# Gram-schmidt procedure

1. Mar 27, 2012

### spaghetti3451

1. The problem statement, all variables and given/known data

Use the Gram-Schmidt procedure to orthonormalize the three-space basis $\left|e_{1}\right\rangle = (1 + i) \widehat{i} + (1) \widehat{j} + (i) \widehat{k}, \left|e_{2}\right\rangle = (i) \widehat{i} + (3) \widehat{j} + (1) \widehat{k}, \left|e_{3}\right\rangle = (0) \widehat{i} + (28) \widehat{j} + (0) \widehat{k}.$

2. Relevant equations

3. The attempt at a solution

1. To obtain the first orthonormal basis vector, normalise $\left|e_{1}\right\rangle$. Therefore, $\left|e^{'}_{1}\right\rangle = (1/2 + i/2) \widehat{i} + (1/2) \widehat{j} + (i/2) \widehat{k}$.

2. The projection of $\left|e_{2}\right\rangle$ along $\left|e^{'}_{1}\right\rangle$ = the scalar multiplication of $\left|e^{'}_{1}\right\rangle$ with the inner product of $\left|e^{'}_{1}\right\rangle$ and $\left|e_{2}\right\rangle$ = $[(1/2 - i/2)(i) + (1/2)(3) + (-i/2)(1)] [(1/2 + i/2) \widehat{i} + (1/2) \widehat{j} + (i/2) \widehat{k}] = (1 + i) \widehat{i} + (1) \widehat{j} + (i) \widehat{k}$.

Subtract the projection from $\left|e_{2}\right\rangle$ and normalise to obtain $\left|e^{'}_{2}\right\rangle = (-1/\sqrt{7}) \widehat{i} + (2/\sqrt{7}) \widehat{j} + ( (1 - i)/\sqrt{7} ) \widehat{k}$

3. The projection of $\left|e_{3}\right\rangle$ along $\left|e^{'}_{1}\right\rangle$ = the scalar multiplication of $\left|e^{'}_{1}\right\rangle$ with the inner product of $\left|e^{'}_{1}\right\rangle$ and $\left|e_{3}\right\rangle$ = $[0 + 14 + 0] [(1/2 + i/2) \widehat{i} + (1/2) \widehat{j} + (i/2) \widehat{k}] = (7 + 7i) \widehat{i} + (7) \widehat{j} + (7i) \widehat{k}$.

The projection of $\left|e_{3}\right\rangle$ along $\left|e^{'}_{2}\right\rangle$ = the scalar multiplication of $\left|e^{'}_{2}\right\rangle$ with the inner product of $\left|e^{'}_{2}\right\rangle$ and $\left|e_{3}\right\rangle$ = $[0 + 56/\sqrt{7} + 0] [(-1/\sqrt{7}) \widehat{i} + (2/\sqrt{7}) \widehat{j} + ( (1 - i)/\sqrt{7} ) \widehat{k}] = (-8) \widehat{i} + (16) \widehat{j} + (8 - 8i) \widehat{k}$.

Subtract the projections from $\left|e_{3}\right\rangle$ and normalise to obtain $\left|e^{'}_{3}\right\rangle = ( (1 - 7i)/\sqrt{130}) \widehat{i} + (5/\sqrt{130}) \widehat{j} + ( (i - 8)/\sqrt{130} ) \widehat{k}$.