# ACFrOgA4n7gOv1 V9w7kGkF9v ySgf6KOAH-u9sCvrBX813-ngarS3jeRs86oPzJX2pIc7hU9z7JLlWfJrNoMVieVSYGjrtKAFPF2Hh1X3YxkGw2IK-kUni TwjGKKJVKvQT5m2xtiYiOkdKB bZ

```Independent University Bangladesh (IUB)
Assignment 02
Course title: Quantum Mechanics 1
Course code: PHY304
Instructor: Dr. Jewel Kumar Ghosh
Problem 01 (5 + 5 = 10 points)
Consider the following Hamiltonian:
Ĥ1 = −
~2 d2
+ V (x).
2m dx2
(1)
Suppose ψ0 (x) is the ground state wave function of this Hamiltonian with ground state energy
0. That means:
~2 d2 ψ0 (x)
Ĥ1 ψ0 (x) = −
+ V (x)ψ0 (x) = 0.
(2)
2m dx2
Now let us write the Hamiltonian as:
Ĥ1 = Â† Â
(3)
where
~ d
~ d
+ W (x), Â = √
+ W (x).
Â† = − √
2m dx
2m dx
W (x) is called the superpotential in the literature.
a) By equating two different expressions of the Hamiltonian:
~ d
~ d
†
√
Ĥ1 ψ0 (x) = Â Âψ0 (x) = − √
+ W (x)
+ W (x) ψ0 (x)
2m dx
2m dx
~2 d2 ψ0 (x)
+ V (x)ψ0 (x),
=−
2m dx2
find the relation between the potential V (x) and the superpotential W (x).
b) Argue that for
~ d
~ d
√
Ĥ1 ψ0 (x) == − √
+ W (x)
+ W (x) ψ0 (x) = 0
2m dx
2m dx
(4)
(5)
(6)
the superpotential is given by:
~ ψ00 (x)
W (x) = − √
.
2m ψ0 (x)
The ground state wave function of an infinite potential well of width a is given by:
r
π 2
Φn (x) =
sin
x where n = 1, 2, 3, &middot; &middot; &middot;
a
a
with the ground state energy E0 =
π 2 ~2
.
2ma2
(7)
(8)
Write down the expression of the superpotential W (x).
1
Problem 02 (5 + 5 + 5 = 15 points)
a) For three operators Â, B̂, Ĉ, show the following:
h
i
h
i h
i
ÂB̂, Ĉ = Â B̂, Ĉ + Â, Ĉ B̂.
(9)
b) For a function f (x) show that
[f (x), p̂] = i~
df
.
dx
c) For simple harmonic oscillator show that
h
i
Ĥ, â&plusmn; = &plusmn;~ωâ&plusmn; .
2
(10)
(11)
```