The perfect quantal gas

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1 The perfect quantal gas Asaf Pe er 1 March 27, Background So far in this course we have been discussing ideal classical gases. We saw that the conditions for gases to be treated classically are that the the mean occupation number of the various states, 1,2,...,r,... is very small, namely n r 1 (for all r). Moreover, we demanded that the gas molecules don t interact with each other. In real life there are various complexities: e.g., The gases are non-ideal (or real classical gases). The gases may be relativistic (moving at speeds close to the speed of light). Quantum mechanical effects may be important in determining the occupation numbers. 2. Quantum statistics When treating the classical gas, we stated that each particle can be in different states, 1,2,...,r,... To each of these states there is an associated energy, E 1,E 2,...,E r,... For N particles, there are n 1 particles in state 1, n 2 particles in state 2, etc. The basic question is: what values can the occupation number n r assume? According to quantum mechanics, the answer to this question depends on the type of the particle. All the known particles in the world can be split into two categories: bosons and fermions Bosons The first type of particles obey Bose-Einstein statistics, and are known as bosons. For these particles: 1 Physics Dep., University College Cork

2 2 There is no restriction on n r, which can take any integer value: n r 0,1,2,... These particles have intrinsic angular momentum(spin) which is an integer number of : Spin,2,...,n. The most important example of such particles are photons. Other examples for such particles are π-mesons and K-mesons Fermions The second type of particles obey Fermi - Dirac statistics, and are known as Fermions. For these particles: The occupation number is restricted to only two options: n r 0,1. These particles have half-integer spin: ( 2, 3 2, 5 2,...). Many basic (massive) particles are fermions. These include electrons, protons, neutrons, positrons and others. It was shown by Pauli that there is a fundamental reason for the connection between the spin (half integer or integer) and the statistics (occupation number that is not restricted vs. occupation number that can be only 0 or 1): this relates for the need of quantum mechanics to be consistent with the special theory of relativity. This is one of the fundamentals of quantum field theory Symmetry properties In quantum mechanics (QM), a particle is described by a wave function (very similar to waves in sea!; hence the name wave mechanics ). The different behavior of bosons and fermions originate from the different symmetry properties of the wave functions. While the wave function Ψ(x) is not directly measured, according to QM treatment, the probability of finding a particle at a point x,,x+dx is related to its wavefunction: P(x)dx Ψ(x)Ψ (x)dx Ψ(x) 2 dx (1) Let us consider two bosons, which are in different states, say r and s. Each particle is described by a wave function: Ψ r (x 1 ), Ψ s (x 2 ).

3 3 The state of the combined two particle system is described by a wavefunction Ψ BE (x 1,x 2 ) Ψ r (x 1 )Ψ s (x 2 ). (2) However, since the bosons are identical, we also have Ψ BE (x 1,x 2 ) Ψ r (x 2 )Ψ s (x 1 ). (3) Thus, the correct wavefunction is symmetric with interchange of the particles, Ψ BE (x 1,x 2 ) Ψ BE (x 2,x 1 ) Ψ r (x 1 )Ψ s (x 2 )+Ψ r (x 2 )Ψ s (x 1 ) (4) Suppose now that both particles occupy the same state, say r. Thus, Ψ BE (x 1,x 2 ) Ψ r (x 1 )Ψ r (x 2 )+Ψ r (x 2 )Ψ r (x 1 ) 2Ψ r (x 1 )Ψ r (x 2 ) (5) The occupation number (in this example, n r 2) thus appears in the expression for the wavefunction of the system. If we add more bosons to the same state, the same thing happens and the occupation number further increases. Let us consider now two Fermions at states r and s. According to QM, the correct wavefunction is anti-symmetric with interchange of the particles, Ψ FD (x 1,x 2 ) Ψ FD (x 2,x 1 ) Ψ r (x 1 )Ψ s (x 2 ) Ψ r (x 2 )Ψ s (x 1 ) (6) As a result of this anti-symmetry property, if the 2 Fermions occupy the same state r, then Ψ FD (x 1,x 2 ) Ψ r (x 1 )Ψ r (x 2 ) Ψ r (x 2 )Ψ r (x 1 ) 0 (7) The wavefunction vanishes, indicating that such a state is impossible. 3. The partition function Consider a gas of N non-interacting identical particles(can be either bosons or fermions). The occupation numbers will be n 1,n 2,...,n r,... The total number of particles is N r n r, (8) and total energy of the system is E r n r E r. (9)

4 4 The possible occupation number differs for fermions and bosons, since Fermions : n r 0,1 Bosons : n r 0,1,2,... (10) Thus, for each type, there are many possible combinations of occupation numbers that satisfy the conditions set by Equations 8 and 9. By definition, the partition function is the sum over all possible sets of these occupation numbers. For a system in equilibrium at temperature T in an enclosure of volume V, it is Z(T,V,N) e β r nrer e β(n 1E 1 +n 2 E ) (11) n 1,n 2,... n 1,n 2,... The mean occupation number of state i, n i is obtained by taking the partial derivative of lnz with respect to E i and divide by ( β): n i 1 β ( ) lnz E i T,E r(r i)const n 1,n 2,... n ie β(n 1E 1 +n 2 E ) n 1,n 2,... e β(n 1E 1 +n 2 E ) (12) 3.1. Photon gas Let us take an example of particles that obey Bose-Einstein statistics: photons (you can think of black body radiation). Each particle (photon) has spin. As photons don t interactwitheachother, theycanbetreatedasidealgas. Afterenoughtime, thereisthermal equilibrium between the photon gas in the cavity and in the wall. Note that the number of photons in a cavity is not constant: thus, the condition N r n r is not fulfilled. Each state is thus occupied independently of the occupation of other states. The partition function can therefore be written as Z(T,V) n 1 0 n e β(n 1E 1 +n 2 E ) (13) In order to proceed, we need a mathematical trick. Let us look at a somewhat simpler, double summation: ( )( ) ( ) X e a 1n 1 +a 2 n 2 Π 2 r1 (14) n 1 0 n 2 0 e a1n1 e a2n2 e arnr n 1 n 2 n r

5 5 and if there are more than 2 components, (e.g., infinity)then ( n 1 n 2...e a1n1+a2n2+... Π r1 n r e arnr ). (15) Since n r takes integer values between 0 and, each term inside the brackets is a geometrical series with the sum 1 e arnr (16) 1 e ar n r0 This converges for a r < 0 (e ar < 1). Since we are interested in the partition function, for which a r βe r, we can use this result in Equation 13, to write Z(T,V) Π 1 r1 (17) 1 e βer The mean occupation number of state i is thus calculate using lnz(t,v) ln ( 1 e βer) (18) r1 and is n i 1 β (ln Z) E i 1 e βe i 1 (19) Finally, recall that the density of state for (classical) ideal gas is f(p)dp V4πp2 dp h 3, (20) where for photon, the momentum p is related to the frequency (E ω) via p ω c. (21) Since a photon have 2 polarizations, the density of states must be multiplied by 2. We thus get f(ω)dω 2 V4π( ω/c)2 /cdω Vω2 dω. (22) h 3 π 2 c 3 Thus, the number of photons at the frequency range ω..ω +dω is dn ω n ω f(ω)dω V π 2 c 3 ω 2 dω e β ω 1 (23)

6 6 and the energy density (energy per unit volume) of radiation in this frequency range is u(ω,t)dω ωdn ω V ω 3 dω π 2 c 3 (e β ω 1) (24) This is known as Planck s radiation law: it gives the distribution of energy density as a function of frequency for radiation in thermal equilibrium (also known as black-body radiation). The total energy density is given by integrating Equation 24 over all frequencies, u(t) 0 u)ω,t)dω at 4, (25) where the proportionality constant, a, known as the radiation constant is given by a ( ) 4 kb x 3 dx π 2 c 3 e x 1 π2 kb 4 (26) 15 3 c 3 (where in the integration we used x ω/k B T). 0 In the low frequency limit, ω k B T, we can expand the exponent in Equation 24 as e ω/k BT 1+ ω/k B T, and get u(ω,t)dω ω2 k B T dω (27) π 2 c2 This is known as Rayleigh-Jeans law. IT implies that at low frequencies, the energy density is linearly proportional to the temperature. In the opposite limit, ω k B T, we have e ω/k BT 1, and Equation 24 becomes u(ω,t)dω ω3 ω k BT dω. (28) π 2 c 3e This is known as Wien s law, which describe the exponential decay of the energy density at high frequencies Thermodynamic properties of a photon gas To conclude, it is nice to show a thermodynamic treatment of the photon gas. For this, we need to relate the pressure to the energy density. This is done using P u(t) 3 (29)

7 7 This result could be derived in more than one way, of course. A simple derivation is to recall that the momentum of photons (or relativistic particles) is related to their energy by E pc. The pressure exerted by N molecules of relativistic gas in a 3-d container of volume V on a wall is P 1 N 3 V p v 1 N 3 V E u 3. Using the fundamental thermodynamic relation TdS de +PdV with E Vu(T), we can write ds 1 T [u(t)dv +Vdu(T)+PdV] 1 T [ 4 u(t)dv +V 3 ( du(t) ) ] (30) From Equation 30 we get ( ) S V T 4u(T) 3T ; ( ) S T V V T du(t) (31) Using now 2 S/( T V) 2 S/( V T), we get [ ] 4u(T) T 3T V or from which we again find u(t) at 4. ( 4 1 du u 3 T 1 1 du 4 u 3 T 3 T 2 du 4u T [ V T ) T 1 2 T du ] du (32) (33) Using Equation 31, we can integrate and find the entropy, S V 4u(T) 3T S 4 3 at3 V +Const 4 3 at3 V (34) where the constant of integration must be equal to zero, for S to be proportional to V.

Ideal gases. Asaf Pe er Classical ideal gas

Ideal gases. Asaf Pe er Classical ideal gas Ideal gases Asaf Pe er 1 November 2, 213 1. Classical ideal gas A classical gas is generally referred to as a gas in which its molecules move freely in space; namely, the mean separation between the molecules