If a general spin state is characterised by 2 complex numbers, that would mean that 4 real parameters characterise it. I'm assuming that they are given by the complex and real parts of said complex numbers. Why would the normalisation equation only give us 3 parameters back? I've just started learning linear algebra, so I'm not too familiar with complex ideas yet.
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1$\begingroup$ The keyword that's missing is that you have 3 independent parameters. $\endgroup$infinitezero– infinitezero2026-02-27 08:36:44 +00:00Commented yesterday
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1$\begingroup$ The fourth isn’t independent because it’s determined by the normalization. $\endgroup$Ghoster– Ghoster2026-02-27 21:35:59 +00:00Commented yesterday
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Everytime you introduce a new equation (or constraint) you remove a degree of freedom. For example, consider the 1D complex "vector" $z=a+bi$. It has two degrees of freedom. If I require normalization I introduce the constraint $|z|^2=1$. So I could write my vector as $$z=\frac{a+bi}{\sqrt{a^2+b^2}}.$$ From this equation it is not clear that it depends on one degree of freedom. We can introduce polar coordinates to make it more clear: $a+bi=r e^{i\theta}$. This gives us $$z=\frac{r e^{i\theta}}{r}=e^{i\theta}.$$
To summarize, introducing an equation removes a degree of freedom. This may not be obvious immediately but generally you can reparemetrize to make it clear that a degree of freedom has disappeared.
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$\begingroup$ This is not true in general but often used. Take a linear system: you could introduce an equation that is a combination of the others, in this case nothing changes, or an incompatible equation making the system has no solution anymore. $\endgroup$JP2– JP22026-02-27 18:23:17 +00:00Commented yesterday