# What is the dimension considered in the Schmidt Decomposition?

In the Schmidt decomposition, is the dimension considered of each Hilbert space the complex or real one? Meaning the complex dimension of $$\mathbb C^2$$ has dimension $$2$$, not $$4$$. If so when you choose the orthonormal vectors for a space like $$\mathbb C^2$$ do they have to be real or can also be complex?

It is the complex dimension. So for an orthonomal basis of $$\mathbb{C}^2$$, you would choose two vectors which need not be real.
Example: If $$\mathbb{C}^2$$ is the Hilbert space of a $$1/2$$-spin, the basis of eigenstates of the $$z$$ and $$y$$ spin component are respetively \begin{align} B_z & = \lbrace(1,0) \, , \, (0,1)\rbrace \\ B_y & = \left\lbrace\left(\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}}\right) \, , \, \left(\frac{1}{\sqrt{2}},\frac{-i}{\sqrt{2}}\right) \right\rbrace \end{align} Note how they are complex in general. It is also physically sensible to speak of dimension $$2$$ since there are $$2$$ posible outcomes of any measurement, rather than $$4$$.