We define an inner product similar to a scalar product:

\begin{align*} \text{scalar product:}& & \vec{a}\cdot\vec{b}&= a_1b_1 + a_2 b_2 + a_3b_3 + \dots\\ \text{inner product:}& & \langle a | b \rangle &= \overline{a_1}b_1 + \overline{a_2}b_2 + \overline{a_3}b_3+\dots \end{align*}

So the inner product is designed to work in $\mathbb{C}$ similar to how scalar product works in $\mathbb{R}$. So the question arizes about a geometrical interpretation of an inner product. For scalar product it is $|\vec{a}||\vec{b}|\cos{\varphi}$ which we interpret as:

˝Projection of a 1st vector's norm to a 2nd vector multiplied by a 2nd vector's norm.˝

Question: What about for an inner product? Does the above quotation hold?

Some background on my question: I am asking this because in many quantum mechanical books they are talking of a state vector $\left|\Psi(t)\right\rangle$ which is an abstraction that defines the state of a system and is only a function of time $t$. Then we have to define a base whose base vectors and the space (we talk about position, momentum space...) we get is dependant of the base vectors we choose. So if i am talking about position space in 1-D some physicists claim that this inner product:

$$\left\langle x \right.\left|\Psi(t)\right\rangle$$

is a projection of a state vector $\left| \Psi(t)\right\rangle$ onto a normalised position base vector $\left|x\right\rangle$ and we can interpret this as $\Psi(x,t)$ which is now a function of position!

  • 6
    $\begingroup$ It's better not to use the term "scalar product" at all. The first thing is a real inner product and the second thing is a complex inner product. If you want to refer to the first thing where the $a_i$ and $b_i$ are complex, call it a symmetric bilinear form. $\endgroup$ May 26, 2013 at 9:40
  • 1
    $\begingroup$ @QiaochuYuan: unfortunately it is not that easy to kill off a French tradition. $\endgroup$ May 26, 2013 at 10:58
  • $\begingroup$ Well thats how they taught us in schools. $\endgroup$
    – 71GA
    May 26, 2013 at 11:07
  • $\begingroup$ what des the bar above $\bar{x}$ stand for in the inner product ? $\endgroup$
    – partizanos
    Dec 17, 2020 at 5:58

1 Answer 1


The quotation above holds in $\mathbb{C}$ as well. It might not be clear what "projection" means in the contest of complex vectors, but algebraically it is exactly as with real vectors (remember the conjugate, though).

In $\mathbb{C}$, the "inner" product of $a+ib$ and $c+id$ is: $$ ac+ bd +i (ad-bc). $$

In a way, its real part is the $\mathbb{R}^2$ scalar product, and its imaginary part is some sort of "cross product" (can you see why?). What matters, anyway, is that this way squared moduli behave exactly like in $\mathbb{R}^2$: $$ \langle(a+ib) | (a+ib)\rangle = a^2 + b^2. $$

Passing from $\mathbb{C}$ to $\mathbb{C}^n$, or any other complex vector space, pretty much the same holds: squared moduli of vectors are exactly the same as the squared moduli of $\mathbb{R}^{2n}$ vectors, while general inner product acquire a skew-symmetrical imaginary part. (This skew-symmetric part should not disturb you, because it is imaginary: think of Hermitian matrices, which generalize real symmetric matrices adding an imaginary skew-symmetric part.)

In linear algebra, you can think of the inner/scalar product as something that can "give you the coordinates". (Geometers please don't be mad at me for saying this - I agree with you but this is what they do in physics!)

For example, to get the first component of the vector $v=(x,y,z)$ we calculate its inner product with the basis element $(0,1,0)$, getting $v_2=y$.

Just as well, a function on the real line $f:x\mapsto f(x)$ is a vector in a vector space (they can be summed and multiplied by scalars). If the functions are suitably integrable (say, $L^1$ or $L^2$), we can define the inner product in the usual way. Now, a function has as its infinite components all its values $f(x)$ for all the $x$! So to know the "$x_0$-th" component, that is $f(x_0)$, we simply take the scalar product $\langle f | x_0\rangle$, where $| x_0\rangle$ is the continuous analogue of $(0,1,0)$, our "basis element". You can think of it as zero everywhere, and a sharp bump in $x_0$ (look at Dirac Delta if you don't know what it is).

Strictly speaking, the basis elements $|x\rangle$ form improper vectors, but this is a subtle, entirely different story (look at the spectral theorem, and at Schwarz distributions to know more about this).


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