ExerClass1

ExerClass1 - PHY4211 Quantum Mechanics I Fall Term of 2005...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
PHY4211 Quantum Mechanics I Fall Term of 2005 Exercise in Page 7 of your Lecture Notes (Chapter 1): Consider the function () ( ) ( ) exp exp f xx A x B = , its first derivative could be obtained as: ( ) () , xA xB xA xB xA xB xA xB x Ax B x B x B x A x B xA xB xA xB df x dd ee e e dx dx dx Ae e e Be Ae e Be e Be e e Be ABee e Be ⎛⎞ =+ ⎜⎟ ⎝⎠ =+−+ ⎡⎤ + ⎣⎦ ; (1) Now consider the term , xA eB : [] {} 00 0 0 1 1 1 1 1 0 11 0 ,, !! ! , ! , ! , ! , ! , 1! , nn n n xA n n n n mn m nm n n mnm n n n n xA x A B x B A B n xAB n x AA BA n x AB A A n x nA n xA n xABe ∞∞ == = = ∞− −− = = ⎢⎥ = ⎧⎫ = ⎨⎬ ⎩⎭ = = = = ∑∑ . (2) Note that in the above calculation we need that n-1>=0 and so n is not smaller than 1. But this is in fact not a problem at all: since if n=0 the corresponding component of exp(xA) is 1 which is a component commuting with B and thus has no contribution to the last result. If you like, you could drop this component at the very first of the whole derivation.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

ExerClass1 - PHY4211 Quantum Mechanics I Fall Term of 2005...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online