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Lecture Notes on

Quantum Mechanics

Jadon

[email protected]

2 Contents

• 1 Introduction Contents
• 2 Hilbert Space
• 3 Postulates
• 4 Density Operator
• 5 Symmetries and Conservation Laws
• 6 Miscellaneous
• 7 Comments

4 2. Hilbert Space

We call this bra and ket hermitian conjugates of each other and write it as ⟨ψ| = (|ψ⟩)†, |ψ⟩ = (⟨ψ|)†

• For operators, we have both linear and antilinear ones. Take an operator A: E → E, we can also define its actions on bras as (⟨ψ|A)|ϕ⟩ = ⟨ψ|(A|ϕ⟩). (4) Dropping the parenthesis gives ⟨ψ|A|ϕ⟩ and we can think of A acting either to the left or right.
• We define an operator called the outer product |α⟩⟨β| from |α⟩ and |β⟩ by

(|α⟩⟨β|)|ψ⟩ = |α⟩⟨β|ψ⟩, (5)

from which we can derive, from simply acting upon another |ϕ⟩,

⟨ψ|(|α⟩⟨β|) = ⟨ψ|α⟩⟨β|. (6)

• A Hilbert space always has a complete basis. It is always possible to choose a discrete, countable basis, although sometimes this is not wanted (such as when we look at the spectrum of an operator which is continuous and wish to use their eigenkets as basis).
• If we also make this basis orthonormal, we have the resolution 1 =

P

n|n⟩⟨n|. This follows if we expand a ket, |ψ⟩ =

X

n

cn|n⟩ =

X

n

|n⟩cn = |n⟩⟨n|ψ⟩ (7)

• An operator’s hermitian conjugate (or adjoint operator) is defined most naturally via scalar products, or the metric, g(u, Aw) = g(A†u, w) (8) This gives some insight to its dirac notation definition since we can always strip away w so

A†|ψ⟩ = (⟨ψ|A)†^ (9)

which satisfies the relationship ⟨ϕ|A|ψ⟩ = ⟨ψ|A†|ϕ⟩∗^ which is the metric definition.

• To take hermitian conjugates of a product of objects, we reverse their order and replace each by its Hermitian conjugate. Intuitively, this rule works because all objects can be regarded as operators.
• A hermitian operator satisfies A†^ = A or ⟨ψ|A|ϕ⟩ = ⟨ϕ|A|ψ⟩∗. An anti-hermitian operator satisfies A†^ = −A.
• Any operator can be decomposed into

A =

A + A†

A − A†

From the above definition, it follows that matrix element ⟨ψ|A|ψ⟩ are all real (imaginary) if A is hermitian (anti-hermitian). We say a Hermitian operator is positive definite if it satisfies

⟨ψ|A|ψ⟩ > 0 , (11)

5 2. Hilbert Space

for |ψ⟩̸ = 0. A is nonnegative definite if

⟨ψ|A|ψ⟩ ≤ 0 (12)

for all kets |ψ⟩. This means all its eigenvalues are positive (nonnegative) if A is positive (nonnegative) definite.

Theorem. If A, B are both hermitian, AB = BA ↔ AB is hermitian

Proof. Assuming they’re both hermitian, (AB)†^ = BA, so it’s clear BA = AB is the only way for (AB)†^ = AB.

• An operator is unitary if U −^1 = U †. The product of two unitary operators are always unitary (think rotations).
• We define the spectrum of an operator as the set of its eigenvalues. With right eigenvalues and eigenkets defined as usual, the left eigenvalue and eigenbra is defined such that

⟨ψ|A = b⟨ψ| (13)

Intuitively, this is simply the complex conjugate of eigenvalues of A†. Then, considering the determinant, we can show the left and right eigenvalues are the same for any finite-dimenisonal operator (but not in infinite dimensions).

• The spectrum of an operator is the set of all its eigenvalues, a subset of the complex plane. We can have continuous spectra.
• If the operator is Hermitian and the spectrum discrete, the eigenvalue is real, eigenkets and eigenbras are hermitian conjugates, eigenvectors for distinct eigenvalues orthogonal.
• For E 1 , E 2 such that E 1 ∩ E 2 = { 0 }, we define the direct sum as

E 1 ⊕ E 2 = {|ψ 1 ⟩ + |ψ 2 ⟩ such that |ψ 1 ⟩ ∈ E 1 , |ψ 2 ⟩ ∈ E 2 }. (14)

This is a vector subspace again, and since there’s no intersection dim(E 1 ⊕ E 2 ) = dim E 1 + dim E 2. For an operator A acting on a finite dimensional space E, we define the nth eigenspace as

En = {|u⟩ such that A|u⟩ = an|u⟩}. (15)

A is complete if E 1 ⊕ E 2 ⊕ · · · EN = E. (16) Any finite-dimensional Hermitian A is complete, but this is not true in infinite dimensions. We define a complete Hermitian operator as an observable, which links to a physical interpretation of observables in qm.

Theorem. To sum up all the following examples, we state: An observable always possesses an orthonormal eigenbasis. This eigenbasis must include states of infinite norm if the observable has a continuous spectrum at some place.

7 2. Hilbert Space

Example. Consider the position operator xˆ which multiplies the wavefunction by x. The eigenvalue equation is xfx 0 (x) = x 0 fx 0 (x) (24) or, (x − x 0 )fx 0 (x) = 0 (25) which implies f = 0 everywhere except at x 0 where it can be any value. We interpret this situation as the Dirac delta function and write fx 0 (x) = δ(x − x 0 ). Again, we choose a real, continuous spectrum, and the eigenfunctions satisfy

⟨x 0 |x 1 ⟩ =

Z

dxδ(x − x 0 )δ(x − x 1 ) = δ(x 0 − x 1 ), (26)

where we wrote fx 0 (x) as |x 0 ⟩ in ket language. We have the identity

ψ(x 0 ) =

Z

dx δ(x − x 0 )ψ(x) =

Z

dx f (^) x∗ 0 (x)ψ(x) = ⟨x 0 |ψ⟩. (27)

This means we can now extract the resolution. If we write ψ(x 0 ) =

Z

dx ⟨x 0 |x⟩⟨x|ψ⟩ = ⟨x 0 |ψ⟩, (28)

both ⟨x 0 | and ψ are really arbitrary, hence we have the resolution 1 =

R

dx |x⟩⟨x|. We can do the same for |p⟩, where we notice

ψ(x) =

Z

dx ⟨x|p⟩⟨p|ψ⟩ = ⟨x|ψ⟩, (29)

throwing away ⟨x 0 | and ψ, we get 1 =

R

dp |p⟩⟨p|. Note throughout these examples only completeness is assumed. We didn’t forcefully invoke the theorem and conclusion given. Theorem. Two observables possess a simultaneous eigenbasis if and only if they commute. This eigenbasis may be chosen to be orthonormal.

Proof. We observe that if A|u⟩ = an|u⟩, AB|u⟩ = BA|u⟩ = an(B|u⟩) (30) Pictorially, if |u⟩ is in the nth eigenspace of A, then B|u⟩ maps |u⟩ to another vector in this space. First, this means if the eigenspace is a ray, |u⟩ is also an eigenvector of B. If not a ray, we know B is still Hermitian on this eigenspace, hence there is a set of orthonormal eigenbasis. The key is since we’re in A’s eigenspace, any orthonormal basis is an eigenbasis. So we can choose the one that matches B’s basis. Formally, this is done by introducing the restriction of B, Bˆ such that on ϵn, there is an orthonormal eigenbasis Bˆ|nr⟩ = bnr|nr⟩ (31) where we fix n here (non-trivial), and vary r. Since fixing n means working in one of A’s instead of B’s eigenspaces, repetitions in bnr show degeneracies. Conversely, if they have a set of simultaneous eigenbasis, wlog, pick any basis vector |u⟩ from the set, AB|u⟩ = anbn|u⟩ = BA|u⟩. (32) Since any vector is a linear sum of basis vectors, this equation is true for any vector, hence AB = BA. QED

8 3. Postulates

3 Postulates

• Every physical system is associated with a Hilbert space E with vectors we call kets.
• Every pure state of a physical system is associated with a definite ray in E.
• Every measurement that can be carried out corresponds to a complete Hermitian operator A(an observable)
• Possible results of the measurement are the eigenvalues an of A, either the discrete eigenvalues a 1 , a 2 , · · · or the continuous ones a(υ).
• The probability of measuring A = an is

p(A = an) = ⟨ψ|Pn|ψ⟩ ⟨ψ|ψ⟩

where Pn is the projector onto the eigenspace En corresponding to eigenvalue an, and |ψ⟩ is any nonzero ket in the ray representing the state of the system. In the continuous case,

p(a 0 ≤ A ≤ a 1 ) = ⟨ψ|PI |ψ⟩ ⟨ψ|ψ⟩

where PI is the projection operator corresponding to interval I. These probabilities depend only on the ray.

• After a measurement with discrete outcome A = an, the system is represented by the ket Pn|ψ⟩ where |ψ⟩ represents the system before the measurement. In the continuous case, with outcome a 0 ≤ A ≤ a 1 , the system is represented by PI |ψ⟩ after measurement.

Note. Just like Newton’s laws, this gives no description to how to find the specific Hilbert Space of a physical system. We continue with questions

Example. Consider the Stern-Gerlach experiment. We measure the magnetic moment of the silver atoms, wlog we first measure the x-component of the magnetic moment μ, and pass the beam with μx = +μ 0 to a second magnet, with μz = ±μ 0 each carry 50% of the atoms that entered. We think

• For a normalized, pure state |ψ⟩, there is a probability fn = ⟨ψ|Pn|ψ⟩ of measuring A = an. Hence, in Hilbert space oprations, the mean of a is

⟨a⟩ =

X

n

fnan = ⟨ψ|

X

n

anPn|ψ⟩ = ⟨ψ|A|ψ⟩. (35)

We define ⟨A⟩ = ⟨a⟩. For the variance ∆a^2 , it can be expressed as

∆a^2 = ⟨a^2 ⟩ − ⟨a⟩^2 = ⟨ψ|A^21 |ψ⟩, (36)

where A 1 = A − ⟨A⟩ which is true because (for later). We write ∆A^2 = ∆a^2.

10 5. Symmetries and Conservation Laws

5 Symmetries and Conservation Laws

This part now follows Griffiths, but I might change it later and continue reading Littlejohn.

• When we say a system has symmetry in QM, it means the Hamiltonian is unchanged after a transformation.
• Translation operator T ψˆ (x) = ψ′(x) = ψ(x − a)
• Parity operator Πˆψ(x) = ψ′(x) = ψ(−x). In three dimensions parity changes sign of all three coordinates.
• Rotation operators will be introduced later.

Example. (GR 6.1) Consider the parity operator in three dimensions. (a) Πˆ is equivalent to mirror reflection followed by a rotation because (b)

• If we Taylor expand Tˆ (a)ψ(x) = ψ(x − a) around x, we get

X^ ∞

n=

n!

ia ℏ pˆ)nψ(x)). (45)

Hence we have Tˆ (a) = exp[− ia ℏ

pˆ] (46)

and say momentum is the generator of translations.

• We can show Tˆ is a unitary operator. In physics intuition, the hermitian conjugate of the above equation replaces −i with i hence Tˆ †^ = Tˆ (−a).
• A transformation acting on an operator is defined to be the operator that gives the same expectation value in the untranslated state ψ as does the operator Qˆ in the translated state ψ′:

⟨ψ′| Qˆ|ψ′⟩ = ⟨ψ| Qˆ′|ψ⟩ (47)

We can do either active (actually moving the function) or passive (moving the coordinate system) transformations to calculate the effect of translation on an expectation value. The operator Q′ is the operator in this shifted coordinate system (?). Since |ψ′⟩ = T |ψ⟩, the translation of any Qˆ is Qˆ′^ = T †^ QT.ˆ (48)

• After translation, ˆx → xˆ + a, pˆ → ˆp. The former we expect because it is equivalent to shifting the origin of our coordinate system to the left by a hence the expectation value of position shifts to the right by a. The momentum shouldn’t change under a shift of x. And from this we can prove: Qˆ(ˆx, pˆ) → Qˆ(ˆx + a, pˆ). (49)

11 5. Symmetries and Conservation Laws

Proof. 1.

x′f (x) = Tˆ †(a)ˆx T fˆ = Tˆ (−a)ˆx Tˆ (a)f (50) = Tˆ (−a)[xf (x − a)] = ˆ(x + a)f (x). (51)

Hence xˆ′^ = ˆx + a. (52)

pˆ′f = T †^ pT fˆ = T (−a)

i

f ′(x − a) =

i

f ′(x) = ˆpf (53)

Hence pˆ′^ = ˆp.

1. If we can write Q(x, p) = amnxmpn, then by using T †T = 1, we have

x′^ = (T †xT )m^ = T †xmT p′^ = (T †pT )n^ = T †pnT (54)

Hence Q′^ = amnT †xmpnT = T †xmT T †pnT = amnx′mp′n^ = Q(x + a, p) (55)

• If we want translational symmetry in H, start with H†^ = T †HT = H (since if the expectation value of an operator acted on the same ket doesn’t change, then it’s the same ket). Then HT = T H, so [H, T ] = 0 ⇔ Translational Symmetry (56) Also H′^ = p

2 2 m +^ V^ (x^ +^ a), so translational symmetry requires V (x + a) = V (x). (57)

If this works for all a as in the case of a constant potential we call it continuous translational symmetry. If this works only for discrete a as in the case of a periodic potential, we call it discrete translational symmetry.

• From the commutating relationship, we know both should share the same eigenstates. This means if Hψ(x) = Eψ(x),

T (a)ψ(x) = e−iqaψ(x) = ψ(x − a). (58)

where the first equality follows from T being unitary hence has a phase as the eigenvalue. qa is a choice for simplicity. If we substitute ψ = eiqxu(x), then we have

ψ(x − a) = eiq(x−a)u(x) (59)

which if we substitute x − a → x, we have

ψ(x) = eiqxu(x + a) = eiqxu(x). (60)

Hence we have the second equality which also requires u to have a period of a. This is Bloch’s theorem in 1D and is easily generalizable. This also means electrons in periodic potential lattices travel with a wave function that of a wave apart from a modulus. This means it doesn’t scatter(?)!

13 7. Comments

7 Comments

• It’s so hard to familiarize myself with the math of Hilbert Space!
• We’re so lucky to only need to work with complete Hermitian operators.
• Under an orthonormal basis, we can write the resolution as 1 =

P

nr|nr⟩⟨nr|,^ n^ specifying the eigenspace and r labeling the orthonormal vectors in the subspace.

• For some operators, they don’t have a discrete spectrum. They might also not have eigenvectors in the Hilbert space (non-normalizable). They are also orthonormal in the δ − f unction sense. Hence we obtain a orthonormal basis outside the Hilbert space which still spans every point on the Hilbert space.