An operator is a symbol which defines the mathematical operation to be cartried out on a function.

Examples of operators:

d/dx = first derivative with respect to x

Ö = take the square root of

3 = multiply by 3

Operations with operators:

If A & B are operators & f is a function, then

(A + B) f = Af + Bf

A = d/dx, B = 3, f = f = x²

(d/dx +3) x² = dx²/dx +3x² = 2x + 3 x²

ABf = A (Bf)

d/dx (3 x²) = 6x

Note that A(Bf) is not necessarily equal to B(Af):

A = d/dx, B = x, f = x²

A (Bf) = d/dx(x× x²) = d/dx (x³) = 3 x²

B (Af) = x (d/dx x²) = 2 x²

In general, d/dx (xf) = f + x df/dx = (1 + x d/dx)f

So d/dx x = 1 + x d/dx

Since A & B are operators rather than numbers, they don’t necessarily commute. If two operators A & B commute, then

AB = BA

and their commutator = 0:

[A,B] = AB -BA = 0

(Numbers always commute: 2× 3 f = 3× 2 f; [2,3] = 0)

What is the commutator of d/dx & x?

[d/dx,x] = ?

Since we have shown that d/dx x = 1 + x d/dx, then

[d/dx,x] = d/dx x - x d/dx = 1

What is the commutator of 3 & d/dx?

[3,d/dx] f = 3 d/dx f - d/dx 3 f = 3 d/dx f - 3 d/dx f = 0 =

[d/dx,3]

Equality of operators: If Af = Bf, then A = B

Associative Law: A(BC) = (AB)C

Square of an operator: Apply the operator twice A² = A A

(d/dx)² = d/dx d/dx = d²/dx²

C = take the complex conjugate; f = e^ix

C f = (e^ix)^* = e^-ix

C²f = C (Cf) = C (e^-ix) = (e^-ix)^* = e^ix = f

If C²f = f, then C² = 1

Linear Operator: A is a linear operator if

A(f + g) = Af + Ag

A(cf) = c (Af)

where f & g are functions & c is a constant.

Examples of linear operators:

d/dx (f + g) = df/dx + dg/dx

3(f + g) = 3f + 3g

Examples of nonlinear operators:

Ö (f + g) is not equal to Ö f + Ö g

inverse (f + g) = 1/(f + g) is not equal to 1/f + 1/g

Cautionary note: When trying to determine the result of operations with operators that include partial derivatives, always using a function as a "place holder". For example, what is (d/dx + x)²?

(d/dx + x)²f = (d/dx + x) (d/dx + x) f

= (d/dx + x) (df/dx + xf)

= d/dx (df/dx + xf) + x (df/dx + xf)

= d²f/dx² + d/dx (xf) + x (df/dx) + x²f

= d²f/dx² + x df/dx + f + x (df/dx) + x²f

= (d²/dx² + 2x d/dx + 1 + x²)f

So (d/dx + x)² = (d²/dx² + 2x d/dx + 1 + x²)

When an operator operating on a function results in a constant times the function, the function is called an eigenfunction of the operator & the constant is called the eigenvalue

A f(x) = k f(x)

f(x) is the eigenfunction & k is the eigenvalue

Example: d/dx(e^2x) = 2 e^2x

e^2x is the eigenfunction; 2 is the eigenvalue

How many different eigenfunctions are there for the operator d/dx?

df(x)/dx = k f(x)

Rearrange the eq. to give: df(x)/f(x) = k dx

and integrate both sides: ò df(x)/f(x) =ò k dx

to give: ln f = kx + C

f = e^kx+C = e^kx e^C = e^kx C’, C’ = e^C

Since there are no restrictions on k, there are an infinite number of eigenfunctions of d/dx of this form.

C’ is an arbitrary constant. Each choice of k leads to a different solution. Each choice of C’ leads to multiples of the same solution.

Any eigenfunction of a linear operator can be multiplied by a constant and still be an eigenfunction of the operator. This means that if f(x) is an eigenfunction of A with eigenvalue k, then cf(x) is also an eigenfunction of A with eigenvalue k. Prove it:

A f(x) = k f(x)

A [cf(x)] = c [Af(x)] = c [kf(x)] = k [cf(x)]

To specify the type of eigenfunction of d/dx more definitively, one can apply a physical constraint on the eigenfunction, as we did with the Particle in a Box:

c e^kx must be finite as x ® +¥

The most general k is a complex number: k = a + ib

Then c e^kx = ce^(a+ib)x = c e^ax e^ibx = c e^ax (cos bx + isin bx)

Since e^ax ® ¥ for x ® +¥ , a must be 0

b can be any number

So c e^ibx is the correct eigenfunction of d/dx.

In the 1-dimensional Schrödinger Eq.

[(-h²/2m) d²/dx² + V(x)] y (x) = E y (x),

y (x) is the eigenfunction, E is the eigenvalue, & the Hamiltonian operator is

(-h²/2m) d²/dx² + V(x)

The Hamiltonian function was originally defined in classical mechanics for systems where the total energy was conserved. This occurs when the potential energy is a function of the coordinates only. this is the type of system to be studied with quantum mechanics.

p_x = m v_x v_x = p_x /m

with m = mass & v_x = velocity in the x-direction

Classical kinetic energy (KE) is defined as

KE_x = (1/2) m v_x² = p_x²/(2m)

The classical Hamiltonian function is defined as the sum of the kinetic energy (a function of momentum) & the potential energy (a function of cordinates)

H = p_x²/(2m) + V(x)

for a 1-dimensional system

Comparison to the Schrödinger Eq. shows that

(-h²/2m) d²/dx² « p_x²/(2m)

Each coordinate operator, q, is replaced by multiplication by the coordinate

operator q = q× q=x,y,z

p_q = (h/i) ¶ /¶ q = -i h ¶ /¶ q, q=x,y,z

So operator x = x× , etc. , p_x = -i h ¶ /¶ x, etc.

Then p_x² = (-i h ¶ /¶ x) ² = (i) ²h^{2 ¶ 2}/¶ x² = - h^{2 ¶ 2}/¶ x²

Potential energy functions are usually functions of the coordinates, such as

V(x) = a x²

In general, the operator V(x) is replaced by multiplication by V(x): V(x) ×

In summary

Classical mechanics (1-dimension)

H = T + V = KE + PE = p_x²/(2m) + V(x)

Quantum mechanics (1-dimension)

H (operator) = T (operator) + V (operator)

= - (h²/2m) d²/dx² + V(x)

H = Hamiltonian energy operator = - (h²/2m) d²/dx² + V(x)

H y _i = E _i y _i i=1,2,.. different states

Measurement of the energy of the system will result in one of the E _i (eigenvalues, observables)

Example: Is Y (x,t) an eigenfunction of the p_x operator for the 1-dimensional particle in a box?

Y (x,t) = e^iEt/h y (x) state function

y (x) = Ö (2/L) sin (np x/L), E_n = n²h²/(8mL²)

p_x = -i h¶ /¶ x

For Y (x,t) to be an eigenfunction of p_x, must have

p_xY (x,t) = c Y (x,t)

But d/dx sin (Ax) = A cos (Ax), so Y (x,t) is not an eigenfunction of p_x

Example: Is Y (x,t) an eigenfunction of the p_x² operator for the 1-dimensional particle in a box?

p_x^{2 Y} (x,t) = - h²(d²/dx²) { e^{iEt/h Ö} (2/L) sin (np x/L)}

= - h² e^{iEt/h Ö} (2/L) (np /L) d/dx cos (np x/L)

= h² e^{iEt/h Ö} (2/L) (np /L) ² sin (np x/L)

= h² (np /L) ²{ e^{iEt/h Ö} (2/L) sin (np x/L)}

= h² (np /L) ^{2 Y} (x,t)

= h² (n²/(4L ²) ^Y (x,t) Yes

Since n=1,2,.., the eigenvalue h² (n²/(4L ²) is quantized.

Find the eigenfunctions of p_x.

p_x g(x) = k g(x)

-ih dg/dx = k g

dg/g = (ik/ h) dx

ln g = (ik/ h)x + C

g = A e^{(ik/ h)x}

To keep g well-behaved as x ® + ¥ , k must be real. So the eigenvalues of p_x are all the real numbers k, - ¥ < k < ¥ .

T = (-h²/2m) (¶ ²/¶ x² + ¶ ²/¶ y² + ¶ ²/¶ z²)

= (-h²/2m) Ñ ² Ñ ² is the Laplacian operator

H y (x,y,z) ={(-h²/2m)Ñ ² + V(x,y,z)}y (x,y,z)=E y (x,y,z)

The probability of finding the particle at time t in a region bounded by (x,y,z) & (x+dx,y+dy,z+dz) is

½ y (x,y,z,t)½ ²dx dy dz dt = dx dy dz

1 = ò _{-¥ ¥ ò -¥ ¥ ò -¥ ¥ ½ y} (x,y,z,t)½ ²dt

Particle i has mass m_i, position (x_i, y_i, z_i) and momentum (p_xi, p_yi, p_zi)

T = (-h²/2m₁) (¶ ²/¶ x₁² + ¶ ²/¶ y₁² + ¶ ²/¶ z₁²) +

(-h²/2m₂) (¶ ²/¶ x₂² + ¶ ²/¶ y₂² + ¶ ²/¶ z₂²) + ... +

(-h²/2m_n) (¶ ²/¶ x_n² + ¶ ²/¶ y_n² + ¶ ²/¶ z_n²)

= S (-h²/2m_i) Ñ _i²

If V depends only on the Cartesian coordinates,

V = V (x₁,y₁,z ₁, ..., x_n,y_n,z_n)

Then y = y (x₁,y₁,z₁, ..., x_n,y_n,z_n) and

H y = {S (-h²/2m_i) Ñ _i² + V (x₁, ...,z_n)}y = E y

The probability of finding the first particle in a region bounded by (x₁,y₁,z₁) & (x₁+dx₁,y₁+dy₁,z₁+dz₁), the second particle in a region bounded by (x₂,y₂,z₂) & (x₂+dx₂,y₂+dy₂,z₂+dz₂), etc. is

½ y (x₁,y₁,z₁,..., x_n,y_n,z_n, t)½ ²dt

dt = dx₁dy₁dz₁ ... dx_ndy_ndz_n

1 = ò _{-¥ ¥} ...ò _{-¥ ¥ ½ y} ( x₁,y₁,z₁,..., x_n,y_n,z_n, t)½ ²dt

V(x,y,z) = 0 0 < x < a, 0 < y < b, 0 < z < c

y = 0 outside the box, as in the 1-dimensional case

Inside the box: H y = E y

(-h²/2m) (¶ ²/¶ x² + ¶ ²/¶ y² + ¶ ²/¶ z²) = E y

Solve by Method of Separation of Variables: Assume that y is a product of functions, each depending only on one variable. This is a reasonable assumption because the potential has no cross terms (i.e. terms including products of variables)

y (x,y,z) = f(x) g(y) h(z)

H y = (-h²/2m) {g h d²f/dx² + f h d²g/dy² + f g d²h/dz²} =

E f(x) g(y) h(z)

Dividing both sides by f(x) g(y) h(z) gives:

(-h²/2m){ (1/f) d²f/dx² + (1/g) d²g/dy² + (1/h) d²h/dz²} = E

Can rewrite so that the left-hand side depends only on x & the right-hand side depends only on y & z:

(1/f) d²f/dx² = (1/g) d²g/dy² + (1/h) d²h/dz² - 2mE/ h²

But this means that the left & right-hand sides must be equal to a constant.

Let k_x = (1/f) d²f/dx²

Could rewrite the eq. so that the left-hand side depends only on y, etc. and get

k_y = (1/g) d²g/dx² k_z = (1/h) d²h/dx²

with k_x + k_y + k_z = -2mE/ h²

Can redefine the energy components as

k_x = -2mE_x/ h², etc.

So that E_x + E_y + E_z = E

and (1/f) d²f/dx² = -2mE_x/ h², etc.

Then d²f/dx² + 2mE_x/ h² f = 0

d²g/dy² + 2mE_y/ h² g = 0

d²h/dz² + 2mE_z/ h² h = 0

Boundary Conditions: Functions must be zero at the walls.

f(x) = 0 at x = 0, a

g(y) = 0 at y = 0, b

h(z) = 0 at z = 0, c

So the solutions are the same as for the 1-dimensional particle in a box:

f(x) = Ö (2/a) sin (n_xp x/a), E_x = (n_x²h²)/(8ma²), n_x=1,2,...

g(y) = Ö (2/b) sin (n_yp y/b), E_y = (n_y²h²)/(8mb²), n_y=1,2,...

h(z) = Ö (2/c) sin (n_zp z/c), E_z = (n_z²h²)/(8mc²), n_z=1,2,...

E = E_x + E_y + E_z = (h²)/(8m) {n_x²/a² + n_y²/b²+ n_z²/c²}

with the quantum numbers n_x, n_y, n_z varying independently

y (x,y,z) = Ö [8/(abc)]sin (n_xp x/a) sin (n_yp y/b) sin (n_zp z/c)

Normalize y :

1 = ò _{-¥ ¥ ò -¥ ¥ ò -¥ ¥ ½ y} (x,y,z,t)½ ²dt

= ò ₀^adx ½ f(x)½ ^{2 ò} ₀^bdy ½ g(y)½ ^{2 ò} ₀^cdz ½ h(z)½ ²

But each function is separately normalized

1 = ò ₀^adx ½ f(x)½ ², etc.

so y is automatically normalized.

Consider a particle in a cube: a = b = c,

E = (h²)/(8m a²) {n_x² + n_y²+ n_z²}

or {n_x² + n_y²+ n_z²} = (E 8m a²)/ h²

Tabulate

n_x n_y n_z 111 211 121 112 122 221 212

{n_x² + n_y²+ n_z²} 3 6 6 6 9 9 9

n_x n_y n_z 113 131 311 222 etc

{n_x² + n_y²+ n_z²} 11 11 11 12

Degeneracy occurs when two or more independent wavefunctions correspond to states with the same energy eigenvalue

Each set of (n_x n_y n_z) corrsponds to an independent wavefunction. Since there are 3 independent wavefunctions which give {n_x² + n_y²+ n_z²}= 6, the corresponding energy level is said to be 3-fold degenerate.

A rectangular box wouldn’t have degenerate energy levels. Degeneracy is related to the symmetry of the system.

The degree of degeneracy of an energy level equals the number of linearly independent wavefunctions corresponding to that value of the energy.

A set of n functions is said to be linearly independent if no member of the set can be written as a linear combination of the others.

y ₁, y ₂, y ₃, etc are linearly independent if

c_{1y 1} + c_{2y 2} + ... + c_{ny n} = 0 only if c₁ = c₂=...= c_n= 0

Example: f₁ = 3x, f₂ = 5x² - x, f₃ = x²

f₂ = 5 f₃ - f₁/3 not linearly independent

Example: f₁ = 1, f₂ = x, f₃ = x² linearly independent

Prove that for

H f = w f

if f = c_{1 y 1} + c_{2 y 2} +...+ c_{n y n} ,

then H y _i = w y _i for i = 1,...,n

Proof: H f = H (c_{1 y 1} + c_{2 y 2} +...+ c_{n y n})

= c₁ Hy ₁ + c₂ H y ₂ +...+ c_n H y _n

= c_{1 w y 1} + c_{2 w} y ₂ +...+ c_{n w} y _n

= w (c_{1 y 1} + c_{2 y 2} +...+ c_{n y n})

= w f

Note that the degree of degeneracy of energy level w is the number of linearly independent eigenfunctions (n) belonging to that level.

For a quantity that depends on discrete changes in the variables, the average value is defined by a sum

F - the physical property

<F> - average value of F

N - the number of systems that are measured

f - a possible value of F

Example: In a class there are 9 (N=9) students. On a quiz the grades are: 0 (f₁), 20 (f₂), 20 (f₃), 60 (f₄), 60 (f₅), 80 (f₆), 80 (f₇), 80 (f₈), 100 (f₉). There are 5 questions & each question is either all right (20 points) or all wrong (0 points). Calculate the average grade.

<F> = S f_i /N = (1/9) [0 + 20 + 20 + 60 + 60 + 80 + 80 +