Exercise 2.8

The Gram-Schmidt procedure converts a non-orthonormal basis \(|w_1\rangle, \dotsc, |w_d\rangle\) in vector space \(V\) into an orthonormal basis \(|v_1\rangle, \dotsc, |v_d\rangle\). I will proof new basis \(|v_1\rangle, \dotsc, |v_d\rangle\) is orthonormal below. The new vector \(|v_{k+1}\rangle\) is given by \(|v_1\rangle = |w_1\rangle/|||w_1\rangle||\) and for \(1\leq k \leq d-1\) then \(|v_{k+1}\rangle\) is defined by

(1)\[ |v_{k+1}\rangle = \frac{|w_{k+1}\rangle - \sum_{i=1}^{k}\langle v_{i}|w_{k+1}\rangle |v_{i}\rangle}{|||w_{k+1}\rangle - \sum_{i=1}^{k}\langle v_{i}|w_{k+1}\rangle |v_{i}\rangle||} \]

I will prove for \(i,j\in [1, d]\) we have \(\langle v_i|v_j\rangle = \delta_{ij}\) by induction. That is,

• For \(|v_1\rangle\), we have \(\langle v_1|v_1\rangle = 1\).

• For \(n=1\) or namely, for \(|v_2\rangle\), \(\langle v_1|v_2\rangle = 0\) and \(\langle v_{2}|v_2\rangle = 1\).

• If for \(n=k\) we have \(\langle v_{j}|v_{k+1}\rangle = \delta_{(k+1)j}\) for \(j \leq i+1 \), then we have for \(n=k+1\), \(\langle v_{j}|v_{k+2}\rangle = \delta_{(k+2)j}\) for \(j\leq k+2\).

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Here is the detail of proof of each statement.

• For \(|v_1\rangle\), we have \(|v_1\rangle = |w_1\rangle / |||w_1\rangle||\), then

  (2)\[ \langle v_1|v_1\rangle = \frac{\langle w_1|w_1\rangle} {|||w_1\rangle||^2} = 1 \]

• For \(n=1\), we have

  (3)\[ |v_{2}\rangle = \frac{|w_{2}\rangle - \langle v_{1}|w_{2}\rangle |v_{1}\rangle}{|||w_{2}\rangle - \langle v_{1}|w_{2}\rangle |v_{1}\rangle||} \]

  Suppose \(|a\rangle = |w_{2}\rangle - \langle v_{1}|w_{2}\rangle |v_{1}\rangle\) then

  (4)\[ \langle v_2|v_2\rangle = \frac{\langle a|a\rangle}{|||a\rangle||^2} = 1 \]

  Meanwhile, we have

  (5)\[ \langle v_1|v_{2}\rangle = \frac{\langle v_1|w_{2}\rangle - \langle v_{1}|w_{2}\rangle \langle v_1|v_{1}\rangle}{|||w_{2}\rangle - \langle v_{1}|w_{2}\rangle |v_{1}\rangle||} = 0 \]

• Suppose \(n=k\) if we have \(\langle v_{j}|v_{k+1}\rangle = \delta_{(k+1)j}\), then when \(n=k+1\), for \(j\neq k+2\),

  (6)\[\begin{split} \begin{align} \langle v_j|v_{k+2}\rangle &= \frac{\langle v_j|w_{k+2}\rangle - \sum_{i=1}^{k+1}\langle v_{i}|w_{k+2}\rangle \langle v_j|v_{i}\rangle}{|||w_{k+2}\rangle - \sum_{i=1}^{k+1}\langle v_{i}|w_{k+2}\rangle |v_{i}\rangle||} \\ &= \frac{\langle v_j|w_{k+2}\rangle - \langle v_{j}|w_{k+2}\rangle \langle v_j|v_{j}\rangle}{|||w_{k+2}\rangle - \sum_{i=1}^{k+1}\langle v_{i}|w_{k+2}\rangle |v_{i}\rangle||} \\ &= \frac{\langle v_j|w_{k+2}\rangle - \langle v_{j}|w_{k+2}\rangle }{|||w_{k+2}\rangle - \sum_{i=1}^{k+1}\langle v_{i}|w_{k+2}\rangle |v_{i}\rangle||} = 0 \end{align} \end{split}\]

  For \(j = k+2\), let \(|a\rangle = |w_{k+1}\rangle - \sum_{i=1}^{k+1}\langle v_{i}|w_{k+2}\rangle |v_{i}\rangle\), then

  (7)\[ \langle v_{k+2}|v_{k+2}\rangle = \frac{\langle a|a\rangle}{|||a\rangle||^2} = 1 \]

  From eq. (6) and eq. (7), we conclude that for \(n=k+1\), \(\langle v_{j}|v_{k+2}\rangle = \delta_{(k+2)j}\) for \(j\leq k+2\).