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Vectors and normalization

Kets are column vectors; probabilities must add to 1

about 5 min

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A ket is a column vector

For one qubit, ∣0⟩\ket{0} and ∣1⟩\ket{1} are just names for the two standard column vectors:

∣0⟩=[10],∣1⟩=[01],α∣0⟩+β∣1⟩=[αβ]\begin{gathered}\ket{0} = \begin{bmatrix}1\\0\end{bmatrix},\qquad \ket{1} = \begin{bmatrix}0\\1\end{bmatrix}, \\[6pt] \alpha\ket{0} + \beta\ket{1} = \begin{bmatrix}\alpha\\ \beta\end{bmatrix}\end{gathered}

The label inside a ket is only a name: ∣ψ⟩\ket\psi (“ket psi”) stands for a state in general, and ∣±⟩=12(∣0⟩±∣1⟩)\ket\pm = \tfrac{1}{\sqrt2}(\ket0 \pm \ket1) are two states that come up constantly. The entries are complex amplitudes. A vector represents a qubit state only if it is normalized: ∣α∣2+∣β∣2=1|\alpha|^2 + |\beta|^2 = 1. To normalize any nonzero vector, divide every entry by its length ∣α∣2+∣β∣2\sqrt{|\alpha|^2 + |\beta|^2}.

Worked example

Normalizing 3∣0⟩+4i∣1⟩3\ket0 + 4i\ket1

  1. Squared magnitudes: ∣3∣2=9|3|^2 = 9 and ∣4i∣2=16|4i|^2 = 16. The ii has magnitude 1, so it contributes +16+16, not −16-16.

  2. Length: 9+16=5\sqrt{9 + 16} = 5.

  3. Divide every entry by 5: 35∣0⟩+4i5∣1⟩\tfrac35\ket0 + \tfrac{4i}{5}\ket1. Check: 925+1625=1\tfrac{9}{25} + \tfrac{16}{25} = 1.

Check yourself

Normalize ∣0⟩+2∣1⟩\ket{0} + 2\ket{1}. Enter the amplitudes.

Hints
  1. What is the vector’s length, ∣α∣2+∣β∣2\sqrt{|\alpha|^2 + |\beta|^2}?
  2. The length is 12+22=5\sqrt{1^2 + 2^2} = \sqrt5.
  3. Divide both entries by 5\sqrt5.
Show the answer

Answer: 15∣0⟩+25∣1⟩\frac{1}{\sqrt{5}}|0\rangle + \frac{2}{\sqrt{5}}|1\rangle

Length 5\sqrt5, so the state is 15∣0⟩+25∣1⟩\tfrac{1}{\sqrt5}\ket{0} + \tfrac{2}{\sqrt5}\ket{1}, and 1/5+4/5=11/5 + 4/5 = 1.

Check yourself

Which of these are normalized qubit states?

Select all that apply.

  1. Ai2∣0⟩−12∣1⟩\tfrac{i}{\sqrt2}\ket{0} - \tfrac{1}{\sqrt2}\ket{1}
  2. B35∣0⟩+45∣1⟩\tfrac{3}{5}\ket{0} + \tfrac{4}{5}\ket{1}
  3. C12∣0⟩+12∣1⟩\tfrac{1}{2}\ket{0} + \tfrac{1}{2}\ket{1}
  4. D∣0⟩−∣1⟩\ket{0} - \ket{1}
Show the answer

Answers: A. i2∣0⟩−12∣1⟩\tfrac{i}{\sqrt2}\ket{0} - \tfrac{1}{\sqrt2}\ket{1}; B. 35∣0⟩+45∣1⟩\tfrac{3}{5}\ket{0} + \tfrac{4}{5}\ket{1}

Add the squared magnitudes: (a) 9/25+16/25=19/25 + 16/25 = 1; (c) 1/2+1/2=11/2 + 1/2 = 1, since ∣i∣2=1|i|^2 = 1.

Why not the others

  • C. 1/4+1/4=1/21/4 + 1/4 = 1/2, not 1.
  • D. Squared magnitudes add to 2. A minus sign doesn’t reduce a magnitude.

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