HW 03: Difference between revisions

Latest revision as of 18:27, 12 November 2008

Problem

If $⟨ ϕ_{n} | ϕ_{m} ⟩ = δ_{m n}$ and $ϕ_{n}$ span the space of functions for which $x (t)$ and $y (t)$ are members and $x (t) = \sum_{n} a_{n} ϕ_{n} (t)$ and $y (t) = \sum_{m} b_{m} ϕ_{m} (t)$ , then show

$⟨ x | y ⟩ = \sum_{n} a_{n} b_{n}^{*}$
$⟨ x | x ⟩ = \sum_{n} {| a_{n} |}^{2}$

Notes

$⟨ x | y ⟩ = \int_{- \infty}^{\infty} x (t) y (t)^{*} d t$

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

Solution

$\int_{- \infty}^{\infty} \sum_{n} a_{n} ϕ_{n} (t) \sum_{m} b_{m} ϕ_{m} (t)^{*} d t$	$= \sum_{n} \sum_{m} a_{n} b_{m}^{} \int_{- \infty}^{\infty} ϕ_{n} (t) ϕ_{m} (t)^{} d t$
	$= \sum_{n} \sum_{m} a_{n} b_{m}^{*} ⟨ ϕ_{n} (t) \| ϕ_{m} (t) ⟩$
	$= \sum_{n} \sum_{m} a_{n} b_{m}^{*} δ_{n m}$
	$= \sum_{n} a_{n} b_{n}^{*}$

$\int_{- \infty}^{\infty} \sum_{n} a_{n} ϕ_{n} (t) \sum_{m} a_{m} ϕ_{m} (t)^{*} d t$	$= \sum_{n} \sum_{m} a_{n} a_{m}^{} \int_{- \infty}^{\infty} ϕ_{n} (t) ϕ_{m} (t)^{} d t$
	$= \sum_{n} \sum_{m} a_{n} a_{m}^{*} ⟨ ϕ_{n} (t) \| ϕ_{m} (t) ⟩$
	$= \sum_{n} \sum_{m} a_{n} a_{m}^{*} δ_{n m}$
	$= \sum_{n} {\| a_{n} \|}^{2}$

@@ Line 2: / Line 2: @@
 If <math> \left \langle \phi_n | \phi_m \right \rangle = \delta_{mn}</math> and <math> \phi_n \,\!</math> span the space of functions for which <math>x(t)\,\!</math> and <math>y(t)\,\!</math> are members and
 <math>x(t)= \sum _n a_n \phi_n (t)\,\!</math> and
-<math>y(t)= \sum _n b_n \phi_n (t)\,\!</math>, then show
+<math>y(t)= \sum _m b_m \phi_m (t)\,\!</math>, then show
 #<math> \left \langle x | y \right \rangle = \sum_n a_n b_n^*</math>
@@ Line 12: / Line 12: @@
 ==Solution==
-#<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \left ( \sum _n b_n \phi_n (t) \right )^*\,dt</math>
+{| border="0" cellpadding="0" cellspacing="0"
-#<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \left ( \sum _n a_n \phi_n (t) \right )^*\,dt</math>
+|-
+|<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \sum _m b_m \phi_m (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>
+|}
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>\int_{-\infty}^{\infty} \sum _n a_n \phi_n (t) \sum _m a_m \phi_m (t)^* \,dt</math>
+|<math>=\sum_n \sum _m a_n a_m^* \int_{-\infty}^{\infty} \phi_n (t) \phi_m (t)^* \,dt</math>
+|-
+|
+|<math>=\sum_n \sum _m a_n a_m^* \left \langle \phi_n (t) | \phi_m (t) \right \rangle</math>
+|-
+|
+|<math>=\sum_n \sum _m a_n a_m^* \delta_{nm}</math>
+|-
+|
+|<math>=\sum_n \left | a_n \right |^2</math>
+|}

$\int_{- \infty}^{\infty} \sum_{n} a_{n} ϕ_{n} (t) \sum_{m} a_{m} ϕ_{m} (t)^{*} d t$	$= \sum_{n} \sum_{m} a_{n} a_{m}^{} \int_{- \infty}^{\infty} ϕ_{n} (t) ϕ_{m} (t)^{} d t$
	$= \sum_{n} \sum_{m} a_{n} a_{m}^{*} ⟨ ϕ_{n} (t) \| ϕ_{m} (t) ⟩$
	$= \sum_{n} \sum_{m} a_{n} a_{m}^{*} δ_{n m}$
	$= \sum_{n} {\| a_{n} \|}^{2}$

HW 03: Difference between revisions

Latest revision as of 18:27, 12 November 2008

Problem

Notes

Solution

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