Chapter 11 Wave Mechanics

Home , Eigenfunction, Energy operator, Operator (physics), Position operator

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 11: Wave Mechanics Wave Function in Position & Wavenumber Representation

In quantum mechanics, we associate wavenumber with particle momentum p⃗ = ℏk⃗

Remember the Fourier transform relation for waves? ∞ ∞ , √1 , −ikx , √1 , ikx a(k t) = ∫ Ψ(x t)e dx and Ψ(x t) = ∫ a(k t)e dk 2휋 −∞ 2휋 −∞

Let’s replace k with p in the Fourier expansions...

P. J. Grandinetti Chapter 11: Wave Mechanics Wave Function in Position and Momentum Representation

Wave function in position and momentum representations become

∞ , √1 , −ipx∕ℏ . Φ(p t) = ∫ Ψ(x t) e dx 2휋ℏ −∞

and ∞ , √1 , ipx∕ℏ , Ψ(x t) = ∫ Φ(p t) e dp 2휋ℏ −∞

We’ll use these expressions to predict a particle’s properties from its wave function.

P. J. Grandinetti Chapter 11: Wave Mechanics Operators in Quantum Mechanics

P. J. Grandinetti Chapter 11: Wave Mechanics Position Operators, x̂ Mean particle position, or expectation value of x, is weighted average calculated from wave function ∞ ∞ ⟨ ⟩ ∗ , , x(t) = ∫ x p(x) dx = ∫ x Ψ (x t)Ψ(x t) dx −∞ −∞ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ p(x)

In QM, we deﬁne x̂ operator: x̂ Ψ(x, t) = x Ψ(x, t) x̂ operates on Ψ(x, t) to give back position times wave function. In formalism of quantum mechanics, we write ⟨x(t)⟩ as

∞ ∞ ∞ ⟨ ⟩ ∗ , ̂ , ∗ , , ∗ , , x(t) = ∫ Ψ (x t) x Ψ(x t) dx = ∫ Ψ (x t) x Ψ(x t) dx = ∫ x Ψ (x t)Ψ(x t) dx −∞ −∞ −∞ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ p(x)

P. J. Grandinetti Chapter 11: Wave Mechanics Position Operators, x̂ Other mean or expectation value quantities, such as ⟨x2(t)⟩, are written

∞ ∞ ∞ ⟨ 2 ⟩ ∗ , ̂2 , ∗ , 2 , 2 ∗ , , x (t) = ∫ Ψ (x t) x Ψ(x t) dx = ∫ Ψ (x t) x Ψ(x t) dx = ∫ x Ψ (x t) Ψ(x t) dx −∞ −∞ −∞ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ p(x)

For any function of x we write

∞ ∞ ∞ ⟨ , ⟩ ∗ , ̂ , , ∗ , , , , ∗ , , f (x t) = ∫ Ψ (x t) f (x t) Ψ(x t) dx = ∫ Ψ (x t) f (x t) Ψ(x t) dx = ∫ f (x t) Ψ (x t) Ψ(x t) dx −∞ −∞ −∞ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ p(x)

As long as the operator f̂(x, t) depends only on x and t, then

f̂(x, t)Ψ(x, t) = f (x, t)Ψ(x, t)

What’s the point of introducing these extra steps with x̂?

P. J. Grandinetti Chapter 11: Wave Mechanics Momentum Operators, p̂ What about mean or expectation value of particle momentum? Using wave function in momentum basis Φ(p, t) we calculate ∞ ∞ ∞ ⟨ ⟩ ∗ , ̂ , ∗ , , ∗ , , p(t) = ∫ Φ (p t) p Φ(p t) dp = ∫ Φ (p t) p Φ(p t) dp = ∫ p Φ (p t) Φ(p t) dp −∞ −∞ −∞ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ p(p)

Here p̂ operates on Φ(p, t) to give p̂ Φ(p, t) = pΦ(p, t) Similarly we have ∞ ⟨ 2 ⟩ ∗ , ̂ 2 , p (t) = ∫ Φ (p t) p Φ(p t)dp −∞ and ∞ ⟨ , ⟩ ∗ , ̂ , , f (p t) = ∫ Φ (p t) f (p t) Φ(p t)dp −∞ Again, what’s the point of introducing these extra steps with p̂? Let’s consider this further. P. J. Grandinetti Chapter 11: Wave Mechanics Momentum Operators If p̂ operates on Ψ(x, t) we should get same result as p̂ acting on Φ(p, t) ∞ ⟨ ⟩ ∗ , ̂ , , ̂ , p(t) = ∫ Ψ (x t) p Ψ(x t)dx but what is p Ψ(x t) =?? −∞ Use Fourier transform to expand Ψ(x, t) in momentum representation ∞ ̂ , √1 ̂ , ipx∕ℏ p Ψ(x t) = ∫ p Φ(p t)e dp 2휋ℏ −∞

Take derivative of Fourier expansion for Ψ(x, t) ∞ , √1 , ipx∕ℏ if Ψ(x t) = ∫ Φ(p t)e dp then 2휋ℏ −∞ ( ) ( ) ( ) 휕 ∞ ip ∞ , √1 , ipx∕ℏ i √1 , ipx∕ℏ i ̂ , 휕 Ψ(x t) = ∫ ℏ Φ(p t)e dp = ℏ ∫ pΦ(p t)e dp = ℏ p Ψ(x t) x 2휋ℏ −∞ 2휋ℏ −∞ ( ) 휕 i Ψ(x, t) = p̂ Ψ(x, t) 휕x ℏ

P. J. Grandinetti Chapter 11: Wave Mechanics Momentum Operators Rearrange and ﬁnd that p̂Ψ(x, t) is given by 휕 p̂ Ψ(x, t) = −iℏ Ψ(x, t) 휕x

Identify momentum operator when applied to wave function in position basis as

휕 p̂ = −iℏ 휕x

Operator notation allows us to work with Ψ(x, t) basis only and obtain expectation values for both x̂ and p̂.

Homework Determine position operator when applied to wave function in momentum basis.

P. J. Grandinetti Chapter 11: Wave Mechanics Eigenfunctions and Eigenvalues Whenever operator acts on wave function and gives back numerical value times original wavefunction then that wave function is an eigenfunction for that operator and numerical value returned is called the eigenvalue.

Af̂ (…) = a f (…)

f (…) is eigenfunction of Â and a is eigenvalue.

Is Ψ(x, t) eigenfunction for x̂? x̂Ψ(x, t) = xΨ(x, t), Yes

Is Φ(p, t) eigenfunction for p̂? p̂Φ(p, t) = pΦ(p, t), Yes dΨ(x, t) Is Ψ(x, t) eigenfunction for p̂? p̂Ψ(x, t) = −iℏ , No dx Is Φ(p, t) eigenfunction for x̂? No. Prove this at home.

P. J. Grandinetti Chapter 11: Wave Mechanics Operators in Quantum Mechanics must be linear Can only lead to linear diﬀerential equations involving Ψ

( ) ̂ ̂ ̂ . A c1Ψ1 + c2Ψ2 = c1AΨ1 + c2AΨ2

Nonlinear operations, such as take square root log cosine sine raise to a power–other than 0 or 1 are not allowed.

P. J. Grandinetti Chapter 11: Wave Mechanics Uncertainty and Commutator Relations

P. J. Grandinetti Chapter 11: Wave Mechanics Commutation Relations Operators may not commute, i.e., order that operators are applied matters. Example What is result when operator below is applied to a wave function?

x̂p̂ − p̂x̂ = ?

Hint: not zero 휕 Recall p̂ = −iℏ . Apply this diﬀerence operator to Ψ(x, t) we ﬁnd 휕x ( ) ( ) 휕Ψ(x, t) 휕 (x̂p̂ − p̂x̂)Ψ(x, t) = x̂p̂Ψ(x, t) − p̂x̂Ψ(x, t) = −iℏ x − −iℏ (x Ψ(x, t)) 휕x 휕x ( ) 휕Ψ(x, t) 휕Ψ(x, t) = −iℏx − −iℏΨ(x, t) − iℏx = iℏΨ(x, t) 휕x 휕x

(x̂p̂ − p̂x̂)Ψ(x, t) = iℏΨ(x, t)

P. J. Grandinetti Chapter 11: Wave Mechanics Commutation Relations

(x̂p̂ − p̂x̂) Ψ(x, t) = iℏΨ(x, t)

Such operator product diﬀerences occur often in quantum mechanics. So often it is called a commutator and given shorthand notation

x̂p̂ − p̂x̂ = [x̂,p ̂] = iℏ, for all Ψ(x, t)

When [Â , B̂ ] = Â B̂ − B̂ Â ≠ 0 we say that Â and B̂ do not commute. Why are commutation relations useful?

P. J. Grandinetti Chapter 11: Wave Mechanics Uncertainty and Commutator Relations

From [x̂,p ̂] = iℏ we obtain the uncertainty relation: ⟨Δx⟩⟨Δp⟩ ≥ ℏ∕2

Commutator gives us more general uncertainty relationships.

Given [Â , B̂ ] = iC, then ⟨ΔA⟩⟨ΔB⟩ ≥ ⟨C⟩∕2

where √ ⟨ΔA⟩ = ⟨A2⟩ − ⟨A⟩2

This general relation, ⟨ΔA⟩⟨ΔB⟩ ≥ ⟨C⟩∕2, quantiﬁes ability to specify precisely and simultaneously two observables for Â and B̂ .

P. J. Grandinetti Chapter 11: Wave Mechanics Kinetic and Potential Energy Operators

P. J. Grandinetti Chapter 11: Wave Mechanics Kinetic and Potential Energy Operators Kinetic energy operator in momentum and position bases are

p̂ 2 ℏ2 휕2 K̂ = and K̂ = − 2m 2m 휕x2

Potential energy operator, V̂ (x), depend on system under study. Solving Schrödinger Eq. is easy or impossible depending on V̂ (x). Examples of some easy ones are... ▶ Electron trapped in 1D box of length L has potential energy operator V̂ (x) = 0 if 0 ≤ x ≤ L, V̂ (x) = ∞ if x < 0 and x < L This is called the inﬁnite well potential. ▶ 휅 1D harmonic oscillator with force constant f has potential energy operator 1 V̂ (x) = 휅 x̂2 2 f

P. J. Grandinetti Chapter 11: Wave Mechanics Total Energy Operator ≡ Hamiltonian Operator Total energy operator is sum of kinetic and potential energy operators,

p̂2 ℏ2 휕2 ̂ = K̂ + V̂ = + V̂ (x) = − + V̂ (x) 2m 2m 휕x2 ̂ is called Hamiltonian operator. Recall Schrödinger equation [ ] ℏ2 휕2 휕 E Ψ(x, t) = − + V̂ (x) Ψ(x, t) = iℏ Ψ(x, t) 2m 휕x2 휕t ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ̂ We can write Schrödinger equation more compactly as 휕 E Ψ(x, t) = ̂ Ψ(x, t) = iℏ Ψ(x, t) 휕t With this substitution we see that ̂ is also given by 휕 ̂ = iℏ 휕t

P. J. Grandinetti Chapter 11: Wave Mechanics Solving the Schrödinger Equation

P. J. Grandinetti Chapter 11: Wave Mechanics Solving the Schrödinger Equation If potential energy operator, V̂ (x), depends only on position and not time then separation of variables can be used Ψ(x, t) = 휓(x)휙(t) Substituting into Schrödinger equation gives [ ] ℏ2 휕2 ℏ2 휕2휓(x) 휕Ψ(x, t) 휕휙(t) − + V̂ (x) 휓(x)휙(t) = − 휙(t) + 휙(t)V̂ (x)휓(x) = iℏ = iℏ휓(x) 2m 휕x2 2m 휕x2 휕t 휕t

Dividing both sides by Ψ(x, t) = 휓(x)휙(t) leads to [ ] 1 ℏ2 휕2휓(x) 1 휕휙(t) − + V̂ (x)휓(x) = iℏ = E 휓(x) 2m 휕x2 휙(t) 휕t ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ depends only on x depends on on t

For equality to remain true for all x and t both sides must equal separation constant, E, which we’ll ﬁnd is total energy and is time independent. P. J. Grandinetti Chapter 11: Wave Mechanics Solving the Schrödinger Equation This gives 2 uncoupled ODEs:

d휙(t) iE d2휓(x) 2m ( ) + 휙(t) = 0 and + E − V̂ (x) 휓(x) = 0 dt ℏ dx2 ℏ2

ODE for 휙(t) has trivial solution: 휙(t) = e−iEt∕ℏ ODE for 휓(x) is called the time independent Schrödinger equation. Solution to full partial diﬀerentiation equation is Wave√ equation for 휓(x) doesn’t contain , 휓 −iEt∕ℏ Ψ(x t) = (x)e i = −1, so its solutions are real functions. Called stationary state as probability density is time independent, [ ][ ] |Ψ(x, t)|2 = 휓∗(x)eiEt∕ℏ 휓(x)e−iEt∕ℏ = 휓∗(x)휓(x)

Operator expectation values for stationary states are time independent. P. J. Grandinetti Chapter 11: Wave Mechanics Particle in Inﬁnite Well – Stationary States Solutions to Schrödinger equation for particle in inﬁnite square well are analogous to string standing waves. Potential takes form { 0 if 0 ≤ x ≤ L V̂ (x) = ∞ otherwise

Outside the well we have Ψ(x, t) = 0. What are the stationary states, Ψ(x, t), inside the well? P. J. Grandinetti Chapter 11: Wave Mechanics Particle in Inﬁnite Well – Stationary States What are stationary states inside well? Look for solutions to time independent Schrödinger equation,

2 2휓 >0 2휓 ℏ d (x) ̂ d (x) 2 − + V(x)휓(x) = E휓(x) or + k 휓(x) = 0 2m dx2 dx2 where 2mE k2 = ℏ2 We’ve seen this ODE and boundary conditions before. Its solutions look like n휋 휓 (x) = B sin k x where k = and n = 1, 2, 3, … n n n L In this case we have

n2휋2 2mE n2h2 k2 = = n or E = ⟵ Stationary State Energies n L2 ℏ2 n 8mL2

P. J. Grandinetti Chapter 11: Wave Mechanics Particle in Inﬁnite Well – Stationary States

Stationary State Energies

In quantum mechanics we describe n as a principal quantum number. On the right are energy levels and spacing for n = 1 to 4, Note lowest energy is not zero. 2 2 Lowest possible energy is E1 = h ∕(8mL ). Every bound quantum particle has zero point energy. This is consequence of particle’s wave properties. Energy level spacings are not equal and increase with increasing n.

P. J. Grandinetti Chapter 11: Wave Mechanics Particle in Inﬁnite Well – Stationary States For wave function to give probability density function, it has to be normalized: ( ) ( ) L L 휋 휋 |L |휓 |2 2 2 n 2 x L 2n x | 2 L ∫ | n(x)| dx = B ∫ sin dx = B − 휋 sin | = B = 1 0 0 L 2 4n L |0 2 √ which leads to B = 2∕L. Normalized wave functions are given by, √ 2 n휋 휓 (x) = sin x n L L

Any wave function inside inﬁnite well potential can be expressed as linear combination of these stationary states √ ∑∞ ∑∞ ( ) 2 n휋 ℏ Ψ(x, t) = a 휓 (x)휙 (t) = a sin x e−iEnt∕ n n n n L L n=1 n=1

P. J. Grandinetti Chapter 11: Wave Mechanics Probability of observing a particle in an inﬁnite well

P. J. Grandinetti Chapter 11: Wave Mechanics Position Measurement of Particle in Inﬁnite Well Wave function magnitude squared is probability density. Zero chance of ﬁnding particle at nodes where amplitude is always zero.

Left are first 5 wave functions for particle in infinite square well. Right are corresponding probability densities. Doesn’t mean particle cannot have well defined position. We can easily measure its position. P. J. Grandinetti Chapter 11: Wave Mechanics Collapsing Wave Function with a Measurement Act of measuring particle’s position changes wave function. particle observed here

Immediately after position measurement wave function would be ∑∞ 휓 훿 휓 after(x) = (x − x0) = an n(x) n=1 In practice, particle located within measurement spatial resolution : Δx. Better wave function description after measurement could be { √ x for x < x < x 휓 (x) = 1∕ Δ −Δ ∕2 Δ ∕2 after 0 otherwise Act of measurement collapses wave function to wave packet centered on spot where particle is found. P. J. Grandinetti Chapter 11: Wave Mechanics Collapsing Wave Function with a Measurement Can we always predict where particle will be found in position measurement? No. Every time you measure particle’s position it appears at random positions consistent with its wave function in box. No one has ever ﬁgured out how to predict exactly where particle will be found in given measurement. It’s always random and no one knows why. Einstein didn’t like this and famously said “God doesn’t play dice.”

To which Niels Bohr replied “Don’t tell God what to do.”

P. J. Grandinetti Chapter 11: Wave Mechanics Predicting Measurement Outcomes

While outcome of given measurement will be random, wave function does give us precise probability for where particle will be observed.

For example, if particle in inﬁnite well has n = 2 stationary state wave function, √ 2 2휋 휓 (x) = sin x 2 L L then... we’re 100% certain that particle will never be observed at x = L∕2. if we measured its kinetic energy we would also be 100% certain that we would obtain 2 2 E2 = 4h ∕(8mL )

P. J. Grandinetti Chapter 11: Wave Mechanics Predicting Measurement Outcomes

Example What is probability of locating particle in 1D inﬁnite well between L∕4 and 3L∕4 when particle is in n = 1 stationary state?

3L∕4 3L∕4 ( ) |3L∕4 2 2 휋 2 x L 2휋 Probability |휓 | 2 | = ∫ | 1(x)| dx = ∫ sin x dx = − 휋 sin x | L∕4 L L∕4 L L 2 4 L |L∕4 1 1 Probability = + ≈ 0.82 2 휋

P. J. Grandinetti Chapter 11: Wave Mechanics Momentum Measurement of Stationary State If we operate on stationary state of particle in 1D inﬁnite well with p̂, ( ) √ √ ( ) 휕 2 n휋 2 n휋 n휋 p̂휓 (x) = −iℏ sin x = −iℏ cos x n 휕x L L L L L 휓 ̂ we ﬁnd that n(x) are NOT eigenstates of p. If we calculate L ⟨ ⟩ 휓∗ ̂휓 p = ∫ n (x)p n(x) dx 0 we get ⟨p⟩ = 0. n2h2 How can average momentum, ⟨p⟩, be zero when kinetic energy, E = , is non-zero? n 8mL2 휓 Makes more sense to rewrite n(x) as √ ( ) 2 eikx − e−ikx 휓 (x) = ⟵ 2 traveling waves going in opposite directions. n L 2i

P. J. Grandinetti Chapter 11: Wave Mechanics Momentum Measurement of Stationary State ̂ If we apply p to √ ( ) 2 eikx − e−ikx 휓 (x) = n L 2i we ﬁnd √ [( ) ( ) ( ) ( )] 2 휕 eikx 휕 e−ikx p̂휓 (x) = −iℏ − −iℏ n L 휕x 2i 휕x 2i √ [ ( ) ( )] 2 ( ) eikx ( ) e−ikx = −ℏk − +ℏk L n 2i n 2i ⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟

1 2

̂ ℏ 1 eigenstate of p with eigenvalue of − kn is right traveling wave ̂ ℏ 2 eigenstate of p with eigenvalue of + kn is left traveling wave. ℏ Think of 1D well stationary states as 2 traveling waves with opposite signs of momenta, ± kn. Any real wave function, i,e., no imaginary part, describes state with no net motion. Why?

P. J. Grandinetti Chapter 11: Wave Mechanics Hermitian Operators

P. J. Grandinetti Chapter 11: Wave Mechanics Hermitian Operators Expectation values associated with physically observable quantities, ∞ ⟶ ⟨ ⟩ ∗ ̂ physical observables must be real (not complex) O = ∫ Ψ OΨdx −∞ While ⟨O⟩ must be real, i.e., ⟨O⟩ = ⟨O⟩∗, or [ ] ∞ ∞ ∗ ∗ ̂ ∗ ̂ ∫ Ψ OΨdx = ∫ Ψ OΨdx −∞ −∞ both and ̂ can be complex. Since ∗ ∗ ∗ ∗, we rewrite right hand side as Ψ O (z1z2z3) = z1z2z3 [ ] ∞ ∞ ∗ ∞ ∞ ( ) ∞ ( ) ∗ ̂ ∗ ̂ ̂ ∗ ∗ ̂ ∗ ̂ ∗ ∫ Ψ (OΨ)dx = ∫ Ψ OΨdx = ∫ ΨO Ψ dx = ∫ Ψ OΨ dx = ∫ OΨ Ψdx −∞ −∞ −∞ −∞ −∞

Deﬁne Hermitian operator as one that satisﬁes a more general (and stronger) condition: ∞ ( ) ∞ ( ) ∗ ̂ ̂ ∗ ∫ Ψ1 OΨ2 dx = ∫ OΨ1 Ψ2 dx −∞ −∞ Operator associated with physically observable quantities must be hermitian. P. J. Grandinetti Chapter 11: Wave Mechanics Hermitian Operators Example Is D̂ = 휕∕휕x is a hermitian operator?

( ) ( ) ∞ ∞ ∞ ∞ ∗ ( ) ? ( )∗ 휕Ψ ? 휕Ψ Check if Ψ∗ D̂ Ψ dx = D̂ Ψ Ψ dx or Ψ∗ 2 dx = 1 Ψ dx ∫ 1 2 ∫ 1 2 ∫ 1 휕 ∫ 휕 2 −∞ −∞ −∞ x −∞ x

Use integration by parts , ∫ udv = uv − ∫ vdu

휕Ψ∗ 휕Ψ take u = Ψ∗ with du = 1 dx and v = Ψ with dv = 2 dx. 1 휕x 2 휕x ∞ 휕Ψ |∞ ∞ 휕Ψ∗ Ψ∗ 2 dx = Ψ∗Ψ | − Ψ 1 dx ∫ 1 휕 1 2| ∫ 2 휕 −∞ x |−∞ −∞ x

P. J. Grandinetti Chapter 11: Wave Mechanics Hermitian Operators Example Is D̂ = 휕∕휕x is a hermitian operator? ( ) ( ) ∞ 휕Ψ ? ∞ 휕Ψ ∗ Check if Ψ∗ 2 dx = 1 Ψ dx ∫ 1 휕 ∫ 휕 2 −∞ x −∞ x

0 ∞ 휕Ψ |¨∞¨* ∞ 휕Ψ∗ Ψ∗ 2 dx = Ψ∗Ψ¨| − Ψ 1 dx ∫ 1 휕 1¨ 2| ∫ 2 휕 −∞ x ¨ |−∞ −∞ x 1st term on right goes to zero as all wave functions should at ±∞. We are left with ( ) ( ) ∞ 휕Ψ ∞ 휕Ψ ∗ Ψ∗ 2 dx = − 1 Ψ dx ∫ 1 휕 ∫ 휕 2 −∞ x −∞ x Right hand side has opposite sign of left hand side. We conclude that D̂ = 휕∕휕x is NOT a hermitian operator. P. J. Grandinetti Chapter 11: Wave Mechanics Hermitian adjoint of an operator For any operator, Â , (not necessarily hermitian) we deﬁne its hermitian adjoint, Â †, with

( ) ∗ ̂ † ̂ ∗ ∫ Ψ1A Ψ2dx = ∫ AΨ1 Ψ2 dx

In previous example we saw that D̂ = 휕∕휕x and D̂ † = −휕∕휕x.

Any operator equal to its hermitian adjoint is a hermitian operator, e.g. p̂† = p̂.

Other helpful theorems about adjoints are ( )† ▶ Adjoint of sum of 2 operators equals sum of their adjoints: Â + B̂ = Â † + B̂ † ▶ Sum of 2 hermitian operators is a hermitian operator. ( ) ▶ Even if Â is not hermitian we ﬁnd that both Â + Â † and i Â + Â † are hermitian. ( )† ▶ Given arbitrary operators Â and B̂ we have Â B̂ = B̂ †Â †

P. J. Grandinetti Chapter 11: Wave Mechanics Eigenfunctions of Hermitian Operators are Orthogonal

휓 휓 If m(x) and n(x) are eigenfunctions of a hermitian operator then

∞ 휓∗ 휓 훿 ∫ m(x) n(x) dx = m,n −∞ 훿 where m,n is Kronecker delta function: { 0 if m ≠ n 훿 = mn 1 if m = n

P. J. Grandinetti Chapter 11: Wave Mechanics Web Video: Double Slit Experiment

P. J. Grandinetti Chapter 11: Wave Mechanics