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Gram-schmidt procedure

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the Gram-Schmidt procedure to orthonormalize the three-space basis [itex]\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}.[/itex]

    2. Relevant equations

    3. The attempt at a solution

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

    2. The projection of [itex]\left|e_{2}\right\rangle[/itex] along [itex]\left|e^{'}_{1}\right\rangle[/itex] = the scalar multiplication of [itex]\left|e^{'}_{1}\right\rangle[/itex] with the inner product of [itex]\left|e^{'}_{1}\right\rangle[/itex] and [itex]\left|e_{2}\right\rangle[/itex] = [itex] [(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}[/itex].

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

    3. The projection of [itex]\left|e_{3}\right\rangle[/itex] along [itex]\left|e^{'}_{1}\right\rangle[/itex] = the scalar multiplication of [itex]\left|e^{'}_{1}\right\rangle[/itex] with the inner product of [itex]\left|e^{'}_{1}\right\rangle[/itex] and [itex]\left|e_{3}\right\rangle[/itex] = [itex] [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}[/itex].

    The projection of [itex]\left|e_{3}\right\rangle[/itex] along [itex]\left|e^{'}_{2}\right\rangle[/itex] = the scalar multiplication of [itex]\left|e^{'}_{2}\right\rangle[/itex] with the inner product of [itex]\left|e^{'}_{2}\right\rangle[/itex] and [itex]\left|e_{3}\right\rangle[/itex] = [itex] [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}[/itex].

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

    I would be grateful if you could please provide comments.
     
    Last edited: Mar 27, 2012
  2. jcsd
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