What are free particles in quantum mechanics
- 1. What are free particles in quantum mechanics The free particle is the simplest situation to apply the Schrodinger equation.However the physics is very interesting and there are many important consequences of the result. 1 What is a free particle? A particle is said to be free particle when it moves in a region where the potential is a constant. i.eV(x)=Constant .In such a case the force acting on the particle is F = dV dx = 0 we are considering here the one dimesnional case. 2 What is the difference between a free particle and a particle in box or a particle in infinite square well ? The particle in a box and particle in infinite square well are both examples where the particles are moving in constant potential V (x) = 0.However the difference is that they are bounded by the walls of the box and and 1
- 2. the well. The free particle is really free so that there are no boundary conditions. 3 How does a free particle behave in the clas- sical case In the classical situation the free particle experiences no force and remains in a state of rest or constant motion. 4 How can we study the behavior in the quan- tum mechanical situation? To study the quantum mechanical situation we need to solve the one dimensional time independent Schrodinger equation. 5 What will be the form of Schrodinger equa- tion for a free particle? The Schrodinger equation will be written with V = 0 and will looklike as shown below − 6h2 2m d2ψ dx2 = Eψ —(1) 2
- 3. 6 How to solve this equation? it looks compli- cated!!! Absolutely not, it is a differential equation and can be solved easily.For making the things simple we need to substitute some parameters. We make the following re- placements k = √ 2mE 6h —(2) so that our equation (1) becomes d2ψ dx2 = −k2 ψ—(3) 7 Are these simply mathematical steps or is there any physics in these steps? Of course these mathematical steps have a physical mean- ing. See carefully the equation (2). Equation (2) is the De-Broglie wavelength of the particle. In order to understand the quantum mechanical behavior of the particle we have to understand the wave nature of the particle. 3
- 4. 8 The solution of the differential equation (1) can be written in exponential form or in the form of sines and cosines ? The solution will be of the form Ψ(x) = ψ(x)e −iEt 6h —(4) where Ψ(x) is the solution and ψ(x) are the eigen functions.It is right that we can use the exponential and sine cosine solutions both. However for the time being we consider the exponential solution because the free particle is unbound. The sine and cosine func- tion are useful for the infinite square well and box po- tential where there are boundaries. In equation (4) the quantity E defines the energy of the particle. The energy values are called eigenvalues. for an unbound solution the acceptable solutions exists for all energy values E ≥ 0.This physically means that there are unrestricted values of the Energy and wave number K. 4
- 5. 9 What physical information can be obtained using the exponential solution ? well let us see what the solution looks like and what in- formation it gives regarding the particle. we write the equation in exponential form Ψ(x, t) = Aeikx + Be−ikx —(5) In equation (5) we have space dependence but the time dependence is missing so we have to include the time dependence Ψ(x, t) = Aeik(x−6hkt 2m ) + Be−ik(x+6hkt 2m ) —(6) Now the quantity 6hkt 2m must be creating a confusion.Lets see what it is The time dependence is included in e −iEt 6h The quantity 6hkt 2m has to be broken to get the meaning 6h 2m × √ 2mE 6h –(7) equation (7) simplifies to v . The velocity of the par- ticle. Now our solution looks like Ψ(x, t) = Aeik(x−vt) + Beik(x+vt) —(8) It can be easily seen that the equation (8) rep- resents a traveling wave. + sign represents wave traveling from left to right and + sign represents wave travelling from fight to left 5
- 6. Figure 1: free particle traveling wave 10 How can I visualize the wave? The wave described above is a traveling wave. A particular point on the wave represents a fixed value of the argument (x − vt). The condition of the travelling wave is that the profile of the wave remains same as it propagates This can be mathematically shown as the argument x − vt is a constant. x ± vt = constant or x = ∓vt + constant The traveling wave is shown in figure (1) 6
- 7. 11 What are the other properties of the wave like speed, wavelength etc Remember the wave is the matter wave or De-Broglie wave that is associated with the particle.However there are some interesting points. First let us consider the direction of the wave, the direction of wave is decided by the sign of the wave vector k k = √ 2mE 6h —(9) k > 0wavemovingright k < 0wavemovingleft thus we can write the solution in terms of k as Ψ(x, t) = Aeik(x−6hk2t 2m ) —(10) Comparing with a general wave equation we can see that that the quantity 6hk2 2m is analogous to frequency of the wave . The wavelength is λ = 2π k –(11) . The momentum carried by the wave is given by the De Broglie equation. p = 6h p But the veloc- ity has a discrepancy. The speed of the wave isgiven by vquantum = 6hk 2m–(12). Equation (12) is a general expression for finding the speed of any travelling wave. In classical case the speed of a moving free particle in terms of its energy which is totally Kinetic is 7
- 8. 1 2mv2 = E—-(12) or vclassical = q 2E m –(13) 12 The result is bizarre the classical and quan- tum velocities are different. How to explain this? Yes the results are different vclassical = 2vquantum.I hope this problem occurred earlier in this section also.Physically this means that the quantum mechanical wave moves at a half a speed than the particle. 13 This means that the solution obtained is of no use? we need to find other solution No it is not so. The solution (8) is still a solution of the Schrodinger equation. ψ(x) = Aeikx + Be−ikx d2ψ dx2 = −k2 ψ(x) = 2mE 6h2 ψ(x) so no doubt equation (8) is the solution but the solution is not normalizable The reason is that the particle is free it has no bound- aries. When the solution is not normalizable then it means that the free particle can not have steady solutions with fixed energy. If we try to normal- ize the wave function by conventional method 8
- 9. R +∞ −∞ Ψ∗ (x)Ψ(x)dx = A2 (∞)—(14) Physically this means that there cannot be a free particle whose energy is fixed or quan- tized or Quantization of energy occurs in bounded sys- tems only 14 So if the solution is not normalizable, then the free particle case is closed here and no further study is required? No, in physics when we have no solution by conventional method then we find new methods with new physics. The separable solutions are still useful.The solution of the time dependent Schrodinger equation is still a linear combination but in a different way. In case of bounded systems we have the index n that characterizes differ- ent energy levels. In this case of a free particle we have traveling waves moving from left to right and right to left. Each such wave is characterized by the quantity k. Soinstead of writing Ψ(x, t) = cnψ(x, t) we write ψ(x, t) = 1 √ 2π R +∞ −∞ φ(k)ei(kx−6hk2 2m t) dk—(14) 9
- 10. 15 What is meaning of this equation and what physical information it carries? The equation contains vital information rich in physics.The main features of the equation (14) are • The factor 1 √ 2π is for mathematical convenience and plays the role of cn • The variable over which the integration is carried over the variable . • The range of k which is also the range of energy and range forms an entity called wave packet The general problem then becomes finding the form of the wave function at a time t when the wave function at t=0 is given . Now concentrate on the physics here . We saw that the integration is carried over the variable k which has different range of energies and momentum. So finding the wave function at a later time involves determinig the wave packet function φ(k) of equation (14).This means we need to evaluate the following φ(k) = 1 √ 2π R +∞ −∞ ψ(x, 0)e−ikx dx—-(15) 16 Where does the equation (15) come from? Equation (15 ) comes from equation (14) 10
- 11. R +∞ −∞ Ψ∗ (x)Ψ(x)dx = A2 (∞)—(14) by the application of a mathematical tool Plancherel’s theorem which is the result from Fourier’s analysis. If we have a function f(x) = 1 √ 2π R +∞ −∞ F(k)eikx dx –(16) equation (16) physically means that we are shifting from x(coordinated space to k space) F(k) is called the Fourier Transform of f(x). f(x) is called the inverse Fourier transform of F(k) 17 Lets solve a problem to see what we have learnt till now? Suppose we have a free particle, which is initially con- fined in the range −a < x < a.The particle is released at time t=0. At time t = 0 the wave function is as follows Ψ(x, 0) = Aif − a < x < a Ψ(x, 0) = 0 if A and a are constants then find Ψ(x, t) Thefirst thing we need to do is to normalize Ψ(x, 0) . This we can do using the conventional method of nor- malization R +a −a |Ψ(x, 0)|2 dx = |A|2 R +a −a dx = 2a|A|2 = A = − 1 √ 2a — -(17) so we have evaluated the constant, which is possible 11
- 12. only because the free particle is confined in a boundary. Now we will apply the Plancherel’s theorem and use equation (15) φ(k) = 1 √ 2a 1 √ 2π R +a −a eikx dx = 1 2 √ pia e−ikx −ik |+a −a = − 1 √ aπ sin(ka) k — (18) with this value of φ(k)we can find the solution Ψ(x, t) as follows Ψ(x, t) = 1 √ 2a R +∞ −∞ sin(ka) k ei(kx−6hk2 2m t) dk—(19) In the equation (19) the quantity of significance which is encountered in many topics of physics is sin(ka) a . This equation is perhaps in optics under the heading of diffraction . 18 What is the significance of this equation and the quantity involving sine? The quantity sin(ka) a is very important in terms of physics. Mathematically the equation is not very easy to solve but let us see some limiting cases. • a very small: If the size of the boundary a is very small then sin(ka) ≈ ka and φ(k) = pa π—(20) the equation (20) is flat curve which is due to uncer- tainty principle. a is position and k is momentum. If spread in a is large then the spread in k is small. • If a is very large,the spread in position in large then 12
- 13. the spread in momentum k is very small. In this case φ(k) = pa π sin(ka) ka .Now from mathematics we know that a function of type sin(x) x is minimum at x=0 and drops to zero at x = ±π. The mathematical result is consistent with the Uncertainty principle which says forlarge a the spread in k will be small. 19 what is the graphical representation of the concept explained above? The diagrammatic representation of the concept dis- cussed above is given below In the figure the rectangle shows the variation of prob- ability density with x at time t = 0 the curve shows the probability density at time t = ma2 6h 20 what is the physical significance of the quan- tity Ψk(x, 0)? The quantity as Ψk(x, 0) as usual has no physical significance,it is the square of Ψ(x, 0) that has physi- cal meaning.What we need is information regarding the particle which cannot be directly obtained from Ψk(x, 0) which is the wave packet.We need to find information regarding the speed of the speed of the particle from 13
- 14. Figure 2: Figure showing the variation of probability density in space at different times 14
- 15. Figure showing the plot of Ψ(x, 0) with x 15
- 16. wave packet. 21 what is the physical meaning of wave packet? Packet means an enclosure that contains some thing inside it. If we have to put something inside a packet then the size of the object should be according to size of the packet. In case the object does not fit inside, we have to modulate or slightly change the size of the object . In this case a wave packet is a su- perposition of sinosodial functions whose ampli- tude is modulated by a factor φ, the sinosodials or ripples are contained in an envelope .The sit- uation looks like as shown in the figure 22 It is better to provide a bit more detailed explanation of the wave packet? Yes it is necessary to have a deeper look into the con- cept of wave packet.It can be emphasized that the con- cept of wave packets is the result of uncertainty principle. It will be very interesting to see where does the uncertainty principle come from. This requires no separate knowledge but some points related to the wave function. Three most important aspects of Ψ are : • • Ψ can interfere with itself • • Ψ is maximum at the point where the particle is present 16
- 17. Figure 3: Figure showing the wave packet 17
- 18. Figure showing Ψ(x, t) as a superposition of waves • • Ψ represent single particle or photon it never rep- resents a large number of particle or photons Thus if we concentrate on these points we find that the quantity Ψ contains the behavior of a simple particle or photon Ψ depends on the factors x and t. Ψ(x, t)isawavethatcarriesin Ψ(x, t) shows that it is a superposition of harmonic waves. The two figures below will illustrate this. In the above figure the Fourier transform of Ψ(x, t) with average wavelength λ0 in a region of space 4x. On the other hand the Fourier transform of Ψ(x, t) is shown below The above figure shows the Fourier transform of 18
- 19. Figure showing the Fourier Transform of Ψ(x, t) with respect to k 19
- 20. Ψ(x, t) against wave number k where k = 2π λ . From basic mathematical tools we have 4k4x ≥ 1—(21) returning to the De-Broglie wavelength 4p = 4(h λ) = h 2π × 4k = h4k–(22) combining equations (21) and (22) we reach to the un- certainty relation 4x4p ≥6 h—(23) 23 Well, what about the velocity of the particle and the velocity of the wave? Let us see the figure below again 20
- 21. wave packet containing the simple harmonic oscil- lations or ripples enclosed in an envelope φ(k) The velocity of the particle is the velocity of the envelope and is called group velocity The velocity of the ripples inside the envelope and is called group velocity 24 How can we prove that group velocity is the velocity of the particles,is there any mathe- matical proof?? Yes there is a specific formalism involving Taylor expansion but we will use a very simple method . we have the relation E = hν–(23) The group velocity is the velocity of the center of the wave and is given by dν = dE = d( p2 2m) = p –(24) 21