Discussion
Given \(|v_{1}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}\) and \(|v_{2}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}\), the tensor product \(|v_{1}\rangle \otimes |v_{2}\rangle\) is
Answer: A
The tensor product is \(\frac{1}{2}(1, -i, i, 1)^{T}\), a column vector.
For column vectors \(a\) and \(b\), the tensor product is the column \((a_{1}b_{1}, a_{1}b_{2}, a_{2}b_{1}, a_{2}b_{2})^{T}\). With \(a = (1, i)\), \(b = (1, -i)\) and the prefactor \(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}\):
\[ \begin{aligned} a_{1}b_{1} &= 1, \quad a_{1}b_{2} = -i \\ a_{2}b_{1} &= i, \quad a_{2}b_{2} = (i)(-i) = 1 \end{aligned} \]
The option \((1, i, -i, 1)\) is \(|v_{2}\rangle \otimes |v_{1}\rangle\), the product taken in the other order. The two 2 by 2 matrices are outer products of the form \(|v\rangle\langle w|\), not the tensor product of two kets, which is a four-component column.
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