Question

Given \( |v_1\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix} \) and \( |v_2\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix} \), the tensor product \( |v_1\rangle \otimes |v_2\rangle \) is

• A. \( \dfrac{1}{2}\begin{pmatrix} 1 \\ -i \\ i \\ 1 \end{pmatrix} \)
• B. \( \dfrac{1}{2}\begin{pmatrix} 1 \\ i \\ -i \\ 1 \end{pmatrix} \)
• C. \( \dfrac{1}{2}\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix} \)
• D. \( \dfrac{1}{2}\begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \)

Hint:

Tensor-product a 2-entry ket with a 2-entry ket to get a 4-entry ket: multiply each entry of the first by the whole second ket and stack the results, don't build a 2x2 matrix.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The tensor product (also called the Kronecker product) of two column vectors builds a bigger column vector. If \(|v_1\rangle\) has 2 entries and \(|v_2\rangle\) has 2 entries, then \(|v_1\rangle \otimes |v_2\rangle\) has \(2 \times 2 = 4\) entries. Each entry of \(|v_1\rangle\) gets multiplied by the whole of \(|v_2\rangle\), and the results are stacked one after another.

Step 2: Key Formula or Approach:
For \(|v_1\rangle = \begin{pmatrix} a \\ b \end{pmatrix}\) and \(|v_2\rangle = \begin{pmatrix} c \\ d \end{pmatrix}\), the tensor product is
\[ |v_1\rangle \otimes |v_2\rangle = \begin{pmatrix} a c \\ a d \\ b c \\ b d \end{pmatrix} \]

Step 3: Detailed Explanation:
Here \(a = 1\), \(b = i\) (from \(|v_1\rangle\), before the common factor \(1/\sqrt{2}\)), and \(c = 1\), \(d = -i\) (from \(|v_2\rangle\)). The two \(1/\sqrt{2}\) factors combine to give \(1/2\).
Now build each of the four entries:
\[ ac = (1)(1) = 1 \]
\[ ad = (1)(-i) = -i \]
\[ bc = (i)(1) = i \]
\[ bd = (i)(-i) = -i^2 = 1 \]
Stacking these in order gives
\[ |v_1\rangle \otimes |v_2\rangle = \frac{1}{2}\begin{pmatrix} 1 \\ -i \\ i \\ 1 \end{pmatrix} \]

Step 4: Why the other options are wrong.
Option (B) has the middle two entries swapped in sign (\(i\) then \(-i\) instead of \(-i\) then \(i\)); this comes from multiplying \(b\) by \(c\) and \(a\) by \(d\) in the wrong order. Options (C) and (D) arrange the four numbers as a \(2\times2\) matrix instead of a single stacked \(4\times1\) column, which is not what the tensor product of two column vectors gives; a tensor product of two kets is itself a ket, not an operator.

Final Answer:
The tensor product works out to a column vector with entries 1, -i, i, 1, all scaled by one half.\[ \boxed{\frac{1}{2}\begin{pmatrix} 1 \\ -i \\ i \\ 1 \end{pmatrix}} \]