Ladder Operators Methods - Outline | PHYS 410, Study notes of Physics

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Lecture 10 Outline - The SHO 2

Solvay talk - Schr¨

odinger and Born

stuck in an SHO: ladder operators method [2.3.1]

Ladder operators warmup

Show that

H

a − ψ ) = (

E

ω )(ˆ a − ψ )

General solution

we can now generate the set of stationary states...

ψ n ( x ) =

A

n (ˆ a + ) n ψ 0 ( x )

while

E

n

( n (^) +

2 1 ) ℏ ω

Given the ground state solution and its energy

ψ 0 =

( mω ℏ

) 41 exp

2 mω ℏ

x 2 )

and

E

0

ω

the first excited state is:

remember also that:

Hˆ

a

ψ ) = (

E

ω )(ˆ a

ψ )

Hermitian conjugates

ˆa ±

≡

1

√ 2 ℏ mω

∓^

(^) i ˆp (^) +

(^) mω

ˆx )

ˆa ±

is the Hermitian conjugate of

ˆa ∓ ...

Since for any two functions

f (^) ( x ) , g

( x ) we have

∞

−∞

f (^) ∗ ( x ) (ˆ

a ± g ( x )) (^) dx

∞

−∞

a ∓ f (^) ( x )) ∗ g^ ( x ) dx

disclaimer: here

f (^) ( x ) , g

( x )

must go to zero at

Orthonormality...

We want to show that:

ψ m∗ ψ n dx

δ mn

First we start with:

ψ m∗ ( ˆa

ˆa − ) ψ n dx

n ∫

ψ m∗ ψ n dx

but we can rewrite the above left hand side as

ˆa − ψ m ) ∗ ˆa − ψ n dx

ˆa

ˆa − ψ m ) ∗ ψ n

dx

m

∫

ψ m∗ ψ n dx

thus

m

∫

ψ m∗ ψ n dx

n ∫

ψ m∗ ψ n dx

thus if

m

(^6) =

n

then must have

ψ m∗ ψ n dx

thus if

m

(^) =

n

then must have

ψ m∗ ψ n dx

So all of the SHO states are orthonormal.