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Quoting from Ballentine's textbook on Quantum Mechanics:

There are situations in which it is important to remember that the primary definition of the bra vector is as a linear functional on the space of ket vectors.

What are those situations? I always thought of bras and kets on an "equal footing", i.e., two vectors which---when assembled into a braket---represented the dot product and its concommittant meanings of angle and length. But now, it seems that bras are fundamentally a different animal than kets, and not merely their Hermitian conjugate.

PS: Please keep the answer intuitive/physical and accessible to someone who hasn't had any graduate-level course in algebra.

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    $\begingroup$ Are you familiar with Riesz representation theorem? $\endgroup$
    – Qmechanic
    Commented Aug 9, 2020 at 10:57
  • $\begingroup$ @Qmechanic Ballentine gave a very cursory overview of Riesz theorem whereby he argued that every functional $F$ can be represented one-to-one by some specific vector $f$ such that $F(\phi) = (f, \phi)$ for any vector $\phi$. I kind of follow the math but I'm afraid I don't gain much insight out it. $\endgroup$
    – Tfovid
    Commented Aug 9, 2020 at 11:07
  • $\begingroup$ The wikipedia article says that the "theorem can be seen as a justification for the popular bra–ket notation". What is there to justify, though? The notion of dot product is self-evident from basic undergraduate algebra. What else is there to it? $\endgroup$
    – Tfovid
    Commented Aug 9, 2020 at 11:09

3 Answers 3

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Warning: Since I don't know what you might find useful, this answer may contain parts that you might think not relevant or too verbose. This may also be too mathy for your taste. Perhaps someone else will come along and give a more physical answer.

TLDR: Kets are column vectors, bras are row vectors. They are not conjugate, but conjugate transposes. They are on equal footing because one can go from one to another, but they are different objects living in different Hilbert spaces. These spaces are isomorphic (via "conjugate transposition") so one can often identify them, but they are not the same.

Let's consider first the case when the state space $H$ is finite-dimensional. That is, it is a finite dimensional vector space over $\mathbb{C}$ with Hermitian inner product. The vectors in this space are called kets, written $|\phi\rangle$. If we pick an orthonormal basis $(|\phi_1\rangle, \ldots, |\phi_n\rangle)$ then any ket can be written uniquely as $|\phi\rangle=c_1|\phi_1\rangle+ \ldots+ c_n|\phi_n\rangle)$ where $c_i$ are complex numbers. This means that we can identify the ket with the column vector $\begin{bmatrix} c_1\\\vdots\\c_n\end{bmatrix}$. Then, because of orthonormailty of the basis $(|\phi_1\rangle, \ldots, |\phi_n\rangle)$ the Hermitian product between two such column vectors $\begin{bmatrix} c_1\\\vdots\\c_n\end{bmatrix}$ and $\begin{bmatrix} z_1\\\vdots\\z_n\end{bmatrix}$ is equal $c_1 \bar{z}_1+\ldots c_n \bar{z}_n$.

Associated to the vector space $H$ there is the vector space of linear functions on $H$, called dual of $H$ and written $H^*$. The elements of $H^*$ are maps $l:H\to \mathbb{C}$ that are linear (meaning that $l$ has the property that $l(z|\phi\rangle+w |\psi\rangle)=zl(|\phi\rangle)+w l(|\psi\rangle)$ for all $|\phi\rangle, |\psi\rangle\in H$ and all $z, w \in \mathbb{C}$).
One can straightforwardly check that the set of such maps is a vector space.

Now, using our representation of elements of $H$ (aka kets) as column vectors, we can see that any row vector $\begin{bmatrix}w_1&\ldots & w_n\end{bmatrix}$ represents an element of $H^*$, via the usual matrix multiplication rule $\begin{bmatrix}w_1&\ldots & w_n\end{bmatrix} \left(\begin{bmatrix} c_1\\\vdots\\c_n\end{bmatrix} \right)=\begin{bmatrix}w_1&\ldots & w_n\end{bmatrix} \begin{bmatrix} c_1\\\vdots\\c_n\end{bmatrix}=w_1 c_1+\ldots _+w_nc_n$.

In fact, one can also check that every element of $H^*$ can be represented in this way: if $l$ is ANY linear map, take $w_i=l(|\phi_i\rangle)$. Then by linearity $l(|\phi\rangle)=\begin{bmatrix}w_1&\ldots & w_n\end{bmatrix}(|\phi\rangle)$.

To recap, so far we have: elements of $H$ are kets, which after choice of basis are represented by column vectors. Elements of $H^*$ are linear functions on kets, represented by row vectors.

Now, using the Hermitian product we can associate to every ket an element of $H^*$. This is done as follows: given a ket $|\phi\rangle$ define the linear map $l:H\to \mathbb{C}$ taking an arbitrary $|\psi\rangle\in H$ to the Hermitian product of $|\psi \rangle$ and $|\phi \rangle$ (in this order). By properties of the Hermitian product this is linear (in other order would've been conjugate-linear), so is indeed an element of $H^*$. This element is called the bra associated to $|\phi\rangle$ and is written as $\langle \phi|$.

Observe that the construction of this bra used only the Hermitian product. It is independent of our choice of basis for $H$ and of representing the kets as column vectors.

Now look to see how this bra looks in our chosen representation. For the ket $|\phi\rangle$ represented as $\begin{bmatrix} c_1\\\vdots\\c_n\end{bmatrix}$ the bra $\langle \phi|$ sends a ket $|\psi\rangle$ represented as $\begin{bmatrix} z_1\\\vdots\\z_n\end{bmatrix}$ to $z_1 \bar{c}_1+\ldots z_n \bar{c}_n$, by the first paragraph. This in turn means that the bra $|\phi\rangle$ must be represented by the row vector $\begin{bmatrix}\bar{c}_1&\ldots & \bar{c}_n\end{bmatrix}$.

So, to summarize: to every (vector) ket there is an associated (linear function) bra, and if the ket is represented by a column vector, the corresponding bra is represented by the conjugate transpose (row) vector. Moreover, every bra is representable by some row vector so it comes from some ket. This means that "transpose conjugate" is conplex-conjugate isomorphism between the space of kets and bras ($H$ and $H^*$).

As a result, we can treat bras as conjugate transposed kets - and kets as conjugate transposed bras. But indeed bras are "linear functions on kets".

Now what does this have to do with Riesz representation theorem? The theorem simply says that similar picture is true in the case of an infinite dimensional $H$. In that case we can not pick a finite basis, so we can't operate with column and row vectors as straightforwardly. We also need to stipulate that our linear functions in $H^*$ are countinuous (which was automatic in finite-dimenasional case). Nonetheless, we still have that to each ket $|\phi\rangle$ there corresponds a bra $\langle \phi|$ in $H^*$, obtained by using the Hermitian product on $H$, as before. The theorem says that in fact every element in $H^*$ is of this form -- something we could easily see before by picking a basis and representing everything as column and row vectors, but which is now less clear. Nonetheless it's true, and we can think of all continuous linear functions on $H$ as bras.

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Ballentine uses the standard (non-rigorous) Dirac-style formulation of QM, but he does mention - on a very elementary level - two approaches that make QM mathematically rigorous, one of them is the use of spectral measures, and another of them is the use of Gelfand triplets (rigged Hilbert space). The author's language clearly marks a preference for Gelfand triplets.

If $H$ is a Hilbert space we use to model QM, then a Gelfand triplet is a triplet $(S,H,S^\ast)$, where $S$ is an open subspace of $H$, and $S^\ast$ is the continuous/topological dual of $S$. We have $H=H^\ast$ (Riesz representation theorem), but because $S$ is open in $H$, it is not a Hilbert space itself, and its dual is larger than $H$, in fact we have the containment relations $S<H<S^\ast$.

Now, if $\xi\in S^\ast$ is an element of $S^\ast$ that is not an element of $H$, then the linear action $\xi(\psi)=\langle\xi |\psi\rangle$ makes sense for any $\psi\in S$, but $\xi$ by assumption is not an element of $H$, thus there is no corresponding "ket vector" $|\xi\rangle$.

In common phyical situations, the "position basis" $| x\rangle$ and the "momentum basis" (if the momentum is not quantized) $|p\rangle$ are elements of $S^\ast$ that are not elements of $H$, thus technically speaking these elements exist only as bra vectors.

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  • $\begingroup$ I think you want \subset command to typeset $S\subset H\subset S^*$. $\endgroup$
    – Ruslan
    Commented Aug 10, 2020 at 20:56
  • $\begingroup$ Could you elaborate why $S$ being open in $H$ implies that $S^*$ is larger than $H$? $\endgroup$
    – Ruslan
    Commented Aug 10, 2020 at 20:58
  • $\begingroup$ @Ruslan They are subspaces of each other so the $<$ symbol should work. It is a relatively general fact that the "smaller" a topological vector space is, the "larger" its dual is. $S$ is open in $H$ means that sequences can leave $S$ so $S$ is not complete and thus its not a Hilbert space. Since elements of $S$ are elements of $H$, any element of $H$ is a (continuous) linear functional on $S$, so $S^\ast\ge H$. But $H=H^\ast=H^{\ast\ast}$, so we must have $S^\ast > H$. $\endgroup$ Commented Aug 11, 2020 at 9:15
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    $\begingroup$ @Ruslan Nevermind, this argument is not sound, but nontheless the fact remains. For intuition maybe consider $H=L^2(\mathbb R)$, $S=\text{Schwartz space}$, $S^\ast=\text{Tempered distributions}$. If $f$ is Schwartz, then $\int f(x)g(x)dx$ is well-defined for a lot of functions $g$ that are not in $L^2$, like say functions that polinomially diverge at infinity. $\endgroup$ Commented Aug 11, 2020 at 11:04
  • $\begingroup$ This example makes it obvious, thanks. $\endgroup$
    – Ruslan
    Commented Aug 11, 2020 at 11:15
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I don't have my copy to hand-- its in my Covid inaccessible office --- but I believe that the best way to understand the necessity of thinking of bra vectors as linear maps is to go to its inventor. Dirac's Principles of Quantum mechanics devotes the first two chapters to his language of "bra" and "ket". If I remember correctly Dirac shows that combining the existence of photons as discrete objects with their interactions with polarizers inevitably requires a vector space of states. He then shows how the existence of observables demands the notion of linear maps from the space of states to the complex numbers. He only then combines this notion with the dagger map ket $\to$ bra to define an inner product that makes the space of states into a Hilbert space.

So for Dirac the linear maps comes first, and the inner product later.

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