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Taking a particle m with box potential (one dimensional) where V(x) = 0 when mod(x) <=a and V(x) = infinity when mod(x) > a and where wave function phi(x) = A (phi1(x) + ph2(x)) where phi1(x) and phi2(x) are normalized wave functions of the ground state and first excited state

How do we determine if the state phi an eigenstate of the hamiltonian.

Is my approach correct.

If the state phi has to be an eigenstate of the hamiltonian it has to satisfy two conditions.

a) H phi = E phi -> schrondinger equation.

b) phi (x) tends to 0 when mod(x) tends to infinity.

If this is correct how can I prove that this satisfies a and b.

if it satisfies condition (a) then

(-h^2/dm d^2/dx^2 + V(x)) phi (x) = E phi(x)

when I insert phi(x) = A (phi1(x) + ph2(x)) where phi1(x) = AcosKx1 + BsinKx1

and phi2(x) = AcosKx2 + BsinKx2

phi(x) = A(cosKx1 + coskx2) + B(sinKx1+sinKx2)

phi(x) = 2A*cos(kx1+kx2)/2*cos(kx1-kx2)/2 + 2B*sin(kx1+kx2)/2*cos (kx1-kx2)/2

How do I proceed further ???

How do I test for condition (b)

Regards.

QP