To be clear on this, I know what is the definition of an inner product space and some properties and theorems about them. What I am asking for is an intuition for this definition in the complex case. In the real case, the intuition (or at least one of them) is geometric: The inner product of two vectors is the length of the projection of the first to the second scaled by the norms of both vectors so that it is symmetric (modulo some details). In particular I (and everybody else) think of "inner product zero" as geometric orthogonality and of orthonormal bases as, well, orthonormal bases and so on. The question is, what should I think about when working with complex (or should I say hermitian?) inner product spaces? what is the "meaning" of the complex number associated to two vectors called their inner product?

I will be happy to hear all kinds of answers. For example, what physical phenomena does it model or in what mathematical situations does in "naturally" appear. Answers that stress the "nice structure" resulting are also welcome, yet I feel that by itself it is a bit unsatisfying.

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    $\begingroup$ As for physical phenomena, complete complex inner product spaces (a.k.a. Hilbert spaces) are the common underlying structure of quantum mechanics. $\endgroup$ Nov 2, 2011 at 17:43
  • $\begingroup$ I am vaguely aware of that, I hoped for a more elementary example, though I would love to hear a more detailed explanation of this. $\endgroup$
    – KotelKanim
    Nov 2, 2011 at 17:46
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    $\begingroup$ I have intuition for real inner product spaces of low dimension from Euclidean geometry. Anything else I understand only algebraically. For example the problem of resolving a vector into $n$ linearly independent "components" requires solving $n$ equations in $n$ unknowns in general, but reduces to computing just $n$ inner products if the components are orthogonal. This is clear from the algebra, and so useful it almost justifies arbitrary inner products all by itself. It is good to ask questions like yours, but I wanted to point out that it is not really necessary to have answers to them. $\endgroup$ Nov 2, 2011 at 22:27

3 Answers 3


Think of a inner product between $v_1$ on $v_2$ as an answer to the question: "How much of $v_2$ can be described using $v_1$".

This intuition corresponds nicely to what you know about orthogonality: if two vectors are orthogonal (inner product $0$), that means you can't describe $v_1$ using a linear combination of vectors that includes $v_2$.

In complex space, this is exactly the same idea - two vectors are orthogonal if one cannot be represented using the second one. For instance in $\mathbb{C^2}$, $v_1=(i,0)$ and $v_2=(0,1)$ are orthogonal for exactly the same reason they $v_1=(1,0)$ and $v_2=(0,1)$ are in $\mathbb{R^2}$.

One common example is the Fourier Series: the Fourier coefficients are a measure of "how much" of a given frequency is in our function. This doesn't change when we move to the complex representation, we just ask "how much" $e^{inx}$ is in $f(x)$ instead of $\sin(nx)$.

A concrete example may help:

An inner product over the space of all square-integrable functions on $[−π, π]$ may be defined : $$\langle f,g \rangle=\int_{-\pi}^{\pi} f(x) g(x)^*dx$$ Now, as an example, take $f(x)=\sin(x), \ g(x) = e^{ix}$. The inner product is: $$\langle f,g \rangle =\frac{1}{2i}$$ This is exactly "how much" of $e^{ix}$ is in $\sin(x)$, since, as you probably know: $$\sin(x) = \frac{e^{ix}}{2i} -\frac{e^{-ix}}{2i}$$

  • $\begingroup$ Great explanation. But for the interpretation of the complex result: The complex inner product is a combination of real inner product (Re<f,g>) and exterior product (Im<f,g>)(en.wikipedia.org/wiki/Exterior_algebra). In your example the Re<f,g>=0, that is, f and g are orthogonal and span the area of -1/2. $\endgroup$ Nov 16, 2016 at 16:14

It's quite tricky to answer this, but worth a quick stab.

Let's start with the exciting space $\mathbb C^1$. Here, the inner product of $(z)$ with the standard basis vector $(1)$ is just (depending on convention) $z$ or $\bar z$. This contains two pieces of information: the length of $z$, or how much $z$-ness there is in this direction, and the argument of $z$, which tells you how well the phase of $z$ agrees with the basis vector.

Since the standard basis for (finite dimensional) real inner product spaces carries over to a basis of the corresponding complex space, it's helpful to start by imagining $\mathbb R^2$ when first trying our hand at 2D complex space. Now complex vectors are vectors in the positive quadrant of this real space plus two extra pieces of information - the phases of the two components along the two axes. These are like like rotated wheels dangling off the two axes. Clearly the inner product with $(1,0)$ extracts the real plane length projected into this axis and also the phase information along this axis.

Note that $\mathbb R^2$ doesn't quite look look itself here when viewed as a subspace of the complex space because it's reflected into the positive quadrant and a forward/backward arrow attached for each axis to indicate $\pm$.

This helps to answer the begged question of how these magical arrow-plus-wheels objects transform under changes of basis. I'll leave you to try to visualize it, but essentially when we take linear combinations of the basis vectors to form new ones, the wheel phases of the components and the phases of the coefficients in the linear combination get smashed together.

For example, consider $(1,i)$ in the standard basis. This looks like the real $(1,1)$ vector plus a 90-degree rotated wheel on the second axis.

Now imagine the projection onto the vector $(1,-1)/\sqrt 2$ with an orthonormal $(1,+1)/\sqrt 2$ completing the change of basis. The inner product with the first is (up to conjugation) $(1-i)/\sqrt 2$, a unit-size quantity with phase of a negative 45-degree wheel. With the second, it's the same except the wheel is positively orientated. This tells you that $(1,i) $ is split equally between the two new vectors, with fairly close phase agreement with each. You can try to visualize this by imagining a gradual change of basis, and watching the wheels spin even as the projected lengths equalize out - however I think the previous sentence is more helpful.

Quantum mechanics is where this is all useful; but even a crude example shows the use. Imagine a wavy photon zooming through 3D space as a sinusoidal line traveling along the x axis, with the sinusoid pointed upwards along the z axis. This has z polarization. Imagine a different one, with the sinusoid zigzagging in the xy plane instead of the xz plane. Say this one has y polarization. Both are plane polarized because their graphs fit in planes. Now you can make things which have polarizations at all intermediate angles (only transverse allowed) by adding together these two in-phase waves by treating the two shapes as two basis elements in a real 2D space.

However, now imagine you lift one of the sinusoids up and put it down a quarter period shifted along. Then y peaks lie with z zeros. Now add the sinusoids. The resulting curve is a helix, and forms a circle when you project onto the yz plane. It is not plane polarized, but circularly polarized. Such light definitely exists (I believe it's how 3D glasses work - the polarization can be anti clockwise or clockwise depending which way we make the shift, giving two different types of light which can be filtered for left and right eyes). In fact, you can achieve this by taking complex combinations of the two basis shapes in the right way. (Quantum mechanically you simply define things like $1 y\pm i z$ where $y,z$ are QM states. In terms of functions you work with plane waves with field $\mathbf A= \mathrm {Re}(\boldsymbol\epsilon e^{i\mathbf{k\cdot x}})$ and allow the $\boldsymbol\epsilon$ polarization vector to be complex but orthogonal to $\mathbf k$.)

Then these circularly polarized states are just like the $(1,i)$ example above, and you can consider their decomposition into phases etc as discussed above; it all fits together. The inner product is still a measure of how much of a thing - and shifted by what phase - you need to make up a general signal in terms of an orthonormal basis.


This is all very belated to the OP, but since the question has gotten bumped, here's a standard algebraic interpretation of the complex (i.e., Hermitian) inner product on $\mathbb{C}^n$.

Let $z_j = x_j + iy_j$ be complex numbers decomposed into real and imaginary parts, and identify $\mathbf{z} = (z_1, \dots, z_n)$ in $\mathbb{C}^n$ with $\phi(\mathbf{z}) = (x_1, \dots, x_n, y_1, \dots, y_n)$ in $\mathbb{R}^{2n}$.

Write the complex inner product in terms of its real and imaginary parts: $\langle\mathbf{w}, \mathbf{z}\rangle = g(\mathbf{w}, \mathbf{z}) + i\omega(\mathbf{w}, \mathbf{z})$. The real part $g(\mathbf{w}, \mathbf{z})$ is easily checked to the the ordinary Euclidean dot product of $\phi(\mathbf{w})$ and $\phi(\mathbf{z})$, while the imaginary part $\omega(\mathbf{w}, \mathbf{z})$ is minus the standard symplectic form on $\mathbb{R}^{2n}$ evaluated on $\phi(\mathbf{w})$ and $\phi(\mathbf{z})$ (modulo a multiplicative factor of $2$, depending on one's definition of the wedge product).

Particularly, the imaginary part is skew-symmetric precisely because $\langle\mathbf{z}, \mathbf{w}\rangle = \overline{\langle\mathbf{w}, \mathbf{z}\rangle}$.

  • $\begingroup$ How does one show that $\omega ( \bf{w}, \bf{z} )$ is non-degenerate? $\endgroup$
    – Siupa
    May 24, 2022 at 11:27
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    $\begingroup$ @Siupa If $\mathbf{z}=(z_j)=(x_j+iy_j)$ and $\mathbf{w}=(w_j)=(u_j+iv_j)$, then (up to possible minor changes of notation) $$\omega(\mathbf{w},\mathbf{z})=\operatorname{im}\sum_j w_j \bar{z}_j=\sum_j(x_jv_j-u_jy_j)=\Bigl(\sum_j dx_j\wedge dy_j\Bigr)(\mathbf{z},\mathbf{w}).$$The $n$th exterior power is therefore proportional to the Euclidean volume form. $\endgroup$ May 26, 2022 at 18:48

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