Regarding the wiki: https://en.wikipedia.org/wiki/Sesquilinear_form#Hermitian_form

you can see that the wiki states that physics defines the inner product for complex vectors as:

$$\langle \, \mathbf{A} \, |\, \mathbf{B} \, \rangle = \sum_{i}a_i^*b_i$$

However, a quick google search would show (e.g. here or here) that mathematics defines the inner product as:

$$\langle \, \mathbf{A} \, |\, \mathbf{B} \, \rangle = \sum_{i}a_ib_i^*$$

So you can see, for physics the conjugate is given to the 1st variable, while in mathematics the conjugate is given to the 2nd variable

Why does physics defines the conjugate to the 1st variable? And isn't it wrong? because clearly:

$$\sum_{i}a_ib_i^* \neq \sum_{i}a_i^*b_i$$

  • $\begingroup$ A comment: the physicist definition is the complex conjugate of the mathematician's, so they agree for any inner product whose answer is real, in particular for $\langle \psi | \psi \rangle$. $\endgroup$
    – jacob1729
    May 24, 2019 at 14:54
  • $\begingroup$ Hi thank you! But is there any reason why physics and math define it differently? Because I find it really confusing why we need 2 different definitions for the same thing $\endgroup$
    – D. Soul
    May 24, 2019 at 14:56
  • $\begingroup$ The mathematicians convention is, in a sense, the more obvious one (the first thing acts 'normally', the second one is 'abnormal') but physicists convention agrees with the notation for taking the duel of vector and applying it to some other vector (e.g. $v^\dagger u$). $\endgroup$ May 24, 2019 at 15:35

1 Answer 1


Firstly, a definition can never be wrong. It can be a bad or inconvenient definition (not the case here), or it can be incompatible with other definitions (which is the case here), or not reflect our intuitive understanding of a word, but a definition is not a statement and as such it can't be wrong.

The physicists' convention follows the idea of taking column vectors as fundamental. If you think a vector in $\mathbb{C}^n$ is a vertical array of complex numbers

$$v = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix},$$

it makes sense that to calculate $\langle u, v \rangle$ you leave $v$ alone, and transpose and conjugate $u$, to get

$$\langle u, v \rangle = u^\dagger v = \begin{pmatrix} u_1^* & \cdots & u_n^* \end{pmatrix} \begin{pmatrix} v_1\\ \vdots \\ v_n \end{pmatrix}.$$

This also leads naturally to the idea of a dual vector or covector, in which for each $u$ we associate a function $f_u$ defined by $f_u(v) = \langle u, v \rangle$.

Of course, nothing stops from using the mathematicians' convention, in which $\langle u, v \rangle = u^T v^*$. But it looks a bit strange to apply operations to both vectors. Also, unless you reverse the definition of the function $f_u$ (to think of $v$ acting on $u$ instead of the other way around), it's antilinear instead of linear.


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