HW 03: Difference between revisions

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<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \sum _n a_n \phi_n (t)^* \,dt</math>
{| border="0" cellpadding="0" cellspacing="0"
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|<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \sum _m b_n \phi_n (t)^* \,dt</math>
|<math>=\sum_n \sum _m a_n b_m^* \int_{-\infty}^{\infty} \phi_n (t) \phi_m (t)^* \,dt</math>
|-
|
|<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^* \delta_{nm^*}</math>
|-
|
|<math>=\sum_n a_n b_n^*</math>
|}

Revision as of 16:26, 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)mbnϕn(t)*dt =nmanbm*ϕn(t)ϕm(t)*dt
=nmanbm*ϕn(t)|ϕm(t)*
=nmanbm*δnm*
=nanbn*


nanϕn(t)mbnϕn(t)*dt =nmanbm*ϕn(t)ϕm(t)*dt
=nmanbm*ϕn(t)|ϕm(t)*
=nmanbm*δnm*
=nanbn*