Operators in Quantum Mechanics. Angular momentum

One axiom of formal quantum mechanics is, that

Important 5.1.3. In quantum mechanics every physical quantity is represented by a linear, self-adjoint (or Hermitian⁶) operator, which acts on complex valued functions.

This operator can be found using the definition of the corresponding quantity from clas-sical mechanics and substituting all of the clasclas-sical phyclas-sical quantities in it with the corresponding operators. As all classical mechanical quantities can be expressed with a combination or as a function of the momentumpand position vectorr, we can determine their operators if we know the operators ˆr and p.ˆ

This means that we have some freedom in selecting operators for the momentum and position, but the operators of other physical quantities must be calculated using these.

We can use the stationary Schr¨odinger equation to determine suitable operators for ˆ

p and ˆx:

Take the classical mechanical formula for the total energy of a particle and replace E, p and V(r) with operators ˆE, pˆ and Vˆ respectively. This will result in an operator equation. Apply both sides of this equation to the wave function ϕ(x) then compare the resulting formula with the stationary Schr¨odinger equation to determine a possible representation of these operators. In one dimension:

E = p²

2m +V(x) classical mechanics Eˆ= pˆ²

2m + ˆV(x) operator equation in quantum mechanics Eˆϕ(x) = pˆ²

2mϕ(x) + ˆV(x)ϕ(x) applied to the wave function E =− ~²

2m

∂²ϕ

∂ x² +V(x)ϕ the stationary Schr¨odinger equation

From this (using the rules of the sum and product of operators) we obtain the definitions of the operators ˆp and ˆx:

6For the meaning of the terms “self-adjoint” and “Hermitian” see (5.1.21).

Using these definitions it is easy to see (c.f. end of previous section) that the operators of p and x do not commute, and

ˆ

xˆp−pˆˆx=i~ (5.1.8)

To characterize the commutativity of operators we introduce a notation:

Important 5.1.4. For any two quantum mechanical operators the quantity

[ ˆO,Pˆ] := ˆOPˆ−PˆOˆ (5.1.9) is called the commutator of Oˆ and Pˆ.

If the two operators commute [ ˆO,Pˆ] = 0.

Commutators are useful. In Appendix 22.6 for instance we used only commutators to derive the possible energies of the linear harmonic oscillator.

The commutator of the position and momentum in 1D is

[ˆx,p] =ˆ i~(6= 0) (5.1.10)

and we know there is an uncertainty relation betweenxandp. This is a general principle.

Important 5.1.5. If the commutator of two operators is not zero then there exists an uncertainty relation between the corresponding physical quantities.

In 3 dimensions we must define operators for all three components of the momentum and position vectors. We have to use partial derivatives in this case. These operators may also be combined into a vector operator

ˆ In this case the operator of pis a constant multiplied symbolic vector, for which there is a special notation, the ∇ symbol, called nabla ordel:

∇ ≡ Although this is not a real vector in many cases we may use it as one. For instance the square of it, which is called the Laplace operator and denoted by ∆ is:

∆ := ∇² ≡ ∇ · ∇=

Similarly the position operator r is a symbolic vector:

ˆr:= (ˆx,y,ˆ z) = (x·, y·, z·)ˆ (5.1.15) The commutators of the components of ˆp, and ˆrthen can be written in a single formula:

Important 5.1.6.

[xk, pl] =i~δkl where k, l= 1,2,3 and e.g. x2 ≡y, p2 ≡py (5.1.16) i.e. different components of the position and the momentum commute, but the same components do not, therefore there is no uncertainty relation between different compo-nents, only between the same components of ˆr and ˆp.

The combination of the operators ˆp and ˆV in the 1D and 3 dimensions Schr¨odinger equation are operators themselves. They are called the (1D and 3 dimensions) Hamilton operator or Hamiltonian of the system:

Hˆ :=− ~²

Therefore the one and three dimensional stationary Schr¨odinger equations both can be written as an operator equation:

H ϕ(x) =ˆ Eϕ(x) and (5.1.18)

H ϕ(r) =ˆ Eϕ(r)

When dealing with the quantum mechanical problem of an electron in a centrally sym-metric potential e.g. in the hydrogen atom it will be much easier to solve the Schr¨odinger equation in an (r, θ, φ) spherical coordinate system than in a Cartesian one. In this case the operator form of the equation will not change, although the formula for the Hamil-tonian will change significantly (see Appendix 22.9):

Hˆ =− ~²

Important 5.1.7. If we use the operator form of an equation it will remain the same independent of the coordinate system, only the representation of the operator will change.

This is another important and very useful property of the operators.

For the use in quantum physics it is important to know the behavior of operators in a scalar product.

Let ˆO an operator that acts on wave functions. We can calculate the scalar product of a function ϕ₂ with ˆO ϕ₁. In 1D:

Let us introduce a new operator ˆO^†, called the adjoint of ˆO with the definition:

hOˆ^†ϕ2|ϕ1i=hϕ2|O ϕˆ 1i (5.1.21a)

Some operators are self-adjoint, which means that they are equal to their adjoint:

Oˆ^† = ˆO. Self-adjoint operators are also called Hermitian⁷.

Example 5.2. Determine the adjoint of the operatorsp,ˆ xˆandH!ˆ Solution a) adjoint of the momentum operator

According to the definition of the adjoint operator:

hˆp^†ϕ₂|ϕ₁i=hϕ₂|p ϕˆ ₁i i.e.

The right hand side can be calculated with integration by parts:

∞

7For the sake of completeness: all self-adjoint operators are Hermitian, but not all Hermitian opera-tors are self-adjoint, but for our purposes the two notions are equivalent.

Because both ϕ₁ and ϕ₂ are physical wave functions they must be square integrable, therefore they must vanish when x→ ∞, so the first term is zero

∞

b) adjoint of the position operator

This is much simpler, becausexˆ≡x·is a multiplication with a real number (or vector in 3 dimensions) and it commutes with the wave functions, therefore

ˆ x^†= ˆx The position operator is self adjoint too.

c) the Hamiltonian

The Hamiltonian is a linear combination of the operators pˆ² =−~^{2 d}

2

dx² and V(x) = V(x)·. It is easy to prove that the product and sum of self-adjoint operators is also a self-adjoint operator.

Because the operator of the potential is a multiplication with a function it is self-adjoint, and pˆ² = ˆppˆis a product of the self-adjoint pˆwith itself, the Hamiltonian is also self-adjoint:

Hˆ^†= ˆH