Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 11 / iv

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- Solution iv

Solution

Using the Hermitian inner product, the scalar product is

1 - i + 2 i^{2} = - 1 - i

. The two squared norms are

6

and

4

. The specified real angle in a complex inner product space therefore has

cos θ = \frac{Re ( - 1 - i )}{6 4} = - \frac{1}{2 6} .

The real part is essential: the complex scalar product itself need not be real.

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