Physical Chemistry
2. Formalism of Quantum Mechanics

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

Table of Content

1. Hilbert space
2. Observables and eigenstates
3. Generalized statistical interpretation

1

2

The Nobel Prize in Physics 1932
Werner Heisenberg• For the creation of quantum
mechanics, the application of whichhas, inter alia, led to the discoveryof the allotropic forms of hydrogen.
The Nobel Prize in Physics 1933
Erwin Schrödinger and Paul A.M. Dirac• For the discovery of new productive
forms of atomic theory.

1. Hilbert space• A Hilbert space is a “complete” “metric” innerproduct space.
• A metric space is a set, in which the distance(𝑑𝑑) between any 2 elements is defined. 𝑥𝑥
𝑑𝑑 𝑥𝑥, 𝑦𝑦 ≥ 0𝑑𝑑 𝑥𝑥, 𝑦𝑦 = 0 ⇔ 𝑥𝑥 = 𝑦𝑦
𝑑𝑑 𝑥𝑥, 𝑦𝑦 = 𝑑𝑑 𝑦𝑦, 𝑥𝑥
𝑑𝑑 𝑥𝑥, 𝑦𝑦 ≤ 𝑑𝑑 𝑦𝑦, 𝑧𝑧 + 𝑑𝑑 𝑥𝑥, 𝑧𝑧𝑧𝑧
• The distance in inner product space can bedefined as 𝑑𝑑 |𝛼𝛼 ⟩, |𝛽𝛽⟩ ≡ |𝛼𝛼 ⟩ − |𝛽𝛽⟩ .
• For example, distance in Euclidean space.

3𝑦𝑦1. Hilbert space• Complete metric space: the limit of every Cauchysequence of points is (converges) in the space.
• Cauchy sequence 𝑥𝑥𝑛𝑛 : if for every positive real
number 𝜀𝜀 > 0, there is a positive integer 𝑁𝑁 suchthat for all positive integers 𝑚𝑚, 𝑛𝑛 > 𝑁𝑁, the distance𝑑𝑑 𝑥𝑥𝑛𝑛 , 𝑥𝑥𝑚𝑚 < 𝜀𝜀
• A complete metric space does not have holes.
• For example, real number space is complete;rational number space is incomplete.
𝑛𝑛1
lim 1 += 𝑒𝑒
𝑛𝑛→∞
𝑛𝑛4

1. Hilbert space• A vector in Hilbert space is represented by a ket ofbra-ket notation, | ⟩.
• The inner product of 2 vectors, |𝛼𝛼 ⟩ and |𝛽𝛽 ⟩, isrepresented by bra-ket notation, 𝛼𝛼 𝛽𝛽 .
• 2 vectors, |𝛼𝛼⟩ and |𝛽𝛽⟩, are orthogonal if 𝛼𝛼 𝛽𝛽 = 0.
• A set of vectors �𝜓𝜓𝑛𝑛 � is orthonormal if𝜓𝜓𝑛𝑛 𝜓𝜓𝑚𝑚 = 𝛿𝛿𝑛𝑛,𝑚𝑚• A set of linearly independent vectors �𝜓𝜓𝑛𝑛 � iscomplete, if any vector in the Hilbert space, |𝛼𝛼 ⟩, canbe expressed as a linear combination of them.

561. Hilbert space|𝛼𝛼 ⟩ = � 𝑐𝑐𝑛𝑛 �𝜓𝜓𝑛𝑛 �𝑛𝑛• If so, �𝜓𝜓𝑛𝑛 � forms a basis of the Hilbert space.

• If �𝜓𝜓𝑛𝑛 � is orthonormal, the coordinates, 𝑐𝑐𝑛𝑛 , arecalculated by
𝑐𝑐𝑛𝑛 = 𝜓𝜓𝑛𝑛 𝛼𝛼• The vectors whose norms are unity form a subsetof vectors in the Hilbert space.
𝛼𝛼 2 = 𝛼𝛼 𝛼𝛼 = 1• The vectors in this subset are called rays.

1. Hilbert space

10• For example, column matrices
andform an
01
orthonormal basis of the 2nd order column matrix
space (2D Hilbert space).
1 21 † 1
10 2
=
= 1 0
=1=000011100 † 1== 0 1=000011𝑎𝑎nd• Any 2 order column matrix,, can be𝑏𝑏
expressed by𝑎𝑎
10+ 𝑏𝑏= 𝑎𝑎
𝑏𝑏017

81. Hilbert space• For example, the sine series, 𝑓𝑓𝑛𝑛 𝑥𝑥 =

2𝑛𝑛𝑛𝑛
sin
𝑥𝑥
𝑙𝑙
𝑙𝑙

𝑛𝑛 ∈ ℤ+ , in 0, 𝑙𝑙 , form an orthonormal basis of an
infinite dimensional Hilbert space.
𝑙𝑙𝑓𝑓𝑚𝑚 𝑓𝑓𝑛𝑛 = �02𝑚𝑚𝜋𝜋
sin
𝑥𝑥
𝑙𝑙
𝑙𝑙
∞

2𝑛𝑛𝑛𝑛
sin
𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝛿𝛿𝑚𝑚,𝑛𝑛𝑙𝑙
𝑙𝑙

𝑓𝑓 𝑥𝑥 = � 𝑐𝑐𝑛𝑛 𝑓𝑓𝑛𝑛 𝑥𝑥

𝑐𝑐𝑛𝑛 = 𝑓𝑓𝑛𝑛 𝑓𝑓 =𝑛𝑛=12 𝑙𝑙𝑛𝑛𝑛𝑛
� 𝑓𝑓 𝑥𝑥 sin𝑥𝑥 𝑑𝑑𝑑𝑑
𝑙𝑙 0𝑙𝑙,9

1. Hilbert space• Dirac-von Neumann axioms: a state, Ψ, of thequantum system is a unit vector (ray) of a fixedcomplex Hilbert space of countably infinitedimension.
• The inner product of 2 states (wavefunctions) isΦ 𝑟𝑟,⃑ 𝑡𝑡 Ψ 𝑟𝑟,⃑ 𝑡𝑡≡ � Φ∗ 𝑟𝑟,⃑ 𝑡𝑡 Ψ 𝑟𝑟,⃑ 𝑡𝑡 𝑑𝑑𝑛𝑛 𝑟𝑟⃑• The integral is over the whole 𝑛𝑛 dimensional space.
• States are normalized.
Ψ 𝑟𝑟,⃑ 𝑡𝑡 2 = Ψ 𝑟𝑟,⃑ 𝑡𝑡 Ψ 𝑟𝑟,⃑ 𝑡𝑡 = 1

1. Hilbert space10

 About the notation in this course.
• Throughout this physical chemistry course, theterm “state” and “wavefunction” are usedinterchangeably in most cases (except for spins).
• When symbols (typically Greek letters) are used forstates, they are used as what they are, without | ⟩.
• For example, Ψ, 𝜓𝜓, Φ, 𝜑𝜑, 𝛼𝛼, 𝛽𝛽, ↑ and ↓, etc.
• When quantum numbers (typically Arabic numbersand Latin letters) are used to represent states, | ⟩are used to avoid ambiguity.
• For example, |𝑣𝑣 ⟩, |𝑛𝑛, 𝑙𝑙, 𝑚𝑚𝑙𝑙 ,⟩,|1,0,0⟩, |±⟩, etc.

2. Observables and eigenstates• The observables (measurable physical quantities,𝑄𝑄) are operators on the Hilbert space.
• A scalar operator is denoted as 𝑄𝑄� (with hat).
� (bold).
• A vector operator is denoted as 𝑸𝑸• The operators act on the states (vectors) as lineartransformations.
� + 𝑏𝑏𝑄𝑄Φ
�𝑄𝑄� 𝑎𝑎Ψ + 𝑏𝑏Φ = 𝑎𝑎𝑄𝑄Ψ
• The expected value of 𝑄𝑄 of state Ψ is given by� 𝑟𝑟,�
𝑄𝑄 = � Ψ∗ 𝑟𝑟,⃑ 𝑡𝑡 𝑄𝑄Ψ
⃑ 𝑡𝑡 𝑑𝑑𝑛𝑛 𝑟𝑟⃑ = Ψ 𝑄𝑄Ψ

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2. Observables and eigenstates

12 The measured values of observables must be real.
𝑄𝑄 = 𝑄𝑄 ∗• That is, for any Ψ,∗�� Ψ�Ψ 𝑄𝑄Ψ = Ψ 𝑄𝑄Ψ = 𝑄𝑄Ψ

• On the other hand,�Ψ 𝑄𝑄Ψ
= 𝑄𝑄� † Ψ Ψ• This yields 𝑄𝑄� = 𝑄𝑄� † .

 Observables are Hermitian operators (self-adjointoperators) in Hilbert space.

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2. Observables and eigenstates• For example, the 1D momentum operator is a
Hermitian operator because for arbitrary states, Ψand Ψ,+∞+∞ℏ𝜕𝜕ℏΨ 𝑝𝑝Ψ̂ = � Φ∗
Ψ𝑑𝑑𝑑𝑑 = � Φ∗ 𝑑𝑑Ψ𝑖𝑖 𝜕𝜕𝜕𝜕𝑖𝑖 −∞−∞+∞+∞+∞ℏ ∗ℏ= Φ Ψ�− � Ψ𝑑𝑑Φ∗ = �𝑖𝑖𝑖𝑖 −∞−∞−∞ℏ 𝜕𝜕
Φ𝑖𝑖 𝜕𝜕𝜕𝜕∗Ψ𝑑𝑑𝑑𝑑= 𝑝𝑝Φ̂ Ψ• Here, the normalization condition is used .
Ψ +∞ = Ψ −∞ = Φ∗ +∞ = Φ∗ −∞ = 0

2. Observables and eigenstates

14• In general, the measured value of an observable ofa certain state is indeterministic.
• If for a state, Ψ, the measured value of anobservable, 𝑄𝑄, is a constant, 𝑞𝑞, then Ψ is one of theeigenstates (eigenfunctions, determinate states) of𝑄𝑄; 𝑞𝑞 is the eigenvalue of 𝑄𝑄 corresponding to Ψ.
• Proof: if the measured value is a constant, 𝜎𝜎 = 0.
22 = 𝑄𝑄 − 𝑄𝑄 2 = Ψ 𝑄𝑄
�− 𝑞𝑞 Ψ𝜎𝜎𝑄𝑄=𝑄𝑄� − 𝑞𝑞 Ψ 𝑄𝑄� − 𝑞𝑞 Ψ = 0� = 𝑞𝑞Ψ
𝑄𝑄Ψ

2. Observables and eigenstates• For example, a plane wave is an eigenstate ofmomentum and Hamiltonian.
Ψ = 𝑒𝑒 𝑖𝑖 𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡ℏ 𝜕𝜕 𝑖𝑖 𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡ℏ𝑒𝑒𝑝𝑝Ψ
̂ == 𝑖𝑖𝑖𝑖𝑒𝑒 𝑖𝑖 𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡 = ℏ𝑘𝑘Ψ𝑖𝑖 𝜕𝜕𝜕𝜕𝑖𝑖
• The eigenvalue is 𝑝𝑝 = ℏ𝑘𝑘.
• It reproduces the de Broglie relation.
𝜕𝜕 𝑖𝑖 𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡
� = 𝑖𝑖ℏ 𝑒𝑒= 𝑖𝑖ℏ −𝑖𝑖𝜔𝜔 𝑒𝑒 𝑖𝑖 𝑘𝑘𝑘𝑘−𝜔𝜔𝑡𝑡 = ℏ𝜔𝜔Ψ𝐻𝐻Ψ
𝜕𝜕𝑡𝑡• The eigenvalue is 𝐸𝐸 = ℏ𝜔𝜔.
• It reproduces the Planck relation.

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2. Observables and eigenstates16 Every measured value of an observable must be
one of its eigenvalues.
• The set of the eigenvalues of an observable iscalled its spectrum.
• For example, the eigenstates and the eigenvaluesof Hamiltonian are the stationary states (𝜓𝜓) andtheir energy levels (𝐸𝐸).
� = 𝐸𝐸𝐸𝐸𝐻𝐻𝜓𝜓• In other words, the energy levels form the spectrumof Hamiltonian.

2. Observables and eigenstates• If the eigenvalues are discontinuous, the spectrum
is discrete.
• Such eigenstates are physically realizable.
• For example, the energies of the infinite quantum
well and the angular momenta of rigid rotator arediscrete.
• If the eigenvalues fill out an entire range, thespectrum is continuous.
• Such eigenstates are not normalizable.
• For example, the momenta and energies of freeparticles (plane wave) are continuous.

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2. Observables and eigenstates• In physical chemistry, only observables withdiscrete spectra are discussed.
 Theorem: the eigenvalues of an observable arereal.
• Proof: suppose Ψ is an eigenstate of 𝑄𝑄� witheigenvalue 𝑞𝑞.
� = 𝑞𝑞Ψ𝑄𝑄Ψ• Then,� = Ψ 𝑞𝑞Ψ = 𝑞𝑞 Ψ Ψ = 𝑞𝑞
𝑄𝑄 = Ψ 𝑄𝑄Ψ

• 𝑞𝑞 is a measured physical quantity. It must be real.

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2. Observables and eigenstates• 2 or more eigenstates corresponding to the sameeigenvalue are degenerate states.
 Theorem: nondegenerate states are orthogonal.
• Suppose� = 𝑞𝑞Ψ
𝑄𝑄Ψ� = 𝑞𝑞′Φ𝑄𝑄Φ• Since 𝑄𝑄� is a Hermitian operator,
� = 𝑄𝑄Φ� Ψ
Φ 𝑄𝑄Ψ𝑞𝑞 Φ Ψ = 𝑞𝑞𝑞 Φ Ψ• And since 𝑞𝑞 ≠ 𝑞𝑞 ′ ,ΦΨ =019

2. Observables and eigenstates• If 2 eigenstates are degenerate,� = 𝑞𝑞Ψ
𝑄𝑄Ψ� = 𝑞𝑞Φ𝑄𝑄Φ• Their linear combination is another eigenstate.
𝑄𝑄� 𝑎𝑎Ψ + 𝑏𝑏Φ = 𝑞𝑞 𝑎𝑎Ψ + 𝑏𝑏Φ
• If Ψ and Φ are orthogonal, 𝑎𝑎Ψ + 𝑏𝑏Φ is notnecessarily orthogonal to Ψ or Φ.
Ψ 𝑎𝑎Ψ + 𝑏𝑏Φ = 𝑎𝑎 Ψ Ψ + 𝑏𝑏 Ψ Φ = 𝑎𝑎
 Degenerate eigenstates may be nonorthogonal.
• The Gram-Schmidt orthogonalization procedurecan be used to construct orthogonal eigenstates.

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2. Observables and eigenstates Axiom: the eigenstates of an observable arecomplete, which form a basis of the Hilbert space.
⇔ Any state, Ψ, can be expressed as a linear
combination of the eigenstates, Ψ𝑛𝑛 , of an�
observable, 𝑄𝑄.
Ψ = � 𝑐𝑐𝑛𝑛 Ψ𝑛𝑛
𝑛𝑛

� 𝑛𝑛 = 𝑞𝑞𝑛𝑛 Ψ𝑛𝑛
𝑄𝑄Ψ The eigenstates of an observable can be used toconstructed an orthonormal basis of the Hilbert
space.

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2. Observables and eigenstates

22• For example, spin in z direction is an observable in
2D Hilbert space.
ℏℏ 1 0𝑺𝑺𝒛𝒛 = 𝝈𝝈𝒛𝒛 =22 0 −1ℏ1• The eigenstate ↑=has eigenvalue .
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ℏ 1 0
ℏ 1
ℏ1𝑺𝑺𝒛𝒛 ↑=== ↑2 0 −1 0
2 02ℏ0• The eigenstate ↓=has eigenvalue − .
21
ℏ 1 0
ℏ 0
ℏ
0𝑺𝑺𝒛𝒛 ↓==−=− ↓2 0 −1 1
2 12

2. Observables and eigenstates• The eigenstates are normalized.
1↑ ↑ = 1 0=100↓ ↓ = 0 1=11•ℏℏThe eigenstates are nondegenerate ( ≠ − ) and22orthogonal.

1=0100↓ ↑ = 1 0=01
• The eigenstates form an orthonormal basis.
↑ ↓ = 023

2. Observables and eigenstates

24• For example, spin in x direction is an observable in
2D Hilbert space.
ℏℏ 0 1
𝑺𝑺𝒙𝒙 = 𝝈𝝈𝒙𝒙 =22 1 01 1ℏ
• The eigenstate 𝛼𝛼 =has eigenvalue .
22 1ℏ 0 1 1 1
ℏ 1 1
ℏ𝑺𝑺𝒙𝒙 𝛼𝛼 === 𝛼𝛼2 1 0 2 12 2 121ℏ1• The eigenstate 𝛽𝛽 =has eigenvalue − .
22 −1ℏ 0 1 1 1
ℏ 1 1
ℏ𝑺𝑺𝒙𝒙 𝛽𝛽 ==−= − 𝛽𝛽2 1 0 2 −12 2 −12

2. Observables and eigenstates•ℏℏThe eigenstates are nondegenerate ( ≠ − ).
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• They form an orthonormal basis.
11 1𝛼𝛼 𝛼𝛼 ==11 122 111 1𝛽𝛽 𝛽𝛽 ==11 −122 −111 1𝛼𝛼 𝛽𝛽 ==01 122 −111 1𝛽𝛽 𝛼𝛼 ==01 −122 125

2. Observables and eigenstates

26• Another example: the 1D quantum well (0 ≤ 𝑥𝑥 ≤ 𝑙𝑙).
� (stationary states) and their• The eigenstates of 𝐻𝐻
eigenvalues (energy levels) are𝜓𝜓𝑛𝑛 =2𝑛𝑛𝑛𝑛
sin𝑥𝑥𝑙𝑙𝑙𝑙

𝑛𝑛2 𝜋𝜋 2 ℏ2𝐸𝐸𝑛𝑛 =2𝑚𝑚𝑙𝑙 2• These nondegenerate stationary states form an
orthonormal basis of the infinite dimensional Hilbert
space.

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3. Generalized statistical interpretation• In classical mechanics, the𝑦𝑦state of a physical system
uniquely determines all thevalues of its measurable
properties.
• Conversely, the values of its
measurable properties
uniquely determine the state.
• Observation does not
influence the state of thesystem.

𝑄𝑄 𝑡𝑡𝑥𝑥283. Generalized statistical interpretation
Before• In quantum mechanics, an
𝑐𝑐1 Ψ1
observable (𝑄𝑄) is an operator.
+𝑐𝑐2 Ψ2• Ψ𝑛𝑛 is an orthonormal basis+
constructed from 𝑄𝑄’s
⋮eigenstates.
+• Before measurement, the state 𝑐𝑐𝑛𝑛 Ψ𝑛𝑛+
is the superposition of Ψ𝑛𝑛 .
⋮• When 𝑄𝑄 is measured, the state+collapses to one its eigenstates. ⋮• The outcome is the eigenvalue(𝑞𝑞𝑛𝑛 ) of the eigenstate.

AfterOutcome
𝑞𝑞𝑛𝑛

1 Ψ𝑛𝑛

293. Generalized statistical interpretation• The expected value of 𝑄𝑄 of a state, Ψ, is� = � 𝑐𝑐𝑚𝑚 ∗ ⟨Ψ𝑚𝑚 | 𝑄𝑄� � 𝑐𝑐𝑛𝑛 �Ψ𝑛𝑛 �
𝑄𝑄 = Ψ 𝑄𝑄Ψ
𝑚𝑚= � 𝑐𝑐𝑚𝑚 ∗ 𝑐𝑐𝑛𝑛 𝑞𝑞𝑛𝑛 Ψ𝑚𝑚 Ψ𝑛𝑛 = � 𝑐𝑐𝑛𝑛 2 𝑞𝑞𝑛𝑛
𝑚𝑚,𝑛𝑛

𝑛𝑛

𝑛𝑛

• According to probability theory, if the probability ofgetting 𝑞𝑞𝑛𝑛 is 𝑓𝑓𝑛𝑛 , the expected value of 𝑄𝑄 is
𝑄𝑄 = � 𝑓𝑓𝑛𝑛 𝑞𝑞𝑛𝑛
𝑛𝑛

30

3. Generalized statistical interpretation• Thus,𝑓𝑓𝑛𝑛 = 𝑐𝑐𝑛𝑛 2• Born rule: the probability ofgetting 𝑞𝑞𝑛𝑛 (simultaneouslycollapsing the state to Ψ𝑛𝑛 )when 𝑄𝑄 is measured is 𝑐𝑐𝑛𝑛 2 .
• The probability is normalized.
� 𝑓𝑓𝑛𝑛 = � 𝑐𝑐𝑛𝑛 2 = 1
𝑛𝑛𝑛𝑛

State
𝑐𝑐1 Ψ1+𝑐𝑐2 Ψ2+⋮+𝑐𝑐𝑛𝑛 Ψ𝑛𝑛+
⋮Outcome
Probability
𝑐𝑐1 2

𝑞𝑞1

𝑐𝑐2 2

𝑞𝑞2

𝑐𝑐𝑛𝑛 2𝑞𝑞𝑛𝑛⋮⋮⋮⋮

3. Generalized statistical interpretation• For example, a spin state, 𝜒𝜒, is𝑎𝑎𝜒𝜒 =𝑏𝑏𝑎𝑎 2 + 𝑏𝑏 2 = 1• When measuring 𝑆𝑆𝑧𝑧 , what are the outcomes andtheir probabilities?
• Answer: the outcomes are the eigenvalues: ± ℏ⁄2.
10𝜒𝜒 = 𝑎𝑎+ 𝑏𝑏01
• The probability of getting ℏ⁄2 is 𝑎𝑎 2 .
• The probability of getting − ℏ⁄2 is 𝑏𝑏 2 .

31

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3. Generalized statistical interpretation• What is the standard deviation of 𝑆𝑆𝑧𝑧 ?
ℏ 1†∗∗𝑆𝑆𝑧𝑧 = 𝜒𝜒 𝑆𝑆̂𝑧𝑧 𝜒𝜒 = 𝝌𝝌 𝑺𝑺𝑧𝑧 𝝌𝝌 = 𝑎𝑎 𝑏𝑏2 0ℏ
𝑎𝑎 2 − 𝑏𝑏 2=2222†
̂𝑆𝑆𝑧𝑧 = 𝜒𝜒 𝑆𝑆𝑧𝑧 𝜒𝜒 = 𝝌𝝌 𝑺𝑺𝑧𝑧 𝝌𝝌20−1𝑎𝑎𝑏𝑏22
ℏ
ℏ
𝑎𝑎0∗
∗
= 𝑎𝑎 𝑏𝑏𝑎𝑎 2 + 𝑏𝑏 2 ==𝑏𝑏−14422ℏℏ−𝜎𝜎𝑆𝑆𝑧𝑧 2 = 𝑆𝑆𝑧𝑧 2 − 𝑆𝑆𝑧𝑧 2 =𝑎𝑎 2 − 𝑏𝑏 2 244ℏ 12 03. Generalized statistical interpretation• After getting ℏ⁄2 from the measurement of 𝑆𝑆𝑧𝑧 , 𝑆𝑆𝑥𝑥 ismeasured.
• What are the outcomes and their probabilities?
• Answer: after getting ℏ⁄2 from measurement of 𝑆𝑆𝑧𝑧 ,1.
the spin collapses to the eigenstate ↑=01 1 1
1 11
11=𝛼𝛼 +𝛽𝛽=+02 2 12 −122••1ℏThere is probability to get .
221ℏ
There is probability to get − .
2233

34

3. Generalized statistical interpretation• Suppose there is an observables, 𝐴𝐴.
𝜎𝜎𝐴𝐴 2 = 𝐴𝐴 − 𝐴𝐴 2 = Ψ 𝐴𝐴̂ − 𝐴𝐴2Ψ• Since 𝐴𝐴̂ − 𝐴𝐴 is Hermitian,𝜎𝜎𝐴𝐴 2 = 𝐴𝐴̂ − 𝐴𝐴 Ψ 𝐴𝐴̂ − 𝐴𝐴 Ψ• For convenience, substitute 𝐴𝐴̂ − 𝐴𝐴 Ψ = 𝛼𝛼.
𝜎𝜎𝐴𝐴 2 = 𝛼𝛼 𝛼𝛼 = 𝛼𝛼 2
• Similarly, suppose there is another observables, 𝐵𝐵.
• Substitute 𝐵𝐵� − 𝐵𝐵 Ψ = 𝛽𝛽.
𝜎𝜎𝐵𝐵 2 = 𝛽𝛽 2

3. Generalized statistical interpretation• The Cauchy-Schwarz inequality states that
𝜎𝜎𝐴𝐴 2 𝜎𝜎𝐵𝐵 2 = 𝛼𝛼 2 𝛽𝛽 2 ≥ 𝛼𝛼 𝛽𝛽 2
• Here,𝛼𝛼 𝛽𝛽 = 𝐴𝐴̂ − 𝐴𝐴 Ψ 𝐵𝐵� − 𝐵𝐵 Ψ= Ψ 𝐴𝐴̂ − 𝐴𝐴 𝐵𝐵� − 𝐵𝐵 Ψ� − Ψ 𝐴𝐴̂ 𝐵𝐵 Ψ − Ψ 𝐵𝐵� 𝐴𝐴 Ψ + Ψ 𝐴𝐴 𝐵𝐵 Ψ= Ψ 𝐴𝐴̂ 𝐵𝐵Ψ
= 𝐴𝐴𝐵𝐵 − 𝐴𝐴 𝐵𝐵• Similarly,
𝛽𝛽 𝛼𝛼 = 𝐵𝐵𝐴𝐴 − 𝐵𝐵 𝐴𝐴• The modulus of a complex number must be equal
or greater than its imaginary part.

35

36

3. Generalized statistical interpretation• Therefore,2𝛼𝛼 𝛽𝛽 − 𝛽𝛽 𝛼𝛼
𝐴𝐴𝐵𝐵 − 𝐵𝐵𝐴𝐴
𝛼𝛼 𝛽𝛽 ≥=2𝑖𝑖2𝑖𝑖
�• Use the commutator of 𝐴𝐴̂ and 𝐵𝐵.
̂ 𝐵𝐵� ≡ 𝐴𝐴̂ 𝐵𝐵� − 𝐵𝐵� 𝐴𝐴̂𝐴𝐴,22• This yields the generalized uncertainty principle.
2122
̂ 𝐵𝐵�
𝐴𝐴,𝜎𝜎𝐴𝐴 𝜎𝜎𝐵𝐵 ≥2𝑖𝑖
̂ 𝐵𝐵� = 0, 𝐴𝐴̂ and 𝐵𝐵� are compatible (commute /• If 𝐴𝐴,
are commutable / are commuting).

3. Generalized statistical interpretation

̂ 𝐵𝐵� ≠ 0, 𝐴𝐴̂ and 𝐵𝐵� are incompatible.
• If 𝐴𝐴,37• For example, the position-momentum uncertainty isℏ 𝑑𝑑 ℏ 𝑑𝑑ℏ 𝑑𝑑𝜓𝜓 𝑑𝑑 𝑥𝑥𝜓𝜓−−𝑥𝑥 𝜓𝜓 = 𝑥𝑥𝑥𝑥,
� 𝑝𝑝̂ 𝜓𝜓 = 𝑥𝑥𝑑𝑑𝑑𝑑
𝑖𝑖 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑑𝑑𝑑𝑑𝑖𝑖
𝑑𝑑𝑑𝑑ℏ 𝑑𝑑𝜓𝜓𝑑𝑑𝜓𝜓
− 𝜓𝜓 − 𝑥𝑥= 𝑖𝑖𝑖𝜓𝜓
=𝑥𝑥𝑖𝑖
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑ℏ∴ 𝜎𝜎𝑥𝑥 𝜎𝜎𝑝𝑝 ≥2• Uncertainty principle: position and momentum
cannot be measured accurately simultaneously.3. Infinite square well

• For example, in the case of 1D quantum well,
112 = 𝑙𝑙 2− 2 2𝜎𝜎𝑥𝑥
12 2𝑛𝑛 𝜋𝜋2 𝜋𝜋 2 ℏ2𝑛𝑛
𝜎𝜎𝑝𝑝 2 =𝑙𝑙 2• Therefore,ℏ 𝑛𝑛2 𝜋𝜋 2ℏ−2>𝜎𝜎𝑥𝑥 𝜎𝜎𝑝𝑝 =
32238

39

3. Generalized statistical interpretation• Suppose 𝐴𝐴̂ and 𝐵𝐵� have a shared set of eigenstates,
Ψ𝑛𝑛 .
̂ 𝑛𝑛 = 𝑎𝑎𝑛𝑛 Ψ𝑛𝑛𝐴𝐴Ψ
� 𝑛𝑛 = 𝑏𝑏𝑛𝑛 Ψ𝑛𝑛𝐵𝐵Ψ
• For an arbitrary state Ψ,̂ 𝐵𝐵� Ψ = 𝐴𝐴̂ 𝐵𝐵� − 𝐵𝐵� 𝐴𝐴̂ � 𝑐𝑐𝑛𝑛 Ψ𝑛𝑛𝐴𝐴,𝑛𝑛= � 𝑐𝑐𝑛𝑛 𝑎𝑎𝑛𝑛 𝑏𝑏𝑛𝑛 − 𝑏𝑏𝑛𝑛 𝑎𝑎𝑛𝑛 Ψ𝑛𝑛 = 0𝑛𝑛• 𝐴𝐴̂ and 𝐵𝐵� are compatible.

3. Generalized statistical interpretation

 Incompatible observables do not have a complete
set of shared eigenstates.
• Incompatible observables cannot be measured
accurately simultaneously.
• For example, 𝑆𝑆𝑧𝑧 and 𝑆𝑆𝑥𝑥 are incompatible.
ℏ 1 0 ℏ 0 1ℏ 0 1 ℏ 1 0𝑺𝑺𝑧𝑧 , 𝑺𝑺𝑥𝑥 =−2 0 −1 2 1 0
2 1 0 2 0 −1
ℏ2 0 2ℏ 0 −𝑖𝑖== 𝑖𝑖ℏ= 𝑖𝑖ℏ𝑺𝑺𝑦𝑦 ≠ 04 −2 02 𝑖𝑖 0
• If fact, angular momenta in 3 directions areincompatible with each other.

40

3. Generalized statistical interpretation

41• The eigenstates of 𝑆𝑆𝑧𝑧 are10↑=↓=01• The eigenstates of 𝑆𝑆𝑥𝑥 are
1 11 1𝛼𝛼 =𝛽𝛽 =2 1
2 −1• 𝑆𝑆𝑧𝑧 and 𝑆𝑆𝑥𝑥 do not have shared set of eigenstate.
• When 𝑆𝑆𝑧𝑧 is measured accurately, e.g., ↑, 𝑆𝑆𝑥𝑥 cannot
be measured accurately.
222ℏℏℏ−𝜎𝜎𝑆𝑆𝑥𝑥 2 =12− 02 2=444

3. Generalized statistical interpretation• Compatible observables may be measuredaccurately simultaneously.
• For example, the energy, total angular momentumand angular momentum in z direction of hydrogen
atom are compatible. They share a set ofeigenstates and can be measured accuratelysimultaneously.
• For example, the 1D infinite quantum well,𝑝𝑝̂ 2� 𝑝𝑝̂ =, 𝑝𝑝̂ = 0𝐻𝐻,2𝑚𝑚

42

3. Generalized statistical interpretation• However, the stationary states are𝜓𝜓𝑛𝑛 =

̂ 𝑛𝑛 =𝑝𝑝𝜓𝜓

2
𝑛𝑛𝑛𝑛
sin
𝑥𝑥
𝑙𝑙
𝑙𝑙

2 ℏ 𝑑𝑑
𝑛𝑛𝑛𝑛
sin
𝑥𝑥 =
𝑙𝑙 𝑖𝑖 𝑑𝑑𝑑𝑑𝑙𝑙

2 𝑛𝑛𝑛𝑛ℏ
𝑛𝑛𝑛𝑛
co𝑠𝑠
𝑥𝑥
3𝑙𝑙 𝑖𝑖𝑙𝑙• The stationary states are not eigenstates ofmomentum.
• 𝐸𝐸 and 𝑝𝑝 cannot be measure accurately
simultaneously.

43