredrik
Staff Emeritus
Gold Member
|x>*=<x|, where x is from H - a Hilbert space
This is a really ugly notation. I wouldn't use it. I assume that it was inspired by the fact that some authors use x,y,z for members of H and x*,y*,z* for members of H*, but the * on x* is just there to remind the reader that we're dealing with a member of H*. x* is unrelated to x.

x* is unrelated to x.

Isn't there a relation between the coordinates of a vector and a covector. In particular the coordinates of the covector in the basis of L*:
$$x_{i} = g_{ij} x^{j}$$,
where g is the metric tensor in L, $$x \in L$$

Anyway this is the thing that bothers me exactly:
http://en.wikipedia.org/wiki/Bra-ket_notation#Hermitian_conjugation
It seems that after Hermitian conjugation the ket becomes a bra (which is essentially the vector becoming covector if I get it right).

Landau
but the * on x* is just there to remind the reader that we're dealing with a member of H*. x* is unrelated to x.
I don't think so. The bra-ket formalism works (or at least the connection with inner products exists) because of Riesz Representation theorem: for a complex Hilbert space H, the map

$$H\to H^*$$

$$h\mapsto \left<h,-\right>$$

is an isometric anti-linear isomorphism. You can call this map (somewhat bad notation, agreed) *, so that $h^*=\left<h,-\right>$.

In bra-ket notation:

If $$h=\left|\phi\right>\in H$$, then $$h^*=\left|\phi\right>^*=\left<h,-\right>=\left<{\phi}\right|\in H^*$$.

Fredrik
Staff Emeritus
Gold Member
In the notation I described, x*,y*,z* are just symbols for arbitrary members of H*. I might as well have written α,β,γ.

Edit: A note for math guys like Landau: Inner products are usually taken to be linear in the second variable in physics books. I'm using that convention below.

I don't recall seeing the claim that the dagger operation takes a bra to the corresponding ket and vice versa before. I'm trying to see a "point" of that definition... The inner product satisfies <x,y>=<y,x>*. A bra acting on a ket is by definition the inner product of the kets, i.e. if I temporarily use (,) instead of <,> for inner products, we have

$$\langle\alpha|\beta\rangle =\langle\alpha||\beta\rangle=(|\alpha\rangle,|\beta\rangle)$$

$$=(|\beta\rangle,|\alpha\rangle)^*=(\langle\beta||\alpha\rangle)^*$$

For operators, we have the identity (AB)*=B*A*. If we insist that that should hold for bras and kets as well, the right-hand side above becomes

$$=|\alpha\rangle^*\langle\beta|^*$$

which only makes sense if * changes bras to the corresponding kets and vice versa. I guess this is the "point" of that definition, but I don't see why we need (AB)*=B*A* to hold for bras and kets.

Let's see if this is consistent with the standard definition of the adjoint. A bra is a linear operator from H into ℂ, and ℂ is a Hilbert space. This means that * acting on a bra is already defined. The inner product on ℂ is defined by <z,w>=z*w. Let B be a bounded bra, i.e. let B →ℂ be linear and bounded. I'll use h for a member of H, and z for a member of ℂ. B*:ℂ→H is defined by

<z,Bh>=<B*z,h>

for all z and all h. So B* isn't a member of H or H**, as a ket should be, it's a function that takes complex numbers to kets. Let's continue anyway. The above means that the adjoint of a linear combination of bras can be evaluated like this

<(aA+bB)*z,h> = <z,(aA+bB)h> = a<z,Ah> + b<z,Bh> = a<A*z,h> + b<B*z,h>

= <a*A*z,h> + <b*B*z,h> = <(a*A*+b*B*),h>

This implies that (aA+bB)*=a*A*+b*B*. If we now use that other definition, we get the identity that Wikipedia wrote as

$$\left(c_1|\psi_1\rangle + c_2|\psi_2\rangle\right)^\dagger = c_1^* \langle\psi_1| + c_2^* \langle\psi_2|$$

Landau
I don't recall seeing the claim that the dagger operation takes a bra to the corresponding ket and vice versa before.
You haven't? What do/did you think the dagger operation 'is'? In my understanding, essentially daggering is just dualizing, the covariant functor which sends a vector space to its dual and maps to the corresponding dual map, but with two caveats: by Riesz, H and H* are canonically isomorphic, and we are working over the complex numbers so some complex scalars have to be replaced by their conjugate.

In R^n, this is just 'taking the transpose', so that the 'dagger' of a column vector is the corresponding row vector.
which only makes sense if * changes bras to the corresponding kets and vice versa. I guess this is the "point" of that definition, but I don't see why we need (AB)*=B*A* to hold for bras and kets.
Now I am even more interested: what is your (understanding of the) definition of * or dagger? I have always taken it to be nothing more than the Riesz isomorphism

$$H\to H^*$$

$$h\mapsto \left<h,-\right>$$

Fredrik
Staff Emeritus
Gold Member
Now I am even more interested: what is your (understanding of the) definition of * or dagger? I have always taken it to be nothing more than the Riesz isomorphism

$$H\to H^*$$

$$h\mapsto \left<h,-\right>$$
The first one, not the second. If you're used to seeing * for the second one too, I'm not going to argue with that. I just don't remember seeing it myself. If you have a notation like $\langle\alpha|$ at your disposal, you don't need the notation $|\alpha\rangle^*$ for the same thing.

Landau
The first one, not the second.
What do you mean? The first what?
If you're used to seeing * for the second one too, I'm not going to argue with that. I just don't remember seeing it myself.
No no, my quantum mechanics classes are more than one year ago, I am not too sure what our notation back then was. I was just wondering what $|\alpha\rangle^*$ means to you. Now I understand that this just wasn't defined at all.

In any case, I think the whole formalism is consistent if we take this map to be the definition of *. If we want to extend the meaning of * to also apply to bra's, then it should assign to it the corresponding ket.

jambaugh
Gold Member
The best way to understand this is to start with a complex vector space V sans the inner product. This vector space has a dual space V* of linear functionals mapping V to C. When V is finite dimensional V* is isomorphic as a vector space (has the same dimension) and even if not finite we can (e.g. with the inner product) define a corresponding isometric dual sub-space.

OK so given any basis $[e_1,e_2, e_3...]$ of V there is a dual basis $[f_1,f_2,f_3...]$ of V* so that
$$f_j(e_k)=\delta_{jk}$$
that is each dual basis covector maps its corresponding basis vector to 1 and the others to 0. (Note that changing just one of the basis vectors can redefine all of the dual basis).

Now we can either define an inner product on V or a conjugation mapping from V to V*. The two are equivalent in that a conjugation defines an inner product and an inner product defines a conjugation.

Now defining a hermitian inner product $$\langle \quad,\quad\rangle$$ on V (thus making V a Hilbert space) the corresponding conjugate mapping, $$\dagger$$ maps an ortho-normal basis to its dual basis but it must be an ortho-normal basis.

So given a Hilbert space vector $$\psi\in V$$ the dual vector $$\psi^\dagger$$ (linear functional on V) is defined as the mapping:
$$\psi^\dagger(\phi) = \langle \psi , \phi \rangle$$
or if you like this equation defines the inner product in terms of the dual mapping.

Now back to the Bra-Ket notation...
Given a vector $$\psi$$ and (ortho-normal) basis, $$e_1,e_2,\cdots$$ the ket $$|\psi\rangle$$ is the column vector of the basis expansion:
$$|\psi\rangle = (\psi_1,\psi_2,\cdots)^T\quad s.t. \quad \psi = \psi_1 e_1 + \psi_2 e_2+\cdots$$

It then must be the case that the dagger dual of psi expanded in the dual basis will have complex conjugate expansion which we express as a row vector:
$$\langle \psi | = (\psi^*_1,\psi*_2,\cdots)$$

We then, thus copy the dagger operation from the hilbert space in question into the row and column spaces of coefficients where it becomes the conjugate transpose. (That is to say we reuse the symbol.) We likewise have the inner product of two Hilbert space vectors in terms of the product of bra (row vector) with ket (column vector):
$$\langle \psi , \phi\rangle \to \langle \psi | \cdot |\phi \rangle \equiv \langle \psi|\phi\rangle$$

And of course everything works nicely as well when we expand operators in the product basis defining their matrix representation. The adjoint operators likewise correspond to the conjugate transpose matrices (provided we work with proper bases.)

Note that once we have our inner product we can dispense with any direct reference to the dual space and dual basis and work always in the hilbert space itself. When we need a dual vector we simply use an appropriate vector and the inner product or what's better work wholly with the the column vectors of coefficients for a given basis and their conjugate transposes.

But IMNSHO to properly understand the notation it is crucial to first understand the relationship between the dagger, inner product and the dual space, and how we resolve these in terms of a basis.

Issues can surface that require careful parsing of the notation, specifically in relativistic quantum mechanics where e.g. spinors will reside not in a hilbert space but rather a space with an indefinite metric/inner product/adjoint (a pseudo-hilbert or philbert space).

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Fredrik
Staff Emeritus
Gold Member
What do you mean? The first what?
Oops, sorry, I misread your post. I meant the first of the two math expressions I quoted, but I see now that they both represent the same thing. For some reason I interpreted the first expression as $A\mapsto A^*$ where A is a linear operator and A* its adjoint, as defined in e.g. Conway. That's what the * symbol means to me, and the $\dagger$ symbol means the same thing.

In any case, I think the whole formalism is consistent if we take this map to be the definition of *. If we want to extend the meaning of * to also apply to bra's, then it should assign to it the corresponding ket.
I agree.

Thanks a lot for the answers I think I am starting to get this :)

Fredrik
Staff Emeritus
Gold Member
The Riesz representation theorem

I need to make a note of this somewhere, and it might as well be here. The first theorem is just a useful observation, and the second is the Riesz representation theorem, which guarantees the existence of a unique bra for each (normalizable) ket.

Theorem 1: Suppose that $\mathcal H$ is a complex Hilbert space, and that $x_0\in\mathcal H$. Define $\phi:\mathcal H\rightarrow\mathbb C$ by $\phi(x)=\langle x_0,x\rangle$ for all $x\in\mathcal H$. Then

(a) $\phi$ is linear.
(b) $\phi$ is bounded.
(c) $\|\phi\|=\|x_0\|$.
(d) $(\ker \phi)^\perp=\mathbb Cx_0$.

Proof:

(a) The inner product is linear in the second variable.
(b) By the Cauchy-Schwarz inequality, $|\phi(x)|\leq\|x_0\|\|x\|$.
(c) Part (b) implies $\|\phi\|\leq \|x_0\|$ (because the supremum is the least upper bound). Since a supremum is an upper bound, we also have

$$\|\phi\|=\sup_{x\in\mathcal H}\frac{|\phi(x)|}{\|x\|}\geq \frac{|\phi(x_0)|}{\|x_0\|}=\|x_0\|$$

(d) $$x\in\ker\phi\Leftrightarrow 0=\phi(x)=\langle x_0,x\rangle\Leftrightarrow x_0\perp x\Leftrightarrow x\in(\mathbb Cx_0)^\perp$$
Since $\ker\phi$ is a closed linear subspace, this implies $(\ker\phi)^\perp=\mathbb C x_0$.

Note that (a) and (b) imply that $\phi\in\mathcal H^*$.

Theorem 2: Suppose that $\mathcal H$ is a complex Hilbert space and that $\phi:\mathcal H\rightarrow\mathbb C$ is linear and bounded (i.e. that $\phi\in\mathcal H^*$). Then there's a unique $x_0\in\mathbb H$ such that $\phi(x)=\langle x_0,x\rangle$ for all $x\in\mathcal H$.

Proof: There can be at most one $x_0\in\mathcal H$ with the required properties, because

$$\forall x\in H\quad \langle x_0,x\rangle=\langle x_0',x\rangle\ \Rightarrow\ \forall x\in H\quad \langle x_0-x_0',x\rangle=0$$

$$\Rightarrow\ 0=\langle x_0-x_0',x_0-x_0'\rangle=\|x_0-x_0'\|^2\ \Rightarrow x_0'=x_0$$

Theorem 1 tells us that if there's an $x_0$ with the required properties, then $x_0\in(\ker\phi)^\perp$ and $\dim(\ker\phi)^\perp=1$. Because of this, our strategy to prove existence will be to pick an arbitrary member of $(\ker\phi)^\perp$, and then show that we can choose $x_0$ to be a multiple of it. (Note that this doesn't mean that we have to use theorem 1 in this proof. It only provides the motivation for the first half of the trick we will use).

We need to consider the possibility that $(\ker\phi)^\perp=\{0\}$. Since $\ker\phi$ is a closed linear subspace, the projection theorem says that any $x\in\mathbb H$ can be uniquely expressed as $x=y+z$ where $y\in\ker\phi$ and $z\in(\ker\phi)^\perp$, so if $(\ker\phi)^\perp=\{0\}$ we must have z=0, and this implies x=y and therefore $\ker\phi=\mathcal H$. This implies $\phi=0$. If this is the case, we can simply choose $x_0=0$.

If $(\ker\phi)^\perp\neq\{0\}$, it contains a non-zero vector y. Now note that

$$x-\frac{\phi(x)}{\phi(y)}y\in\ker\phi$$

Define $z=y/\phi(y)$ (just to simplify the following expression a little). We have

$$0=\langle z,x-\phi(x)z\rangle=\langle z,x\rangle-\phi(x)\|z\|^2$$

$$\phi(x)=\left\langle\frac{z}{\|z\|^2},x\right\rangle$$

and this means that we can choose $x_0=z/\|z\|^2$.

Theorem 3: The map $B:\mathcal H\rightarrow H^*$, defined by $B(x)=\langle x,\cdot\rangle$ for all $x\in\mathcal H$, is an antilinear bijection. (The expression $\langle x,\cdot\rangle$ denotes the map $y\mapsto \langle x,y\rangle$ from $\mathcal H$ into $\mathbb C$, and theorem 1 tells us that this map is a member of $\mathcal H^*$).

Proof: For any $a,b\in\mathbb C$ and any $x,y\in\mathcal H$, we have

$$B(ax+by)z=\langle ax+by,z\rangle=a^*\langle x,z\rangle+b^*\langle y,z\rangle=(a^*B(x)+b^*B(y))z$$

for all $z\in\mathcal H$, so B is antilinear. B is also injective, because

$$B(x)=B(y)\ \Rightarrow\ \forall z\in\mathcal H\quad \langle x,z\rangle=\langle y,z\rangle\ \Rightarrow\ \forall z\in\mathcal H\quad \langle x-y,z\rangle=0$$

$$\Rightarrow\ 0=\langle x-y,x-y\rangle=\|x-y\|^2\ \Rightarrow\ x=y$$

We don't need to prove that B is surjective here, because that's precisely what theorem 2 says, and we have already proved that.

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Fredrik
Staff Emeritus
Gold Member
I forgot to prove that B (the antilinear bijection from $\mathcal H$ into $\mathcal H^*$) is an isometry. This means that it's continuous as well. I should have made this a part of theorem 3.

$$\|B(x)\|=\sup_{y\in\mathcal H}\frac{|B(x)y|}{\|y\|}=\sup_{y\in\mathcal H}\frac{|\langle x,y\rangle|}{\|y\|} \begin{cases} {\displaystyle\leq\sup_{y\in\mathcal H}\frac{\|x\|\|y\|}{\|y\|}=\|x\|}\\ \\ {\displaystyle\geq \frac{|\langle x,x\rangle|}{\|x\|}=\|x\|} \end{cases}$$

Landau
$$H\to H^*$$
$$h\mapsto \left<h,-\right>$$"