130_notes.dvi

6.1.2 The Energy Operator

We can deduce and verify (See section 6.6.2) theenergy operatorin the same way.

E(op)=i ̄h

∂

∂t

6.1.3 The Position Operator

What about theposition operator,x(op)? The answer is simply

x(op)=x

when we are working inposition spacewithup 0 (x,t) =√ 21 π ̄hei(p^0 x−E^0 t)/ ̄h(as we have been above).

6.1.4 The Hamiltonian Operator

We can develop other operators using the basic ones. We will use theHamiltonian operator
which, for our purposes, is the sum of the kinetic and potential energies. This is the non-relativistic
case.

H=

p^2

2 m

+V(x)

H(op)=−

̄h^2

2 m

∂^2

∂x^2

+V(x)

Since the potential energy just depends onx, its easy to use. Angular momentum operators will
later be simply computed from position and momentum operators.

6.2 Operators in Momentum Space

If we want to work inmomentum space, we need to look at the states of definite position to find
our operators. The state (in momentum space) with definite positionx 0 is

vx 0 (p) =

1

√

2 π ̄h

e−ipx^0 / ̄h

The operators are

x(op)=i ̄h

∂

∂p