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The dot product in 2D or 3D Euclidean space between vectors $\mathbf{a}$ and $\mathbf{b}$ with magnitudes $|\mathbf{a}|$ and $|\mathbf{b}|$ can be written as $$ \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|} $$ where $\theta$ is the angle between the vectors. So far so good.

Now using vectors in $\mathbb{C}^2$, we can find the inner product $\langle \mathbf{a},\mathbf{b} \rangle$ by finding the complex conjugate of $\mathbf{a}$ to get $\overline{\mathbf{a}}$ and multiplying that with $\mathbf{b}$ (I understand mathematicians conjugate the second term), giving us $$ \langle\mathbf{a},\mathbf{b}\rangle = \sum_{i=0}^{n-1} \overline{a_i}\,b_i $$ If we think of this as a generalization of the dot product, then we can think the above as also $|\overline{\mathbf{a}}|\,|\mathbf{b}|\,\cos\theta$ for the angle $\theta$ between $\overline{\mathbf{a}}$ and $\mathbf{b}$. This is not the angle between $\mathbf{a}$ and $\mathbf{b}$ because $\overline{\mathbf{a}}$ is not a scaled version of $\mathbf{a}$ - it points in an entirely different direction.

I find this confusing! We're not really generalizing one of the most useful properties of the dot product because we're not getting back something that can give us the angle between the two vectors. We're getting back a different angle, involving $\overline{\mathbf{a}}$ rather than $\mathbf{a}$. Yet I frequently see the inner product used as a generalization of the dot product in this respect.

Can anyone please straighten me out? Definitions are less useful than explanations, as I want to know where this line of thinking is in error.

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3 Answers 3

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Your first formula:

$$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}$$

should actually be:

$$\cos\theta = \frac{\overline{\mathbf{a}}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}$$

We don't normally write it this way, because for Real vectors $\overline{\mathbf{a}}=\mathbf{a}$, but this form is more correct, and more general.

To see why we would want to do it this way, consider the angle between the Complex vector $\pmatrix{i \\ 0}$ and itself.

$$\cos\theta=\frac{\pmatrix{-i \\ 0}\cdot \pmatrix{i \\ 0}}{\left|\pmatrix{i \\ 0}\right|\left|\pmatrix{i \\ 0}\right|}=\frac{-i\times i+0\times 0}{1\times 1}=1$$

The angle between this vector and itself is zero, as expected. If we didn't conjugate one of the vectors, we would get a vector at an angle of $180^\circ$ to itself, which is not a property we expect angles between vectors to have! Conversely, there are distinct vectors that would have an angle of zero between them, which is also undesirable.

Complex numbers, despite the nice Argand plane picture, cannot be treated as Euclidean vectors. The Complex plane is not isotropic: $1$ is algebraically different from $i$. They have different squares, for a start. So squaring and adding components to get the length squared (using Pythagoras' Theorem) won't work. To get a positive definite 'length' with Complex numbers, we need to take the product with the complex conjugate. And then 'angle' follows from 'length' by using the cosine rule for triangles in the way hyportnex indicates.

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Since $d^2=|a-b|^2=|a|^2+|b|^2-\bar a b -a \bar b$ and $d$ is the positive distance between the vectors $a$ and $b$ it is natural to interpret the angle $\theta$ defined by $\bar a b + a \bar b = 2|a||b|cos(\theta)$ as the angle between the vectors $a$ and $b$.


Per @BySymmetry's suggestion, note too that $\bar a b + a \bar b = 2 \Re[\bar a b]$ and, since $|\Re[z]| \le |z|$ for any complex $z$, we also have $\frac{|\bar a b + a \bar b |}{2 |a||b|}\le 1$ and we can define a true (real) angle $\theta$ between them by $cos(\theta) = \frac{\Re[\bar a b]}{|a||b|}$.

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  • $\begingroup$ I don't understand this answer because it tells us to "interpret" the result in one way that doesn't seem to match what's actually being computed. Let me rephrase this with a picture. link shows how I'm thinking about this. When I find the inner product $\langle \mathbf{a},\mathbf{b}\rangle$ I'd expect the angle $\theta$ in the figure, corresponding to treating the vectors as real and computing $\mathbf{a}\cdot\mathbf{b}$. But it's really computing the equivalent of $\overline{\mathbf{a}}\cdot\mathbf{b}$ which is $\delta$ in the figure - a different angle! $\endgroup$
    – Eleeza
    Oct 4, 2022 at 0:12
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    $\begingroup$ set $b=\bar a$ and calculate the distance $d^2=|a-\bar a|^2 = 2|a|^2-2|a|^2cos(\theta)$, does it make sense now? $\endgroup$
    – hyportnex
    Oct 4, 2022 at 0:26
  • $\begingroup$ I'm sorry, I appreciate your help, but that doesn't clarify things for me. And I don't see where cosine is appearing from in your expressions. Can you tell me where that comes from, but more importantly, what is specifically wrong with my argument that we're calculating $\delta$ (see the picture linked the previous comment) rather than $\theta$? $\endgroup$
    – Eleeza
    Oct 4, 2022 at 3:10
  • $\begingroup$ It might be helpful to explicitly note that $\frac{\bar{a}b+\bar{b}a}{2} = \mathrm{Re}[\bar{a}b]$ so it is obvious exactly how this result relates to the real case $\endgroup$ Oct 4, 2022 at 10:09
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To get inner product of vectors a and b we need to take conjugate transpose of a and then multiply it to vector b. But taking conjugate transpose of a is just an intermediate step; it does not mean that we get inner product between conjugate transpose of a and b. What we get is actually the inner product of a and b. So the angle defined by the inner product is the angle between vectors a and b.

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