Note, $\langle \vec{a}\mid\vec{b}\rangle = \vec{a} \bullet \vec{b}$

What does the asterisk mean in the below properties?

I am learning vector math[1]. I have come across the following property:

The inner product has the following properties:

$$\ (i) \ \langle \vec{a}\mid \vec{b} \rangle = \langle\vec{b}\mid\vec{a}\rangle^{*}$$ $$\ (ii) \ \langle \vec{a}\mid \lambda\vec{b} + \mu\vec{c} \rangle= \lambda \langle\vec{a}\mid\vec{b} \rangle + \mu \langle\vec{a}\mid\vec{c} \rangle $$

Literature states that the asterisk means: "When used as a superscript, the asterisk is commonly voiced "a-star." A raised asterisk is used to denote the adjoint $a^*$, or sometimes the complex conjugate."[2]

We note that in general, for a complex vector space, (i) and (ii) imply that

$$\langle \lambda \vec{a} + \mu \vec{b}\mid\vec{c} \rangle = \lambda^{*} \mu \langle \vec{a}\mid \vec{c} \rangle + \mu^{*}\langle \vec{b}\mid\vec{c} \rangle$$ $$\langle \lambda \vec{a}\mid\mu \vec{b} \rangle = \lambda^{*} \mu \langle \vec{a}\mid\vec{b} \rangle$$

I am assuming it is the complex vector space, but I also know, from my math book[3]

$$ (a) \ \langle \vec{a}\mid\vec{b} \rangle = \langle \vec{b}\mid\vec{a} \rangle$$

without the asterisk.

What does the asterisk mean in the context of (i)?


[1] Riley, K. F.. Mathematical Methods for Physics and Engineering: A Comprehensive Guide (Kindle Locations 6288-6293). Cambridge University Press. Kindle Edition.

[2] MathWorld. (2018) Asterisk. Retrieved (2018, May 13). Available from: http://mathworld.wolfram.com/Asterisk.html

[3] Anton, Howard. (1992). Calculus with analytic geometry-4th Edition. Anton Textbooks, Inc.

  • $\begingroup$ Note that $\langle \vec{a}|\vec{b}\rangle$ is a number, so $*$ represents the complex conjugate. $\endgroup$
    – user230944
    May 13, 2018 at 23:20

2 Answers 2


The notation $\langle \vec{a}|\vec{b}\rangle$ is Dirac's `bra-ket' notation for the scalar product between a vector $\vec{b}$ and the dual of a vector $\vec{a}$. It is just a number, so $\langle \vec{a}|\vec{b}\rangle^*$ is just the complex conjugate of $\langle \vec{a}|\vec{b}\rangle$.

With regard to your math book in [3], I'd posit that he is talking about a real vector space, rather than a complex one.


In general on a vector space $V$ over a field $F$, the inner product is a map $V \times V \to F$, satisfying linearity, positive definiteness and conjugate-symmetry.

For a complex vector space, the inner product takes on complex values in general, and must satisfy $\langle \vec{a} \mid \vec{b} \rangle = \langle \vec{b} \mid \vec{a} \rangle^*$, where ${}^*$ indeed denotes the complex conjugate. For a real vector space (which is presumably what your book [3] is dealing with), the inner product only takes on real values, so the conjugate is omitted.


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