HW 03: Difference between revisions

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|<math>=\sum_n \sum _m a_n b_m^* \left \langle \phi_n (t) | \phi_m (t)^* \right \rangle</math>
|<math>=\sum_n \sum _m a_n b_m^* \left \langle \phi_n (t) | \phi_m (t) \right \rangle</math>
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|<math>=\sum_n \sum _m a_n b_m^* \delta_{nm^*}</math>
|<math>=\sum_n \sum _m a_n b_m^* \delta_{nm}</math>
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|<math>=\sum_n a_n b_n^*</math>
|<math>=\sum_n a_n b_n^*</math>
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*Note <math>\delta_{nm^*}=\delta_{nm}</math>


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|<math>=\sum_n \sum _m a_n a_m^* \left \langle \phi_n (t) | \phi_n (t)^* \right \rangle</math>
|<math>=\sum_n \sum _m a_n a_m^* \left \langle \phi_n (t) | \phi_m (t) \right \rangle</math>
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Latest revision as of 18:27, 12 November 2008

Problem

If ϕn|ϕm=δmn and ϕn span the space of functions for which x(t) and y(t) are members and x(t)=nanϕn(t) and y(t)=mbmϕm(t), then show

  1. x|y=nanbn*
  2. x|x=n|an|2

Notes

x|y=x(t)y(t)*dt

  • This notation is called the Bra ϕ| Ket |ψ, or Dirac notation. It denotes the inner product.

Solution

nanϕn(t)mbmϕm(t)*dt =nmanbm*ϕn(t)ϕm(t)*dt
=nmanbm*ϕn(t)|ϕm(t)
=nmanbm*δnm
=nanbn*
nanϕn(t)mamϕm(t)*dt =nmanam*ϕn(t)ϕm(t)*dt
=nmanam*ϕn(t)|ϕm(t)
=nmanam*δnm
=n|an|2