supersymmetry report

8/11/2019 Supersymmetry Report

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Supersymmetry in Quantum Mechanics

4/22/2014

Jayant Krishan

12232

M.Sc Physics(F)

Mentor: Prof. Debajyoti Choudhury


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Table of Contents

Acknowledgements ............................................................................................................................. 3

Introduction ........................................................................................................................................ 4

Formalism ........................................................................................................................................... 5

Hierarchy of Hamiltonians. ................................................................................................................. 9

Examples: .......................................................................................................................................... 11

1 . Particle in a box .......................................................................................................................... 12

2. Harmonic oscillator ..................................................................................................................... 17

3. w = a x 3....................................................................................................................................... 19

4 . Spherically symmetric case ( Hydrogen atom ) .......................................................................... 21

Supersymmetry - scattering potentials............................................................................................. 24

1. Particle in a box. ........................................................................................................................ 26

2. w = A tanh x............................................................................................................................ 27

3. w = A ............................................................................................................................... 28Broken Supersymmetry .................................................................................................................... 29

1. w = a x2n.................................................................................................................................... 30

2. w = a xn..................................................................................................................................... 33

3. w = a x-n..................................................................................................................................... 36

4. w = a cos x............................................................................................................................... 385. Delta function............................................................................................................................ 41

Shape invariance ........................................................................................................................... 45

Supersymmetry, shape invariance and the polynomials. ................................................................. 48

Bibliography ...................................................................................................................................... 49


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Acknowledgements

I am thankful to Prof. Debajyoti Choudhury and Prof. T.R Seshadri for their guidance and immense

support.


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Introduction

In particle physics ,we say that by some transformation a boson can go to a fermion which

will be partner of the former and vice versa. This symmetry is known as Supersymmetry. If

this is true then each of these partners should have same quantum numbers except spin

which would differ by half integer. But this would lead to degeneracy between energy levels

of fermions and bosons which is not observed experimentally. I am not going to talk about

particle physics.

This idea of supersymmetry was extended to solve potentials analytically and determine

various solvable potentials. It also helps to relate and solve certain potentials exactly using

shape invariance and idea of supersymmetric partner potential for bounded and unbounded

systems.

Bound systems have discrete energy spectra and unbounded system have continuous

energy spectra. Using SUSY, we can determine S matrix for the partner potentials and one to

one correspondence between discrete energy spectra of the partner potentials.


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Formalism

To solve for any quantum mechanical system, we solve Schrodinger equation for the

potential corresponding to system and evaluate eigen values and eigen functions.

Schrodinger equation is a second order differential equation and it is not easy to solve it for

every potential. So, we reduce this second order differential equation to first orderdifferential equations using factorization. And try to solve for various potentials using the

first order differential equations.

= E

H vx

vx E (2)For ground state , if E = 0

(x) = v(x) (x) (3) v = (4)

H1 = A+A (5)

Where,

(6)+ (7)

+

=

(8)From 1 and 8,

v wwx (9)Similarly,

For


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AA (10)AA+ wx wx

wx wx (11)v wwx (12)

NOTE: Eigen values of both H1 and H2are positive.

En 0 for n 0

|

+|

|

||A||

So norm is a positive definite and hence energies are positive.

A and A are ladder operators.A|n |n-1

A|n |n1A = 0 (13) is the ground state wave function.

v and v are partner potentials.

Let n(1) and n(2) be eigen function corresponding to H1and H2respectively.H1

n(1) =

+

n(1) = En(1)

n(1) (14)

+ n(1) = En(1) (n(1)) H n(1) ) = En(1) (n(1)) (15)Hm(2)= +m(2) = Em(2) m(2) (16)

++m(2)) = Em(2) (+m(2) )H+m(2)) = Em(2) (+m(2) ) (17)

From (15) and (16)


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m(2)= A n(1)n(2)= N A n+1(1) (18)

where N is some normalization constant.

En(2) = En+1(1) (19)

From (14) and (17)

n+1(1) = +n(2) (20)En+1(1) = En(2) (21)

E(1) = 0 (22)

To determine normalization constant N. Let us assume we start with normalised eigen

functions corresponding to H1.

So,

dx 1N*N dx 1|N|2 A A dx 1

|N|2 H dx 1|N|2 En(1) dx 1

|N|2 = ( En(1) )-1N = ( E

)-1

n(2)= ( E )-1A n+1(1)From 18, 19, 20 ,21 ,22 we conclude:

Energy spectra of H1 and H2 is degenerate except that H1has an extra state ie, there

is no eigen state in H2 corresponding to ground state of H1. If the original

Hamiltonian had k bound states, partner Hamiltonian has k-1 bound states.

A converts eigen function of H1 to H2 and destroys a node in the wavefunction.

A converts eigen function of H2 to H1 and creates a node in the wavefunction.


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As ground state of H1annihilated by A, so it has no SUSY partner.

So, given a wave function corresponding to any one of the two Hamiltonians, we can

determine the spectra for the other Hamiltonian. ( Except for ground state of H1 from H2 )

This degeneracy in spectra can be visualized from SUSY algebra. ie,

[ H , Q ] = 0 [ H , Q] = 0If we have an exactly solvable potential with at least one bound state, we can find its

corresponding partner potential which too will be exactly solvable.( Will have identical

energy spectra ).

A= 0 wx 0 (23) wx wx (24)

w dx

ln

w dx

e (25)


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Hierarchy of Hamiltonians.

Given a ground state wave function corresponding to a potential, we determined a

superpotential, the two partner potentials and their corresponding Hamiltonian. We started

with zero ground state energy for SUSY to be unbroken. Now we argue that if we shift

energies and make ground state energy corresponding to the second Hamiltonian zero, wecan further determine the superpotential and the partner potential vcorresponding topotential v. We can do some transformation and make the ground state energies zero andkeep on finding partner potentials, relate the corresponding Hamiltonian till we exhaust all

the bound states. We can determine the complete spectra of the given potential given its

ground state wave function.

Let E(1) be ground state energy of Hand (1)be its ground state eigen function.

w

v = w w+ E(1)A = w A= wH = AA + E(1)

= vxH = AA + E(1)

= v x)v

=

w

w

+ E

(1)

= v(x) + w= v(x)

En(2) = En+1(1)

n(2)= (En+1(1) - E(1))-An+1(1)ground state energy of H

is:

E0(2) = E1(1)


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Now , we can construct Hamiltonian Hby writing Has:H= AA + E1(1)

Where

A = w A w w

We do scaling and make ground state energy of Hzero and determine the partnerHamiltonian Hto H.Therefore,

v= w w+ E(1)H= AA E1(1)

=

vx)

vx) = w w+ E(1)= v(x) + 2 w= v(x) = v(x) = v(x) 2

En(3)= En+1(2)= En+2(1)

n(3)= (En+1(2) - E(2))-An+1(2)= (En+2(1) - E1(1))-(En+2(1) - E(1))-AAn+2(1)

We can keep determining the next partner Hamiltonian till all the bound states are

exhausted. So energy spectra of the partner Hamiltonians can be expressed in terms of

solutions of original Hamiltonian. If there are k bound states, we can find k-1 partner


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Hamiltonians.the energy spectra of the ith Hamiltonian will be same as original one with i-1

less eigen states.

Hm= AmAm + Em-1(1)Am = wAm= w

wx En(m) = En+1(m-1)= En+2(m-2 )= En+m-1(1)

n(m)= (En+m-1(1) - Em-2(1))-....(En+m-1(1) - E0(1))- Am-1.....An+m-1(1))vx = v(x) 2 .

Energy spectra of a large number of potentials can be determined using this technique but

we need to ensure that the ground state energies are zero for SUSY to be unbroken. To

determine the actual energy, we need to add the shift in energy we introduced.

Now, let us try to solve some potentials using SUSY.Examples:


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1 . Particle in a box .

V = 0 0 < x < L

= - x 0 , x LSchrodinger eqn:

vx EFor 0 < x < L

E

ELet

E k k = 0

Solution of this eqn is given by :

A cos kx Bsin kxA and B are constants determined usind boundary conditions.

Boundary conditions:

x 0 x L 0At x=0

A = 0

At x=L,

0 = B sin kL

But B 0

sin kL = 0 kL = n

k =

E k


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E E

x N sin n = 1,2,3,4,N is some normalization constant .

dx 1 s i n nxL sin nxL

1

|| 1cos = 1 N =

E For a Hamiltonian to be factorizable, it should be defined so as to have zero ground state

energy. Thus,

H1 = HE0

En(1) =

1 n = 1,2,3,.........Here n is starting from 1 so n=1 corresponds to ground state. Generally we take n=0 for

ground state so let

n n-1now normalized energy eigen values and eigen functions are given by:

En(1) = 2 n = 0,1,2,3,....... sin1 for 0 x L

1 cos1


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sin

cos wx

wx cot Now ,

v

w

x

w

cot cosec 1 cosec cosec

v wx w cot cosec 1 cosec cosec 1 2 cosec

Though these two partner potentials differ in shape but from SUSY QM , both have the same

energy spectra except that v2has one less energy eigen state as compared to v1.

The wave functions of the two potentials are related as:

n(2) = ( E )-1A n+1(1) sin1 sin

2 sin


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n(2) = E n+1(1)0(2) = cot 2 sin

2 sin this will be the ground state corresponding to potential v2.

En(2) = En+1(1)

E0(2) =

So the energy spectra is just shifted by one energy level.

8 ____________ ____________ ____________ ____________ ____________

0 ____________

E E E(After scaling of ground state of original Hamiltonian H

)

To obtain higher order potentials, we can start with vand as my initial potential andground state wave function respectively, do scaling so as to make energy corresponding to

this potential equal to zero.

H AAH AA

2

sin

2 sin cos Now, we can get the second superpotential partner:

w

w


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cot v wx w

cot cosec 2 cot cosec n(3) = ( E E )-1A n+1(2)0(3) sin

we can get w , vconsidering 0(3) as new ground state and keep repeating the processtill exhaust all the bound states to construct the whole family of solvable potentials.

v m 1 cot m 2cosec 0(m) sin H AAH+ AA

The ground state energy of the each next partner potential will be greater than of theprevious one. The ground state wave function of each partner potential will be

nodeless.

Fig : Two partner potentials have different shape but their energy spectra is same except that there

is no eigen state corresponding to Hin H. Note : ground state of both potentials is nodeless.Figtaken from Supersymmetry in Quantum Mechanics, Cooper, Khare, Sukhatme)

v cosecxv 0


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2. Harmonic oscillator

m x

v m xv k xwhere,

k m Schrdinger equation :

vx E

m x E k x E

E n E H1 = HE0

En(1) = n n

Ground state wave function is given by:

/ e

/

e

wx

wx

v

w

x

w


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mx m

v wx w

mx m

v v mvand v are same in shape but one is shifted wrt other. So both these potentials havesame energy spectra.

n(2) = ( E )-1A n+1(1)1(1) = / xe E1(1) =

0(2) =

/

xe

= / xe = / e x

E0(2) =

____________ ____________ ____________ ____________ ____________ ____________

E E E


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3. w = a x 3

w = 3 a x2

w2 = a2x6

v1 = a2x6- 3 a x2

v2 = a2x6+ 3 a x2

The potential corresponding to v will be equal to v1uptosome additive constant, so we can

determine the ground state energy upto some additive constant by solving Schrdingerequation for ground state energy.

Schrodinger eqn:

vx E

vx Evx Edx vx a2x6- 3 a x2 + constant

ax6 3 a x constant E ax6 3 a x constant E ax6 3 a x constant E

Let 1

E 2axconstantFor a = 1

v1 = x6- 3 x2


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v x6 + 3 x2

The potential v2is a single well potential and v1is a double well potential , still both have

same energy spectra though their shape entirely different.

So if we can get solution of v2we can easily find out the energy eigen functions and eigenvalues of the potential v1.


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4 . Spherically symmetric case ( Hydrogen atom )

V(r) = -

Schrodinger equation:

vr E (4.1)

r sin (4.2)

RrY, (4.3)Due to spherical symmetry, we can separate radial and angular part .

Substitute values from (4.2) and (4.3) in (4.1)

r sin - R(r)Y(,ER(r)Y(,

r

sin

R(r)Y(,0

by R(r)Y(, r sin =0

r sin Since LHS is a function of R only and RHS function of (

,

), must be equal to some constant.

1R r r Rrr 2mr 1 l l 1 Let us do a transformation to reduce this equation further.

R

r 1

ru

r ur

r


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r Rr r ur ur

r R

r

r u

r ur

r ur r r Rrr 2mr

1 ll 1 R r 0

r urr 2mr 1 ll 1 ur 0

2m

urr 1 ll 1 ur E ur

This is identical to 1-D Schrdinger equation with potential:

1r ll 1r states having same l value are degenerate.

A wr A+ wrLet

1w- w = (r) (upto additive constant)let us choose :

w =+ +

w l 1r 14l 1 1rw =

+ w- w =

+ constantA

r l 1

r 1

2l 1


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A+ r l 1r 12l 1Hl(1) =+ = + + + +Hl(1) = + +Hl+1(1) = ++ +Hl(2) =+ = + + + +

= ++ +Hl(2) = Hl+1(1)+ + +

So eigen values of Hl(1)and Hl(2) are related.

___________________ ___________________

___________________ ___________________En(1) En(2)

There is also a shift in the energy levels of the two partner potentials.


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Supersymmetry - scattering potentials.

Potentials having no bound states have continuous energy spectra and finite reflection and

transmission coefficients.These coefficients for the partner potentials can be related using

supersymmetry.

For scattering, potentials should be finite and corresponding wave function also non zero as

x.Let us define superpotential w such that

W (x) = wLet us consider an incoming plane wave of energy E coming from -. v1and v2 be thepartner potentials.

Reflected and transmitted waves corresponding to v1 are given by :

R1(k) T1(k)

respectively.

Reflected and transmitted waves corresponding to v2 are given by :

R2(k)

T2(k) respectively. Now,

(k,x -) + R1(k) (k,x ) T1(k) (k,x -) + R2(k) (k,x ) T2(k)

The wave functions corresponding to partner potentials are given by:

n+1(1) = +n(2)+ R1(k) = N ( Rk )

Let 1+ R1(k) N[ Rk ]


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Equating coefficients of powers of exponentials,

N =+

R1(k) = + + RkT1(k) = N w+T2(k) T1(k) =

+ + T2(k)

Remarks:

|R1|2= |R2|2|T1|2= |T2|2

All partner potentials have same reflection and transmission probabilities.

If w w+T1(k) = T2(k)

ie the transmission coefficients of partner potentials are identical.

If w 0R1(k) = -

Rk

Let us solve some examples and try to extract information about nature of potentials

and their corresponding partner potentials.


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1. Particle in a box.

V = 0 0 < x < L

=

-

x 0 , x L

wx cot v= w -

w

= -

v w - w= 1 2 cosec

Since v1is a constant potential, so particle will be free and hence we can conclude that

potential vis also reflectionless.


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2. w = A tanh x

w = A tanh xw = A sechxv= w - w= A tanh x - A sechx= A - A ( A +

)sechxv w - w

= A - A ( A -

)sech

x

For A =

v = A=

Since

v is a constant potential, therefore particle under its influence is free.It is

not easy to solve for V1and predict its properties,but using SUSY we, can

conclude that corresponding partner potential :

v = 12sech x

is also reflectionless and will have one bound state.

__________________ No bound states.

En(1) En(2)


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3. w = Aw = Aw= Av = A-

A

v= A+ A

w= W (x-) = 0R1(k) =

+ + Rk R1(k) = -Rkie reflection coefficient of one partner is opposite to that of other. There will be a phase

difference of between the waves reflected by the partner potentials.


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Broken Supersymmetry

Given a ground state wave function corresponding to a potential, we can factorize the

Hamiltonian and find the SUSY partner Hamiltonian corresponding to partner potential.

If we start with some superpotential, we can determine its ground state using :

Ne Ne

Convention: if there exists a normalizable zero energy ground stateeigen function, it will

correspond to H. So choose w such that:

w > 0 for x >> 0

w < 0 for x


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1.

w = a x2n n is some integer

a is some constant which can be positive or negative.

w = a x2n (some even function)

w= 2n a x 2n-1 (some odd function)

w = a x4n (some even function)

Ne Ne

At x 0At x -

So, is not normalizable.Similarly,

Ne Ne +

At x - 0

At x Hence, is also not normalizable.

|0 (x) = xx

v= w - w


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v =a x4n -2n a x 2n-1

v= w + wv= a x

4n + 2n a x 2n-1Let n=1, a=1

v= x4 -x

v= x4+ x

v(x) = v(-x)

v(x)

x

v(x)


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The two partner potentials are mirror images of each other, hence have exactly same

energy spectra and SUSY is broken.

En(1)= En(2)

n(2)= [ En(1)](-)A n(1)

n(1)= [ En(2)](-)An(2)

since wave functions are not normalizable, there are continuous energy eigen states.

Reflection and transmission coefficients due to scattering are given as :

R1(k) = + + Rk

T1(k) =+ +

T2(k)

If the wavefunction limited to x 0Potential vwill have one extra bound state as compared to v2 andrest of the spectra forboth partner potentials is same. By confining ourselves in the limit x > 0, one of the bound

states of v2is reduced. The ground state of the original Hamiltonian is nodeless but ground

state of the partner potential will have one node.

3(1) ________________ ________________ n(2)

2(1) ________________ ________________ n(2)

1(1) ________________ ________________ n(2)

0(1) ________________

En(1) En(2)


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Therefore is not normalizable.

Ground state wavefunction corresponding to v

is zero while that of v is non zero. 0 SUSY is unbroken. Eigen states and eigen values of partner potentials are related as:

En+1(1)= En(2)

n(2) = [ En(1)](-)A n+1(1)

n+1(1)= [ En(2)](-)An(2)

Let a=1, n=3

v= x6 -x2

v x6 + x2


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Though the partner potentials have different shape but the energy spectra is same

except that the ground state of one potential has an extra eigen state ie there is no eigen

state in energy spectra of v corresponding to ground state of v.We can relate the double well and a single well potentials whenever,

v, a x2n n a x n-1n can be any odd integer


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3. w = a x-n n is some odd integer

A is some constant which can be positive or negative.

w = a x-n (some odd function)

w= -n a xn-1 (some even function)

w = a x-2n (some even function)

v= w -w

v =a x-2n +n a x -n-1

v= w + wv= a x-2n - n a x -n-1

Ne Ne

At x

1At x -

1So, is not normalizable.Similarly,

Ne Ne

At x - 1

At x


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4. w = a cos x

w = a cos xw = -a sin xv= w - w

= a cos x a sin x

v

= w +

w

= a cos x - a sin x

Ne

Ne


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at x 0

w

0

Ne Ne

At x 0

Since potential is finite at and wave functions are not normalizable, but herenormalizability is not a problem.

We will get scattering states.

|0 (x) = xxAs the potentials are periodic with period 2,

V(x) = v(x + 2)

The solution can be written in the form:

( x + a ) = e2iq(x)

The ground state wavefunction are nodeless and has to be periodic with same period as that

of the potential ie 2. SUSY predicts that the partner Hamiltonian wavefunction will too

have the same period. We can normalize the ground state wave functions for a finite

interval.

Let

1x Ne x Ne Ne x

For these to be acceptable solutions, they should satisfy above periodicity conditions.

x 2 Ne e

acosxdx+ 0


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Therefore,

x 2 Ne xSimilarly,

x 2 Ne xNow let us try to find a partner Hamiltonian to H2.

w

wx

= - w(x)

vx ) = w w= w w= v(x)

So v is partner potential to v and vice versa.vx v -x)

Since the partner potentials are mirror images of each other, In bound states, partner

potentials have identical energy spectra. Both are isospectral.

En(1)= En(2)

n(2)= [ En(1)](-)A n(1)

n(1)= [ En(2)](-)An(2)

NOTE: This formalism valid iff wxdx+ 0


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5 Delta function

v = - a (x) = x 0 0 elseThough delta function tends to infinity, but the integra of delta function has finite

constant value.

II

vx E

5.1

a x E In the region to Region II 5.2 E Else (region I and III) 5.3

For region I and III

E 0

k 5.4For region II v E 0k

Solutions of the eqn will be:

= C ekx + D e-kx Region I

IIII

x

-


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= E ekx + F e-kx Region IIIfor x0 = F e-kx 5.6Integrate equation 5.2

a x xdx Edx a 0 Edx

The term on RHS tends to zero as it is area of infinitesimal region. Second term on theleft is some constant term, which is equal to k. first term is the difference in slope of the

wavefunction at the boundaries.

The wavefunction should be continuous at the boundaries but the derivative is

discontinuous. So it reduces to :

- a 0k =

5.7

from 5.4 and 5.7

E - Since this energy is n independent, there is only one bound state in the system.

Ground state wave function will be:

(x) = N e || for x < 0,

(x) = N e || w-(x)=

||

||


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Derivative of w in the limit x0+will be delta peak.v= a x

x

So the potential has one bound state and the partner potential has no bound state.


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Shape invariance

Assuming SUSY to be unbroken, potentials can be solved analytically if the partner

potentials have same shape and only differ by parameters. ie,

Partner potentials are said to be shape invariant if:

vx,a vx,a Raa are set of parameters and a are functions of a a fa

and the remainder Ra is independent of x. If we know the energy eigen values andeigen functions of one of the potentials, we can determine the energy spectra of the

other potential if SUSY is unbroken. If SUSY broken, both the partner potentials have

same energy spectra. Shape invariance and SUSY can also be used to determine the

energy spectra for a given potential. Say if H has k bound states, we can construct k-1Hamiltonians having same energy spectra except first m-1 states will be missing in mthHamiltonian. Any scaling in the parameters will leave this conditions invariant as it canbe absorbed in fa a. We start with

Ea 0 x , a = e

H vx,aH vx,a vx,a Ra

We can shift H by Rato get zero ground state energy as it is condition forfactorization of Hamiltonian. Similarly we can write,

H

vx,a vx,a Raa fa

H vx,a Ra RaWe can generalize this:

H vx, a Ra=


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H x , a , vx,a x , a Ra x ,a Ra x ,a

Since ground state energy of H is zero , vx,ax , a Hx ,a 0Therefore,

E RaSo, we can express the ground state energy of H as the remainder. This is also the firstexcited state of H. If we shift the energy and make ground state of H zero, we can find

partner potential H whose ground state energy will be Ra. We can reduce theground state of each partner potential in terms of constant remainder terms....Generalized equations will be:

H x ,a v x , a Ra= x , a Ra= x , a

E(m) = Ra= This is the ground state energy of the m

th

Hamiltonian or the (m-1)th

energy state of thefirst Hamiltonian. This gives the energy spectra of first Hamiltonian and (m-1) partnerpotentials Hamiltonians with one state less in each successive partner Hamiltonian.

E E(m) = Ra= E 0

Similarly, we can relate the wave functions of the partner Hamiltonians.

x , a E-

A x , a x , a

n(2)= (En+1(1) )-An+1(1) x , a En-1-A x , a n-1 x


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Shifted Harmonic oscillator

w = x b

w

=

now we can find two partner potentials using the superpotential.

v x b - =

x - v x b +

= x + H vx,H vx,v v R

x + = x

- R R

Therefore, this remainder is the ground state energy of partner Hamiltonian or firstexcited state of H.E E R f

We can find all the partner Hamiltonians and determine the complete energy spectraand all the ground state of partner potential and excited state of original potential can

be expressed in terms of remainder terms. Therefore,

E E(n) = R= n Ne Ne Ne


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Supersymmetry, shape invariance and the polynomials.

The solution to harmonic oscillator is given by hermite polynomials. Harmonic

oscillator has shape invariance symmetry, so we can use this formalism to study is there

any shape invariance symmetry of these polynomials responsible for the former.

We can again factorize second order differential equations and extend this technique to

study them.


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Bibliography

Supersymmetry in Quantum Mechanics. by Cooper, Khare, Sukhatme .

Supersymmetry,shape invariance and the Legendre equations. byD. Bazeia,Ashok Das
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