Lecture 1529.10.2019

Helmholtz free energy and partition function

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Plan of the day...• Review the definitions of Boltzmann distribution and partition function• Why they are basic concepts in statistical mechanics

• Relate the Boltzmann distribution with entropy

• Relate partition function with Helmholtz free energy

• Apply these concepts in two examples: • Paramagnetic spins

• Quantum harmonic oscillators

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Boltzmann distributionA system in contact with a thermal bath at temperature T can be in many microstates, all corresponding to the same temperature. However, unlike the equally probably microstates ofan isolated system (because they have the fixed energy), the available microstates at constanttemperature are not equally likely (because they have different energies). The probability to be in a given microstate at temperature T is described by the Boltzmann distribution

𝑃 𝑠 =1𝑍𝑒'

( )*+ , 𝑠 ≡ 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑎𝑡𝑒, 𝐸 𝑠 ≡ 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑎𝑡𝑒 𝑒𝑛𝑒𝑟𝑔𝑦

This probability is normalized, meaning that ∑) 𝑃(𝑠) = 1. This determines the definition of thepartition function Z as

𝑍(𝑇) = ∑) 𝑒'(())/*+

𝑍(𝑇) «counts» the available microstates at a given temperature by weighting each microstate

with the Bolztmann factor 𝑒'> ?@A

Boltzmann distribution 𝑃 𝑠 = BC𝑒'

> ?@A

• There is higher probability that the system will occuly a microstate with the lowerenergy (ground state has the highest probability)

• The probability that the system is in a high-energy microstate increases withincreasing temperature

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Boltzmann distribution, so far---So far, we have discussed the Boltzmann distribution of a system composed of a single particle in a thermalbath

In this case, the microstates are typically labelled by the discrete energy levels that the particle can occupy. And the energy levels are determined by the specific Hamiltonian of the particle.

𝑃 𝑠 =1𝑍B

𝑒'( )*+ , 𝑠 ≡ 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑒𝑣𝑒𝑙, 𝜖 𝑠 ≡ 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑙𝑒𝑣𝑒𝑙

The one-partition function 𝒁𝟏 is defined as

𝑍B(𝑇) = ∑) 𝑒'(())/*+

𝑍(𝑇) «counts» the available microstates at a given temperature by weighting each microstate with the

Boltzmann factor 𝑒'> ?@A

Boltzmann distribution 𝑃 𝑠 = BC𝑒'

> ?@A

• There is higher probability that the particle will occuly an energy level with the lower energy(ground state has the highest probability)

• The probability that the system is in an high-energy level increases with increasingtemperature

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Why is Boltzmann distribution important? We want to derive thermodynamic variables and thermodynamic relations from this statistical microscopic

description. So, Boltzmann distribution 𝑃 𝑠 = BC𝑒'

> ?@A does this in two important ways:

1. It allows us to identify some thermodynamic variables as statistical averages. For example: the internal energy 𝑈 of the system at constant temperature is identified as the average energy ⟨𝐸⟩ of all the availablemicrostates

𝑈 𝑇 ≡ ⟨𝐸⟩ (𝑇) =N)

𝐸 𝑠1

𝑍(𝑇) 𝑒' ( )

*+

This average quantity can be directly computed from the partition function 𝑍(𝑇)

⟨𝐸⟩(𝑇) = −1

𝑍 𝛽𝑑𝑍𝑑𝛽

, 𝛽 =1𝑘𝑇

2. As the probability of a microstate, it defines the entropy in a more general way as

𝑆 = −𝑘 N)

𝑃 𝑠 ln𝑃(𝑠)

OBS! The Boltzmann definition of entropy applies only for isolated systems, where 𝑃 𝑠 = BV(W)

𝑆(𝑈) = −𝑘 N)

1Ωln1Ω= 𝑘 lnΩ(𝑈)

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Why is partition function 𝒁 important? The partition function 𝑍(𝑇) = ∑) 𝑒'(())/*+ is a fundamental link between the microscopic states and the macroscopic (thermodynamic) state. It is important in two way:

1. If we know how the partition function depends on temperature, we can determine the internalenergy as average energy, and estimate the importance of fluctuations by computing the mean-square energy

𝑈 𝑇 = −1

𝑍 𝛽𝑑𝑍𝑑𝛽 , ⟨𝐸Y⟩(𝑇) =

1𝑍 𝛽

𝑑Y𝑍𝑑𝛽Y

1. The partition function Z(𝑇) defines the Helmholtz free energy 𝐹(𝑇), which is the thermodynamicpotential that is minimized when the system is at equilibrium with a thermal bath. The Helmholtzfree energy determines all the thermodynamic properties of the system.

𝑆 = −𝜕𝐹𝜕𝑇 ],^

, 𝑃 = −𝜕𝐹𝜕𝑉 +,^

, 𝜇 =𝜕𝐹𝜕𝑁 +,]

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• Foranisolated system,we know everything about the thermodynamic state of thesystemif we know its entropy.Andits entropy isdetermined bythe multiplicity of themacrostate through Boltzmann’s formula

𝑆(𝑈) = 𝑘 lnΩ(U)

• Byanalogy,the Helmholtz free energy F T would depend on ln Z T . We what that 𝐹 𝑇 decreases towards equilibrium, while ln 𝑍 𝑇 increases towards equilibrium. Hence, by an appropriate match of dimensions, we can define

𝐹 𝑇 = −𝑘𝑇 ln 𝑍(𝑇)

This can be rigurously derived using statistical ensembles theory and Laplace transform.

Let’s check this equation

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𝑀𝑎𝑐𝑟𝑜𝑠𝑡𝑎𝑡𝑒𝑇, 𝑉, 𝑁𝑇

Partition function and Helmoltz free energy

From the Legendre transform, we determine the entropy as:

𝐹 = 𝑈 − 𝑇𝑆 → 𝑆 =𝑈 − 𝐹𝑇

(1)

On the other hand the entropy is also defined as a derivative of the Helmholtz free energy withrespect to temperature (conjugate variable)

𝑆 = − ���+ ],^

(2)

From (1) and (2): ���+ ],^

= �'�+

• Verify that this equation is also satisfied by the free energy �𝐹 = −𝑘𝑇 ln 𝑍 :

𝜕 �𝐹𝜕𝑇

= −𝑘 ln 𝑍 − 𝑘𝑇𝜕 ln 𝑍𝜕𝑇

=�𝐹𝑇+ 𝑘𝑇

1𝑘𝑇Y

𝜕 ln 𝑍𝜕𝛽

=�𝐹𝑇−𝑈𝑇→ 𝐹 = �𝐹 + 𝐶𝑜𝑛𝑠𝑡.

At 𝑇 = 0𝐾: 𝐹 0 = 𝑈�, Z� = e'� �� → �𝐹 0 = 𝑈� = 𝐹 0 → 𝐶𝑜𝑛𝑠𝑡 = 0𝑭 𝑻 = −𝒌𝑻 𝒍𝒏𝒁(𝑻) √

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Consistency check of the formula 𝐹 𝑇 = −𝑘𝑇 ln 𝑍(𝑇)

Consistency check: Bolztmann distribution, entropy and Helmholtz free energy

Given the probability of a microstate 𝑃 𝑠 , the entropy is defined in general by

𝑆 = −𝑘 N)

𝑃 𝑠 ln𝑃(𝑠)

For the Boltzmann distribution

𝑆 = −𝑘 N)

𝑃 𝑠 ln1𝑍𝑒'

( )*+

= 𝑘 N)

𝑃 𝑠 ln 𝑍 +1𝑇N)

𝑃 𝑠 𝐸(𝑠)

= 𝑘 ln 𝑍 +𝑈𝑇= −

𝐹𝑇+𝑈𝑇→ 𝐹 = 𝑈 − 𝑆𝑇 √

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Entropy

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𝑇, 𝑉, 𝑁𝑇

𝑃(𝑠) = BVisconstant

Ω(U) = 𝑡𝑜𝑡𝑎𝑙 # 𝑜𝑓 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑎𝑡𝑒𝑠 𝑎𝑡 𝑈

𝑆 = −𝑘 N)

𝑃 𝑠 ln𝑃 𝑠

𝑆 = 𝑘 N)

1Ω lnΩ = 𝑘

1Ω lnΩN

)

(1)

𝑆 = 𝑘 lnΩ

𝑃 𝑠 =1𝑍 𝑒

'�(()), 𝑍 =N)

𝑒'�(())

𝑆 = −𝑘 N)

𝑃 𝑠 ln𝑃 𝑠

𝑆 =1𝑇 𝐸 −

1𝑇 𝐹 → 𝐹 = 𝐸 − 𝑇𝑆

Isolated system System in a thermal bath

𝑃 𝑠 isthe probability that the systemisinagivenmicrostate with energy 𝐸)

𝑺 = −𝒌 ∑𝒔𝑷 𝒔 𝐥𝐧𝑷(𝒔)

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Statistical mechanics

Equilibrium thermodynamic state

Ω� 𝑈 number of equally-likelyaccessible microstates

𝐒 𝐔 = 𝐤 𝐥𝐧 𝜴𝑵 𝑼

𝑇 =𝜕𝑆𝜕𝑈

'B

],^

𝐹 = 𝑈 − 𝑇𝑆

𝑍 𝑇 = ∑) 𝑒'�(? counts the microstates at T (not equally-likely)

𝐅 = −𝐤𝐓 𝐥𝐧𝒁 𝑻

𝑈 ≡ 𝐸 = −𝜕 ln𝑍𝜕𝛽 ],^

𝑆 =𝑈 − 𝐹𝑇

𝑈 𝑇 , 𝑃(𝑉, 𝑇)𝑇

U isfixed𝑆 𝑈 , 𝑇, 𝑃(𝑉, 𝑇)

S, U, T, P, V..

Isolated system System in a thermal bath

Example 1: Paramagnet spin in a thermal bath

• 1spin attemperature Tcan beintwo energy levels 𝜖↑ = −𝜇𝐵 or𝜖↓ = 𝜇𝐵.Hence,the one-particle partition function is

𝑍B 𝑇 = 𝑒'©↑*+ + 𝑒'

©↓*+

𝑍B 𝑇 = 𝑒ª«*+ + 𝑒'

ª«*+ = 2 cosh

𝜇𝐵𝑘𝑇

= 2 cosh 𝜇𝐵𝛽 , 𝛽 =1𝑘𝑇

• Average energy of 1 spin

𝜖 =1𝑍B

𝜖↑𝑒'�©↑ + 𝜖↓𝑒'�©↓ = −1𝑍B𝜕𝑍B𝜕𝛽

= −𝜇𝐵 tanh(𝜇𝐵𝛽)

• Average energy of N independent spins (additive function)

𝑈 = 𝑁 𝜖 = −𝜇𝐵𝑁 tanh(𝜇𝐵𝛽)𝑇 ↗ ∞: 𝑈 ∼ −𝑁𝜇𝐵 𝜇𝐵/𝑘𝑇.Spinsare pointing randomly (average energy dominted bythe exited states)

𝑇 ↘ 0: 𝑈 ∼ −𝑁𝜇𝐵Spinsare pointing inthe samedirection (average energy dominated bythe ground state)

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𝑇 𝐵

Example 1: Paramagnet in a thermal bath

Helmholtz free energy of one spin

𝐹B 𝑇 = −𝑘𝑇 ln 𝑍B = −𝑘𝑇 ln 2 𝑐𝑜𝑠ℎ 𝛽𝜇𝐵

Helmholtz free energy of N independent spins(additive function)

𝐹 𝑇 = 𝑁𝐹B 𝑇 = −𝑁𝑘𝑇 ln 2^ cosh^ 𝛽𝜇𝐵

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Example 1: Paramagnet in a thermal bath

Entropy

𝑆 =𝑈 − 𝐹𝑇

𝑆(𝑇) = −𝑁𝛽𝜇𝐵 tanh 𝜇𝐵𝛽 + 𝑁𝑘 ln 2 𝑐𝑜𝑠ℎ 𝛽𝜇𝐵

𝑇 ↗ ∞: 𝑆 ∼ 𝑁𝑘 ln 2.Spinsare pointing randomly (randomstate,no effect of the magnetic field)

𝑇 ↘ 0: 𝑆 ∼ 0.Spinsare pointing inthe samedirection (ordered state)

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Example 2: harmonic oscillator in a thermal bath (model of thermal vibrations in solids)

• One harmonic oscillator has the available energy levels: 0, 𝜖, 2𝜖, 3𝜖,⋯ , where 𝜖 = ℏ𝜔

• One-particle partition function

𝑍B =N)

𝑒'�©? = 1 + 𝑒'�© + 𝑒'Y�© + 𝑒'µ�© ⋯ =1

1 − 𝑒'�©

• Average energy of one harmonic oscillator at energy 𝑇𝐸B = −� ¶· C¸

��= � ¶· B'¹º»¼

��= ©¹º»¼

B'¹º»¼

• Average energy of N harmonic oscillators at energy 𝑇𝑈 = 𝑁 𝐸B = ^©

¹»¼'B

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Example 2: harmonic oscillator in a thermal bath

Helmholtz free energy of one oscillator 𝐹B = −𝑘𝑇 ln 𝑍B = 𝑁𝑘𝑇 ln 1 − 𝑒'�©

Helmholtz free energy of N independent oscillators 𝐹 = 𝑁 𝐹B = 𝑁𝑘𝑇 ln 1 − 𝑒'�©

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Example 2: harmonic oscillator in a thermal bath

Entropy

𝑆 =𝑈 − 𝐹𝑇

𝑆 = 𝑁𝑘𝛽𝜖

𝑒�© − 1− ln 1 − 𝑒'�©

𝑇 ↗ ∞: 𝑆 ∼ 𝑁𝑘 ln𝑘𝑇𝜖

∼ 𝑁𝑘 ln𝑈𝑁𝜖

Entropy increases logarithmically with the number of energy units.Multiplicity of the macrostate isalso dominated bythe energy units

𝑇 ↘ 0: 𝑆 ∼ 0.Allthe oscillatorsare inthe ground state(ordered state)

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What we have learned...

• Boltzmann distribution :

• the probability that the system is an given microstate, 𝑃 𝑠 = BC𝑒'

>?@A.

• determines the thermodynamic variables as statistical averages, 𝐸 = ∑) 𝐸) 𝑃 𝑠

• defines entropy generally as 𝑆 = −∑) 𝑃 𝑠 ln 𝑃(𝑠)

• Partition function

• Counts the microstates weighted by the Boltzmann’s factor 𝑍 𝛽 = ∑) 𝑒'�( ) , 𝛽 = B*+

• Determines thermodynamic variables as appropriate derivates, e.g. 𝑈 𝑇 = − BC �

¿C¿�

• Determines the equilibrium Helmholtz free energy, F T = −𝑘𝑇 ln 𝑍(𝑇)

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