Semiconductor Physics Study Material (BBS00015) PDF

B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Study Material (Semiconductor Physics, BBS00015)...

B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Study Material (Semiconductor Physics, BBS00015) Module I part 3 Contents 1 Basics (Fundamental) Postulates of Quantum Mechanics.................................................................................... 1 2 Derivation of time independent Schr𝑜̈ dinger’s equation...................................................................................... 2 3 Derivation of time-dependent Schr𝑜̈ dinger’s equation........................................................................................ 3 4 Particle in a one-dimensional potential box with rigid walls................................................................................. 4 5 Annexure................................................................................................................................................................ 7 5.1 General solution of second order differential equation:................................................................................... 7 6 Exercise.................................................................................................................................................................. 8 6.1 Multiple-choice questions................................................................................................................................. 8 6.2 Short answer type questions............................................................................................................................. 9 6.3 Long answer type questions.............................................................................................................................. 9 1 Basics (Fundamental) Postulates of Quantum Mechanics I: Every physically-realizable state of the system is described in quantum mechanics by a wave function Ψ(𝑟⃗,t) that contains all accessible physical information about the system in that state. II: If a system is in a quantum state represented by a wave function Ψ(𝑟⃗,t) , then |Ψ(𝑟⃗,t)|2 dV = P dV (1) is the probability that in a position measurement at time t, the particle will be detected in the infinitesimal volume dV. III: Every observable in quantum mechanics is represented by an operator which is used to obtain physical information about the observable from the wave function. IV: If a system is in a state described by a normalized wave function, then the average value (or, expectation value) of the observable A corresponding to an operator 𝐴̂ is given by: ∞ ∗ ∫ ∞ Ѱ 𝐴Ѱ𝑑𝑉 〈𝐴〉 = −∞ ∗ (2) ∫−∞ Ѱ Ѱ𝑑𝑉 Note: In quantum mechanics, we talk in terms of the probability of finding the particle in a given element as it implies that the result of two successive measurements of the position of the particle may not yield the Prepared by the faculty members of the Physics department 1 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD same value. So we define the expected average or the expectation value of a dynamic variable (such as position coordinate), by considering a large number of independent measurements. V: The time development of the state functions of an isolated quantum system is governed by the time dependent Schr𝑜̈ dinger’s equation 𝜕Ѱ(𝑟⃗,𝑡) 𝑖ℏ ̂ Ѱ(𝑟⃗, 𝑡) =𝐻 (3) 𝜕𝑡 where, Hˆ is known as Hamiltonian operator. 2 Derivation of time independent Schr𝒐̈ dinger’s equation According to de Broglie’s theory, a particle of mass m moving with velocity v is always associated with a wave. If the particle has wave properties, it is expected that there should be some sort of wave equation that describes the behaviour of the particle. We consider one-dimensional wave propagation say along x and also assume that the wave function does not depend on time (so, time t is constant). If Ψ(x) be the wave function, we may write, Ψ(x) = exp{i(kx − ωt)}, (4) 2𝜋 where, k is the wave number(𝑘 = ) and ω is angular frequency. 𝜆 If λ is the wave length then from de Broglie theory, momentum ℎ 2𝜋 ℎ 𝑝= = ℏ𝑘 [𝑎𝑠, 𝑘 = ,ℏ = ] (5) 𝜆 𝜆 2𝜋 For a particle moving in a potential V (x), the total energy is given by, 1 𝑝2 ℏ2 𝑘 2 𝐸 = 2 𝑚𝑣 2 + 𝑉(𝑥) = 2𝑚 + 𝑉(𝑥) = 2𝑚 + 𝑉(𝑥) (6) Now we can write that, ℏ2k2 = 2m[E − V (x)] 2𝑚 or, 𝑘 2 = ℏ2 [𝐸 − 𝑉(𝑥)] (7) Now, if we differentiate the wave equation Eqn.4 w.r.t. x, 𝑑Ѱ = 𝑖𝑘𝑒𝑥𝑝{𝑖(𝑘𝑥 − 𝜔𝑡)} 𝑑𝑥 𝑑2 Ѱ or, 2 = −𝑘 2 exp{𝑖(𝑘𝑥 − 𝜔𝑡)} = −𝑘 2 Ѱ 𝑑𝑥 𝑑2 Ѱ 2𝑚 or, 𝑑𝑥 2 = − ℏ2 [𝐸 − 𝑉(𝑥)]Ѱ (8) ℏ2 𝑑 2 Ѱ or, − 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)Ѱ = 𝐸Ѱ ℏ2 𝑑 2 or, [− 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)]Ѱ = 𝐸Ѱ ℏ2 𝑑 2 or, [− 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)]Ѱ = 𝐸Ѱ So, the time independent Schr𝑜̈ dinger’s equation has the form, Prepared by the faculty members of the Physics department 2 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD ℏ2 𝑑 2 − 2𝑚 𝑑𝑥 2 𝑉(𝑥)]Ѱ(𝑥) = 𝐸Ѱ(𝑥) (9) In operator form, this can be written as, HˆΨ(x) = EΨ(x) (10) ℏ2 𝑑 2 where, ̂ = 𝑇̂ + 𝑉̂ = 𝐻 [− 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)] (For 1 dimensional case, along x- direction). 3 Derivation of time-dependent Schr𝒐̈ dinger’s equation Similar to the previous section, we are assuming that a particle of mass m is moving with velocity v and it must be associated with a wave. We consider one-dimensional wave propagation say along x and also assume that the wave function depends on time. If Ψ(x,t) be the wave function, we may write, Ψ(x,t) = exp{i(kx − ωt)}, (11) 2𝜋 where, k is the wave number(𝑘 = 𝜆 ) and ω is angular frequency. If λ is the wavelength then from de Broglie theory, momentum ℎ 2𝜋 ℎ 𝑝 = 𝜆 = ℏ𝑘 [𝑎𝑠, 𝑘 = 𝜆 ,ℏ = 2𝜋 ] (12) For a particle moving in a potential V (x), the total energy is given by, 1 𝑝2 ℏ2 𝑘 2 𝐸 = 2 𝑚𝑣 2 + 𝑉(𝑥) = 2𝑚 + 𝑉(𝑥) = 2𝑚 + 𝑉(𝑥) (13) Also, from, Planck’s postulate we get, ℎ 𝐸 = ℎ𝜈 = ℏ𝜔 [𝑎𝑠, 𝜔 = 2𝜋𝜈, ℏ = 2𝜋 ] (14) So, putting the value of E from Eqn.14 into Eqn.13, we get, ℏ2 𝑘 2 ℏ𝜔 = 2𝑚 + 𝑉(𝑥) (15) Now, if we differentiate the wave equation (Eqn.11) w.r.t. x, 𝜕Ѱ = 𝑖𝑘𝑒𝑥𝑝{𝑖(𝑘𝑥 − 𝜔𝑡)} (16) 𝜕𝑥 Again differentiating, 𝜕2 Ѱ 𝜕𝑥 2 = −𝑘 2 exp{𝑖(𝑘𝑥 − 𝜔𝑡)} = −𝑘 2 Ѱ (17) ℏ2 𝜕2 Ѱ ℏ2 2 − 2 = 𝑘 Ѱ (multiplying in both sides) 2𝑚 𝜕𝑥 2𝑚 Prepared by the faculty members of the Physics department 3 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Fig. 1: a particle in potential box extending from x = 0 to x = l. If we differentiate the wave equation (Eqn.11) w.r.t. t, 𝜕Ѱ = −𝑖𝜔 exp{𝑖(𝑘𝑥 − 𝜔𝑡)} = −𝑖𝜔Ѱ 𝜕𝑡 𝜕Ѱ 𝑖ℏ 𝜕𝑡 = ℏ𝜔Ѱ (18) or, Also, ℏ2 𝑘 2 ℏ2 𝑑 2 Ѱ [ 2𝑚 + 𝑉(𝑥)] Ѱ = − 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)Ѱ (19) Now, using eqn. 13, eqn. 18 and eqn. 19, we can write, 𝜕Ѱ ℏ2 𝑑 2 𝑖ℏ 𝜕𝑡 = [− 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)]Ѱ (20) So, the time-dependent Schr𝑜̈ dinger’s equation has the form, ℏ2 𝑑 2 𝜕Ѱ(𝑥,𝑡) [− 2𝑚 𝑑𝑥2 + 𝑉(𝑥)] Ѱ(𝑥, 𝑡) = 𝑖ℏ 𝜕𝑡 (21) In operator form, ̂ Ѱ(𝑥, 𝑡) = 𝑖ℏ 𝜕Ѱ(𝑥,𝑡) 𝐻 (22) 𝜕𝑡 𝜕 [Note: in time-dependent case, the operator form of 𝐸̂ = 𝑖ℏ 𝜕𝑡 4 Particle in a one-dimensional potential box with rigid walls Consider a particle of energy E is restricted to a region between x = 0 and x = l on the x-axis by infinitely hard walls (Fig.1). At x = 0 and x = l, there are two absolutely rigid, impenetrable potential walls of infinite height. This means, 0 (0 ≤ 𝑥 ≤ 𝑙) 𝑉(𝑥) = { ∞ elsewhere To leave the region from x = 0 to x = l, the particle would have to perform an infinitely large quantity of work. Since this is not possible, the probability of finding the particle in the region outside the box, i.e., x ≤ 0 and x ≥ l, must be zero. Since the probability is measured by the modulus squared of the wave amplitude, this means that the wave function Ψ(x) must be zero in this region. Thus, Ψ(x = 0) = 0 = Ψ(x = l) (23) Prepared by the faculty members of the Physics department 4 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Also, Ψ is zero outside the box. Also, we assume that this particle does not lose its energy when it collides with a such wall, so that its total energy remains constant within the box. Here, the potential energy is time-independent. So, we are using time-independent Schr𝑜̈ dinger’s equation (Eqn. 9) for the region 0 < x < l. ℏ2 𝑑 2 [− 2𝑚 𝑑𝑥 2 + 𝑉(𝑥)] Ѱ(𝑥) = 𝐸Ѱ(𝑥) (24) ℏ2 𝑑 2 − 2𝑚 𝑑𝑥2 Ѱ(𝑥) = 𝐸Ѱ(𝑥) 𝑎𝑠 𝑉(𝑥) = 0 Therefore, 𝑑2 Ѱ 𝑑𝑥 2 + 𝑘2Ѱ = 0 (25) 2𝑚𝐸 2𝑚𝐸 Where, 𝑘 2 = ℏ 𝑜𝑟, 𝑘 = √ ℏ (26) Now, the eqn. 25 has the solution of the form, (see the annexure at the later section) Ψ = Asin(kx) + B cos(kx) (27) where, A and B are arbitrary constants. Now one of the boundary (or initial) conditions is, Ψ = 0 at x = 0. Therefore, 0 = A × 0 + B × 1 as, sin0 = 0, cos0 = 1 or, B = 0 (28) Therefore, the general dorm of the wavefunction reduces to Ψ = Asin(kx) Another boundary condition is, Ψ = 0 at x = l, we get, 0 = Asin(kl) + 0 as, B=0 (29) or, Asin(kl) = 0 Now, A can not be zero, hence, sinkl = sinnπ, (30) where, n is an integer; n = 1,2,3,... etc. The value n = 0 is excluded as this will make Ψ = 0 everywhere. So, 𝑛𝜋 kl = nπ ; or, 𝑘 = 𝑙 𝑛𝜋 Ѱ𝑛 (𝑥) = 𝐴𝑛 sin ( 𝑙 𝑥) (31) and, From eqn.26, we get, the energy of the particle as, ℏ2 𝑘𝑛 2 ℏ2 𝑛2 𝜋2 𝐸𝑛 = = (32) 2𝑚 2𝑚𝑙 2 Prepared by the faculty members of the Physics department 5 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Fig. 2: different energy and wave functions up to n=3. So, for different values of n, we will get different values of energy (Fig. 2)(b). Now, the wave functions can be normalised following the procedure. 𝑙 ∫ |Ѱ𝑛 |2 𝑑𝑥 = 1 0 𝑙 𝑛𝜋𝑥 or, |𝐴𝑛 |2 ∫0 𝑠𝑖𝑛2 𝑑𝑥 =1 𝑙 𝑙 1−𝑐𝑜𝑠2𝑛𝜋𝑥/𝑙 or, |𝐴𝑛 |2 ∫0 𝑑𝑥 = 1 (33) 2 2 𝑥 𝑠𝑖𝑛2𝑛𝜋𝑥/𝑙 𝑙 or, |𝐴𝑛 | [2 − 4𝑛𝜋/𝑙 ] =1 0 𝑙 or, |𝐴𝑛 |2 2 = 1 2 or, |𝐴𝑛 |2 = 𝑙. So, the normalised wave function has the form, 2 𝑛𝜋𝑥 Ѱ𝑛 = √ sin ( ) (34) 𝑙 𝑙 The wave functions for the first three values of n are shown in Fig.2. It is evident that for Ψ1 has two nodes at x = 0 and x = l. Ψ2 has three nodes at x = 0 and x = l/2 and x = l. Ψ3 has four nodes at x = 0, x = l/3 x = 2l/3 and x = l. The probability of finding the particle over a small distance dx at x is given by, 2 𝑛𝜋𝑥 𝑃(𝑥)𝑑𝑥 = |Ѱ𝑛 |2 𝑑𝑥 = 𝑠𝑖𝑛2 𝑑𝑥 (35) 𝑙 𝑙 Thus the probability density for one dimension is Prepared by the faculty members of the Physics department 6 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD 2 𝑛𝜋𝑥 𝑃(𝑥) = 𝑙 𝑠𝑖𝑛2 𝑙 (36) 𝑚𝜋 The value of the sine function is maximum at , where m is odd. So, we 2 𝑛𝜋𝑥 𝜋 3𝜋 5𝜋 can write, 𝑙 = 2 , 2 , 2 … 𝑙 3𝑙 5𝑙 Or, 𝑥 = 2𝑛 , 2𝑛 , 2𝑛 … (37) 𝑙 Thus for n = 1, the most probable position is at 𝑥 = 2 𝑙 3𝑙 For, n = 2, the most probable position is at 𝑥 = 4 , 𝑥 = 4 𝑙 3𝑙 5𝑙 For, n = 3, the most probable position is at 𝑥 = 6 , 𝑥 = 6 and 𝑥 = 6. 5 Annexure 5.1 General solution of second order differential equation: Here, an example of second order differential equation is given, we need to find out its general solution: 𝑑2 Ѱ 𝑑𝑥 2 + 𝑘2Ѱ = 0 (38) To solve the equation we consider a trial solution, Ψ = 𝑒 𝑚𝑥 , (39) where m is a constant. 𝑑Ѱ Now, differentiating the equation w.r.t. x, we get, 𝑑𝑥 = 𝑚𝑒 𝑚𝑥 𝑑2 Ѱ 𝑑𝑥 2 = 𝑚2 𝑒 𝑚𝑥 (40) 𝑑2 Ѱ Now, putting the values of 𝑑𝑥2 and Ψ in eqn.38, we get, 𝑚2 𝑒 𝑚𝑥 + 𝑘 2 𝑒 𝑚𝑥 = 0 (m2 + k2) 𝑒 𝑚𝑥 = 0 (41) 𝑚𝑥 Now, 𝑒 ≠ 0 otherwise solution will be zero. Hence, m + k2 must be zero 2 or, m2 = −k2 (42) or, m = ±ik (where,i is imaginary number, i =√−1) So, the general solution will be, Prepared by the faculty members of the Physics department 7 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD Ψ = 𝐶1 𝑒 𝑖𝑘𝑥 + 𝐶2 𝑒 −𝑖𝑘𝑥 , (43) where, C1, C2 are arbitrary constants. This solution can also be written as, Ψ = C1(coskx + isinkx) + C2(coskx − isinkx) (44) or, Ψ = Acoskx + B sinkx where, we put A = C1 + C2 and B = i(C1 − C2) are arbitrary constants. 6 Exercise 6.1 Multiple-choice questions 1. Choose the correct statement about the stopping potential of particles and of waves. This phenomenon can be explained as — a) wave-particle duality b) mixing c) confusion d) entanglement 2. Max Planck’s great discovery was that radiation energy is emitted in packets. That packet of energy of is also recognised as — a) photons b) gamma rays c) quanta d) wave functions 3. Choose which pair of observables of a particle cannot be measured precisely at the same time. a) spin and color b) energy and torque c) position and momentum d) size and speed 4. The square of the modulus of a particle’s wave function describes the probability of that — a) it will disappear b) it will decay c) it is at a particular place d) it has a specific spin 5. Identify the correct relation between total energy [E] and temperature [T] in Stefan-Boltzmann law. a) E ∝ T 4 b) E ∝ T − 4 c) E ∝ T d) E ∝ T−1 6. Identify the correct momentum-energy relation. a) 𝐸 = √𝑝2 𝑐 2 + 𝑚02 𝑐 4 b) 𝐸 = √𝑝2 𝑐 2 − 𝑚02 𝑐 4 𝑐) 𝐸 = √𝑝2 𝑐 2 + 𝑚𝑜 𝑐 4 d) 𝐸 = √𝑝𝑐 + 𝑚𝑜 𝑐 4 7. In Relativistic case, as the velocity of the particle approaches the speed of light, the predicted kinetic energy will approach — a) Zero b) kinetic Energy as in Non-Relativistic case c) rest energy d) infinite Prepared by the faculty members of the Physics department 8 Brainware University, Kolkata B.Tech CSE-DS, B.Tech CSE-AIML, B.Tech CSE-CYS 2024 and Semester-I Semiconductor Physics (BBS00015) Class 2024-25 ODD 8. A particle of rest mass m0 moves with speed 0.8c, where c is the speed of light in vacuum. Calculate the relativistic kinetic energy of the particle? a) 1.66m0c2 b) m0c2 c) 0.32m0c2 d) 0.66m0c2 9. Calculate the energy equivalent of mass of an electron. a) 0.51 MeV b) 0.51 meV c) 0.51 J d) 9.11 J 10. A perfectly black body can be classified as a body that — a) absorbs all the incident radiation b) allow all the incident radiation to pass through it c) reflects all the incident radiation d) has its surface coated with lamp black or graphite 6.2 Short answer type questions 1. Construct Wien’s law for blackbody radiation from Planck’s radiation formula in the short wavelength limit. 2. Construct Rayleigh-Jean’s law for blackbody radiation from Planck’s radiation formula in the short wavelength limit. 3. Describe the terms threshold frequency and stopping potential in connection to the photoelectric effect. 4. Explain wave-particle duality. Compute the expression of de Broglie wavelength associated with a particle of mass m moving with velocity v. 5. Describe the physical interpretation of wave-function in quantum mechanics. 6. How does phase velocity differ from group velocity? 7. What is the physical significance of group velocity? 8. With the help of uncertainty principle, infer that no electron exists in the nucleus of an atom. 9. Establish Stefan’s law from Planck’s radiation formula. 10. Derive Wien’s displacement law from Planck’s radiation formula for blackbody radiation. 6.3 Long answer type questions 1. Describe the fundamental postulates of quantum mechanics. 2. Deduce time independent form of Schr𝑜̈ dinger equation. 3. Deduce time dependent form of Schr𝑜̈ dinger equation. 4. Write down Schr 𝑜̈ dinger’s equation for a free particle in a one-dimensional potential box. Applying appropriateboundary conditions determine the wavefunction of the particle. 5. Write down Schr 𝑜̈ dinger’s equation for a free particle in a one-dimensional potential box. Applying appropriateboundary conditions calculate its eigen energies. Answers for multiple-choice questions 1.a 2.c 3.c 4.c 5.a 6.a 7.d 8.d 9.a 10.a Prepared by the faculty members of the Physics department 9 Brainware University, Kolkata