Thiscanberightonlyif eachtermisaconstant 2 x1 2

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I
 Two
non‐interac0ng
par0cles
in
a
box
 2 d 2 2 d 2 H =− − ≡ H1 + H 2 2 2 2 m1 dx1 2 m2 dx2 If
the
hamiltonian
is
wriGen
as
a
sum
of
independent
terms
then
the
wavefunc0on
 is
a
product
of
the
independent
components.
i.e.
 ψ ( x1 , x2 ) = φ ( x1 ) χ ( x2 ) ⎡ 2 d 2 2 d 2 ⎤ Hψ = − ⎢ + φ ( x1 ) χ ( x2 ) = Eψ 2 2 m1 dx12 2 m2 dx2 ⎥ ⎣ ⎦ 2 2 − φ '' ( x1 ) χ ( x2 ) − φ ( x1 ) χ '' ( x2 ) = Eφ ( x1 ) χ ( x2 ) 2 m1 2 m2 dividing by ψ =φχ 2 φ '' ( x1 ) 2 χ '' ( x2 ) − − =E 2 m1 φ ( x1 ) 2 m2 χ ( x2 ) Mathema0cal
detour
II
 The
final
equa0on
derived
is
a
sum
of
two
terms
one
depends
only
on
x1
 and
only
on
x2
and
their
sum
is
equal
a
constant.
This
can
be
right
only
if
 each
term
is
a
constant
 2 φ '' ( x1 ) 2 χ '' ( x2 ) − − =E 2 m1 φ ( x1 ) 2 m2 χ ( x2 ) 2 φ '' ( x1 ) 2 χ '' ( x2 ) ⇒− = E1 − = E2 2 m1 φ ( x1 ) 2 m2 χ ( x 2 ) We
therefore
have
two
independent
one
dimensional
par0cle
in
a
 box
problem
to
solve
 2 2 − φ '' ( x1 ) = E1φ ( x1 )1 − χ '' ( x2 ) = E2 χ ( x2 ) 2 m1 2 m2 & E1 + E2 = E Back...
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This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas at Austin.

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