Without invoking Hermitian conjugation (which, to my mind at least, would beg the question), what is it that makes the bra $\langle\psi\rvert$ the unique dual of the ket $\rvert\psi\rangle$?

For instance I know that provided $\rvert\psi\rangle$ is not null then $\langle\psi\rvert$ maps $\rvert\psi\rangle$ to the positive reals. However there can exist some $$\langle\phi\rvert\neq\langle\psi\rvert$$ which also happens to map $\rvert\psi\rangle$ to the positive reals, so that is not the defining property of $\langle\psi\rvert$. Is that correct?

  • 3
    $\begingroup$ I think that your answers lie in the Riesz representation theorem, but I'm not clear enough on either your question or the theorem to make this more than a comment. $\endgroup$ Commented Oct 22, 2019 at 2:40
  • 1
    $\begingroup$ A tiny side point: their are actually infinitely many choices of "bra" that will, when "matched" with you ket produce a positive real number. eg. $\sqrt{1-\epsilon} <Not \psi| + \sqrt{\epsilon} <\psi|$ (for any $\epsilon$). Its "mirror" is unique only in that this real number is unity. $\endgroup$
    – Dast
    Commented Oct 22, 2019 at 15:51
  • $\begingroup$ I wonder if the Dirac notation is causing unnecessary trouble. The Hilbert space $\mathcal{H}$ has an inner product $\langle a,b\rangle$ defined for all $a,b\in\mathcal{H}$. We don't need to define a dual space. It might be a useful tool for some things, but the general principles of quantum theory don't need it. The notation $\langle a|X|b\rangle$ is a synonym for $\langle a, Xb\rangle$. Your question is still perfectly legitimate, I'm just clarifying that it isn't necessary for understanding the general principles of quantum theory, in case that was the reason for the question. $\endgroup$ Commented Oct 23, 2019 at 2:19

3 Answers 3


Two complex topological vector spaces $X$ and $Y$ are said to be in duality if there is a sesquilinear map $$b:X\times Y\to \mathbb{C}\; .$$

The idea is that, given such map, a dual action of $X$ on $Y$ and viceversa is defined: for any $x\in X$, $$b(x,\cdot): Y\to \mathbb{C}$$ is a linear functional.

There are two natural spaces in duality with a given topological vector space $X$. One is the space of all linear functionals, called the algebraic dual $X^*$; the other is the space of all continuous (w.r.t. the topology on $X$) functionals, called the continuous dual $X'$.

The continuous dual has some special properties, and it is often easy to characterize explicitly, given the space $X$. For example, if $X$ is a Banach space, then also on the dual there is a complete norm "inherited" from the one on the original space. The Lebesgue space $L^p$, for example, has the Lebesgue space $L^{\frac{p}{p-1}}$ as continuous dual, provided $1\leq p<\infty$.

For Hilbert spaces $\mathscr{H}$, the continuous dual has an even nicer property: it is the space itself. In fact, the inner product is a sesquilinear form, that makes the Hilbert space in duality with itself. In addition, given $\psi\in \mathscr{H}$, $$\langle \psi,\cdot\rangle: \mathscr{H}\to \mathbb{C}$$ is continuous. Therefore, $\mathscr{H}\subseteq \mathscr{H}'$ (with the suitable identification of the ket $\psi$ with the bra $\langle \psi, \cdot\rangle$).

On the other hand, as proved in the Riesz's representation theorem cited in the comments, the converse is also true: $\mathscr{H}'\subseteq \mathscr{H}$, that is every continuous linear functional can be written as the bra of a given vector in the space.

Therefore, $\mathscr{H}=\mathscr{H}'$. The algebraic dual, on the other hand, is bigger and there are thus functionals that are not continuous and cannot be written as the bra of an element in the Hilbert space.


This is actually one of the instances where bra-ket notation is a bit confusing, because the exact nature of the duality map is built into the notation.

So instead, let us just denote by $\psi$ an element of your (complex) Hilbert space $H$. We can define the dual of $H$ as $H^*:=\text{Hom}(H,\mathbb{C})$, which is mathematician-speak for “the space of linear transformations $f:H\to\mathbb{C}$.” One can show, under certain technical assumptions physicists don’t worry about, that $H$ and $H^*$ are isomorphic as complex Hilbert spaces. (As pointed out by yuggib, this fails to be true when the Hilbert space is infinite-dimensional.) However, just as you noted, this isomorphism isn’t unique.

There is a particularly nice choice of isomorphism, though. Recall that Hilbert spaces are equipped with a complete Hermitian inner-product, which we will denote $\langle\cdot,\cdot\rangle$. That is, given two vectors $\psi$ and $\varphi$, then $\langle\psi,\varphi\rangle$ is their inner-product (this is already very similar to bra-ket notation). It turns out that the inner-product gives us a natural isomorphism between $H$ and $H^*$. We associate to every $\psi\in H$ a corresponding $\psi^*\in H^*$ which acts on $H$ as


for all $\varphi$ in $H$. This uniquely defines the desired isomorphism.

This is already built into bra-let notation, where we would denote $\psi$ as $|\psi\rangle$ and $\psi^*$ as $\langle\psi|$. In this notation, the definition of $\langle\psi|$ is that it acts as


which we might as well write as $\langle\psi|\varphi\rangle$. The fact that the inner-product-respecting ismomorphism is naturally built in is really where the power of bra-ket notation comes from, but is a detail that is really often skipped over in undergraduate courses, opting for the “complex conjugate transpose”, basis-dependent definition.

In the words of Fredric Schuller:

A gentleman only chooses a basis when he must.

  • 1
    $\begingroup$ If you define $H^*$ as the space of linear functionals, then it is not isomorphic to $H$. Also, what do you mean by isomorphic? The algebraic dual is not a Hilbert space. One should restrict to continuous linear functionals. $\endgroup$
    – yuggib
    Commented Oct 22, 2019 at 10:35
  • 1
    $\begingroup$ @yuggib This falls under the umbrella of “ under certain technical assumptions physicists don’t worry about.” $\endgroup$ Commented Oct 22, 2019 at 12:04
  • $\begingroup$ @yuggib Since the goal of my answer was to give an understandable answer at OP’s level, I felt it wasn’t necessary to crowd the answer with technical difficulties. However, I’ll add in an edit reflecting your comment. $\endgroup$ Commented Oct 22, 2019 at 12:14
  • $\begingroup$ Personally, I do not find a notation involving homomorphisms, and in general abstract algebraic/categorical language as particularly basic. However, my point is only that, given your definition using the algebraic dual, the isomorphism holds if and only if $H$ is finite dimensional. From a physical perspective, you're just saying that the identification holds only for spin systems, while it actually holds for any quantum theory. $\endgroup$
    – yuggib
    Commented Oct 22, 2019 at 12:20
  • $\begingroup$ If I recall, the notation in the math world is flipped relative to what you have. I.e., $\psi^*(\varphi) = \langle \varphi, \psi \rangle$, so that $\langle\psi|(|\varphi\rangle) = \langle \varphi, \psi \rangle$. $\endgroup$
    – march
    Commented Oct 22, 2019 at 22:01

Of course if you just define it as you said then you can have many such functionals which map this ket to real numbers. That is why we should define such functionals which when acting on a basis give ones and zeroes. That would be the basis for the space of functionals. Such space is called dual and the basis is dual basis of the dual space. Now you can prove that dual of this dual is Just the original space basis, the one on which functionals were defined to act. I hope this helps a bit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.