Chem 373- Harmonic oscillator...Quantum mechanically

8/3/2019 Chem 373- Harmonic oscillator...Quantum mechanically

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Harmonic oscillator...Quantum mechanically

We shall now turn to a quantum

mechnical treatment of the onedimensional harmonic oscillator

We have

H = Ekin + Epot

Hh

2

2m

d2

dx2

1

2

kx2

Epot1

2kx2

V(x)1

2kx2

Mass

Displacement

Force constant


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The parabolic potential energy V =

1

/2kx2

a harmonic oscillator, where x isthe displacement from equilibrium. Thenarrowness of the curve depends onthe force constant k: the larger the

value of k, the narrower the well.



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=1

tcycle

1

2

k

mcycles per

time unitor

k 42 2m

Thus

H = - h2

2md2

dx22 2 2mx2

= -h

2

2m(

d2

dx24 2 2m2

h2

x2 )

= -h

2

2m(

d2

dx22x2 )

where

= 2 mh

We must solveH (x) = E (x)

-h

2

2m(

d2

dx22x2 ) (x) = E (x)

or

d2 (x)

dx22x2 (x) = -

2mE

h2

(x)

d2 (x)

dx2(2mE

h2

2x2) (x) = 0


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d2 (x)

dx2(2mE

h2

2x2) (x) = 0

Let us look at a solution of

the form

(x) = exp(-2

x2 )f(x)

let us further try to obtain

a power expansion of f(x)

of the form

f(x) = cnm=0

n=xn


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We have

d

dx exp 2 x2

f ( x ) x exp 2 x2

f(x)

exp2

x2

f'(x)

We have

d2

dx2

exp

2

x2

f (x) exp

2

x2

f ( x )

2x2 exp2

x2 f ( x ) x exp2

x 2 f'(x)

xexp

2

x2

f' ( x ) exp

2

x2

f " ( x )



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We have

d2

dx2exp

2x

2f (x) exp

2x

2f ( x )

2x2 exp2

x2 f ( x ) x exp2

x 2 f'(x)

xexp 2 x

2

f' ( x ) exp 2 x

2

f " ( x )

Combining terms

" (x ) exp

2

x2

f " (x ) 2 xf '( x) f ( x )

2x

2f ( x )

Substituting into

d2 (x)

dx2

(2mE

h2

2x

2) (x)=0


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Combining terms

" ( x) exp2

x2

f "( x) 2 xf '(x) f ( x )

2x

2f ( x )

Substituting into

d2 (x)

dx2

(2mE

h2

2x

2) (x)=0


We get

exp 2 x2 f " ( x ) 2 xf' (x) f ( x ) 2x2f ( x )

(2mE

h2

2x

2)exp

2x

2f ( x ) 0


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We get

exp2

x2

f "( x) 2 xf '(x ) f ( x )

2x

2f ( x )

(2mE

h2

2x

2)exp

2x

2f ( x ) 0

Eliminating the common factor exp2

x2

f" (x) 2 xf' (x) f(x) 2x2f(x)

(

2mE

h2

2

x

2

)f(x) 0and combine affords

f"(x) 2 xf'(x) (2mE

h

2)f(x) 0

H i ill Q h i ll


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f"(x) 2 xf'(x) (2mE

h2

)f(x) 0

We have

f(x) = cnm=0

n=xn f' (x) = ncn

n=0

n=xn 1 f" (x) = n(n - 1)cn

n=0

n=xn 2

Introducing n = j + 2 or j = n - 2 f"(x) =j=-2

j=

(j 1)(j 2)cj 2x j

or since index name does not matter and j = -2,-1

contributions are zero

f"(x) =n=0

n=

(n 1)(n 2)cn 2xn


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f" (x) 2 xf' (x) (2mE

h2

)f(x) 0


Substituting

f(x) = cnm=0

n=xn ; f' (x) = nc

nn=0

n=xn 1 ; f" (x) = (n + 2)(n +1)c

n+2n=0

n=xn

int o

affords

(n+2)(n+1)cn+2n=0

n=xn 2 x ncn

n=0

n=xn 1 (

2mE

h2

) cnm=0

n=xn 0

or

(n+2)(n+1)cn+2 2 ncn (2mE

h2

)cn n=0

n=xn 0


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(n+2)(n+1)cn+2 2 ncn (2mE

h2

)cn n=0

n=xn 0

This equation is satisfied for any x only if

(n+2)(n+1)cn+2 2 ncn (2mE

h

2)cn 0 for all n

or cn+2

2 n-2mE

h2

(n+ 2)(n+1)

cn

We can see that given a value co and c1 we can determine

the additional constants cn in the power series


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or cn+2

2 n-2mE

h2

(n+ 2)(n+1)

cn

Setting c1 to zero and providing a value for co affords

1(x) exp 2 x

2

f(x) exp 2 x

2

cnx

n

n 0,2,4

n

exp2

x2 c2lx2l

l 0,1,2

l

We note that this function is even in x, with 1(x) 1( x)

H i ill t Q t h i ll


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On the other hand, setting c0 to zero and providing a value

for c1 affords

1(x) exp2

x2 f(x) exp2

x2 cnxn

n 1,3,5

n

exp2

x2 c(2l 1x2 1l

l 0,1,2

l

We note that this function is odd in x, with 1(x) 1( x)

Our general solution has the form

(x) A exp2

x2 c2lx2l

l 0,1,2

lB exp

2x2 c(2l 1x

2 1l

l 0,1,2

l


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(x) A exp2

x2 c2lx2l

l 0,1,2

lB exp

2x2 c(2l 1x

2 1l

l 0,1,2

l

It turns out that this solution is divergent for | x |

if we take the total sum l = 0 to l =

If we , however, can have that cn+2 for a certain value n= vbecome zero cv-2 = 0,than we have a finite series since according tothe general recursion formular

cn+2

2 n - 2mEh

2

( n +2)(n+1)cn

Harmonic oscillator Quantum mechanically


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for n = v + 2

cv+42 v -

2mE

h2

(v + 4)(v + 3)cv 2

or

cv+4; cv+6;cv+8,... all would be zero if cv+2 is zero


How can we make C2v zero if Cv is different from zero ?

cv+2

2 v -2mE

h2

(v +1)(v + 2) cv

For cv+2 to be zero when cv is not we must have

2 v -2mE

h2

(v +1)(v + 2) = 0



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2 v -2mE

h2

(v +1)(v + 2)

= 0 or E =h

2

m

(1

2

v)

v = 1, 3, 5, 5

v = 0, 2, 4,6

Since =2 m

h; where = frequency (cycles per time unit)

or =m

hwhere = 2 angular frequency

E = h (12

v) v = 1, 3, 5, 7 ; v = 0, 2, 4, 6

We note that energy always different from

zero.



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E =h2

m(1

2v) v = 1, 3,5, 7 ; v = 0, 2, 4,6

E = h (1

2v)

(x) A exp2

x2 c2lx2l

l 0,1,2

lB exp

2x2 c(2l 1x

2 1l

l 0,1,2

l

let v be even, then all terms with 2l > v will be zero in the series

cv+2

2 v -2mE

h2

(v + 2)(v +1)cv = 2 v - 2 v

(v + 2)(v +1)cv 0

However all terms in the odd series will still

contribute unless B = 0




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Thus for v even

(x) A exp2

x2 c2lx2ll 0,

lv

2


E = h (12

v) v= 0, 2, 4, 6

We note that this function is even in x, with (x) ( x)



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E = h (1

2v) v = 1,3, 5, 7 ; v = 0,2, 4, 6

E =h

2

m

(1

2

v)

(x) A exp2

x2 c2lx2l

l 0,1,2

lB exp

2x2 c(2l 1x

2 1l

l 0,1,2

l

let v be odd, then all terms with 2l+1 > v will be zero in the series

cv+2

2 v -2mE

h

2

(v + 2)(v +1)cv = 2 v - 2 v

(v + 2)(v +1)cv 0

However all terms in the even series will still

contribute unless A = 0




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Thus for v odd

(x) A exp2

x2 c2 1lx2 1ll 0,

lv 1

2


E = h (12

v) v=1,3,5,7

We note that this function is odd in x, with (x) ( x)

E can never be zero

Chem 373- Harmonic oscillator...Quantum mechanically

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