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For a real valued vector $x(t)$ that depends on some parameter $t$ and when requiring that $$\left|x(t)\right| = \mathrm{const}$$ it is easy to show that $$\dot{x}(t) \perp x(t)$$ where the dot denotes the derivative with respect to $t$. For a derivation see for example this math.stackexchange question. The basic idea goes like this:

$$x(t)\cdot\dot{x}(t) = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}(x(t)\cdot x(t)) = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \left|x(t)\right|^2 = 0$$

My question is, can this be generalized to complex vectors $x \in \mathbb{C}^n$? Additionally, can you provide a textbook that deals with this problem?

HerpDerpington
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We have $|x|^2=x^H x$, where $x^H$ is the conjugate transpose of $x$. Then, constant norm implies that

$$0=\frac{\mathrm{d}}{\mathrm{d}t}\left(x^H x\right)=\dot{x}^Hx+{x}^H\dot{x}=2\Re\left(x^H\dot{x}\right)$$

because $\dot{x}^Hx$ is the conjugate of ${x}^H\dot{x}$. So the equivalent condition for complex-valued $x(t)$ is

$$\Re\left(x^H\dot{x}\right)=0.$$

For example, $x=e^{it}$ is a one-component vector with constant norm, and we have

$$\Re\left(x^H\dot{x}\right)=\Re\left(e^{-it} i e^{it}\right)=\Re\left(i\right)=0.$$

wimi
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  • Is there a reason why the dot product can be purely imaginary, but of arbitrary absolute value? From my intuition I would say that there must be some condition regarding the imaginary part... – HerpDerpington Jan 04 '20 at 12:53
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    The imaginary part can be anything. Consider $x=\exp(i a t)$ for arbitrary $a$. It has constant norm, and the dot product $x^H\dot{x}=i a$ has arbitrary imaginary part. – wimi Jan 04 '20 at 12:54
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    @HerpDerpington The"intuition" is that the $|\cdot|$ operator transforms a complex number (2 real dimensions) into a real number (one dimension), thereby losing information about the phase. After the derivative, the information about the magnitude goes to the real part of the dot product. The imaginary part has (in some much more more convoluted way) information about the phase of $x$. – wimi Jan 04 '20 at 13:05
  • Is that in the sense that the large this imaginary part is, the stronger are the the phase oscillations of $x$ with changing $t$? – HerpDerpington Jan 04 '20 at 22:42
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    @HerpDerpington for a one-component vector, if you write $x=re^{i\phi}$, you can check that the dot product is $x^*\dot{x}=r\dot{r} + ir^2\dot{\phi}$. So the imaginary part is the squared magnitude times the derivative of the phase. If you divide the imaginary part by $r^2$, then it does tell you how fast the phase changes. For vectors with several components it gets complicated, but you can try a similar approach and see if you arrive anywhere. – wimi Jan 05 '20 at 07:30