Quantum Mechanics Notes: Particle in a Potential Box

Particle in a potential Box Schrodunger equation can be applied to one - dimengional problems which enables us to compare the predictions of classical and quantim mechanics Consider a particle of mass 'm' moving in one - dumensional potential well / box. Let the particle be restricted to move along x axis with potential O, nxo and XLL V(W) = 00, XLO and x>L ie; b/w the leng! th x=0 to x=L, the pol particle is in a potential free region V=0 free particle and beyond XIL and x 40, the potential is infinite, , so the particle cannot classically exist there We have to calculate the wave Function 4 within the box The time - independent Schrodinger equation for 10 mohin, 224 am (E-V)4 If 0 1 + dx2 h2 1 2 + 2m E 4 =0 But within the box V=0, dx2 h2 or 2 d24 + 8TT E4 = 0 (: h = 12/217) dx2 h 2 I 2 Let 8TT mE 2 k2 = 2mE = h2 h2 2 d k24=0 (2nd order differential equation) + dx 2 or 4(x) = Asin kx + B COS kx 3

The arbitary constant A and B can be calculated by conditions At x=0 , 4=0 : applying the boundary x=L 4=0 case At x=0, 4=0 (3) 0 = B cos (Kxo) + A Sin (Kxo) B=0 case : 2 At x= L, 4=0 eq(3) 0 = A Scn K.L sinkL=0 KL = DIT or K = nTC L - 4(x) = A Sin (DIE) n n=1,2,3, Note : n=0 is not possible because at n=0, 4=0 but wave Function can't be within the box If zero, of it says that particle is nowhere within the box which is not possible :. warefunction for a particle is I-D box of Length,L is, 4(00) = A Sin (nxx) Now we have to determine the value of normalization constant 'A' we know that the particle is somewhere inside the box within x=0 to x =L, J 1 41 ax =1 's A² sin2 (DIX) dx = / 0 A 2 J (1-cos dx=| 0 2

= [[L- Sin ( 2Thnx) = / 2TT L = A L =1 or A = 2/L 2 Thus the wave Function of a particle in a 1-D box of width extending from H=O to x= L is, 4(x) = 2/L Sin n= 1,2,3, * when h= 1, 4,cx) = V 2/2 Sin (Tn) (4) when D=2, 42(2) = V 2/2 Sun (2TT)) (5) when = 3, 4g(x) = 2/L Sin (3Th) (6) $ so on Probability density eq q(A),(5),(6) represents the legen functions 4, 42 43 (x), which can be represented graphically as Fig (a) In classical physics the probability of finding the particle in a box is same everywhere inside the box. But in quantim mechanics , goves the probability of fending the particle on a particular point The density of 1st three eigin value probability 14210012 $ (43(00) 2 is given by hg (b) 43LX) figca) 4,(x) 42(x) x=L x= 0 L L "Ly2 42. n=1 n=2 exted n=3

Fug (b) 142(x)) 14,10012 2 43(x)|2 0 L 4/2 O b/2 L 1/2 n=3 n=1 n=2 From fig (b) 2 we can see that for n=1, the Probability of finding the particle in maximum at L/2 ie; at middle of the box. . But n=2, probability of finding particle at middle of the box is zero and maximum at 4/4 and 3L/4 . But in all three cases, the particle cannot exist at x=0 and 2=L. This is general case Expression For energy eigen values : we know 8K m K E = h2k2 = h 2 8TT2m or E b2k2 = 2m But K= nII :. E = hin22 2 h n 2 , = L SII 2mL2 8m 2 :. En = n2's ,n = 1,2,3, 8mL2 u, E = Eo , , E2 = 4Eo, E3 = 9E $ so on le, Particle cannot take all energy values. can have only certain discrete energy level, which

are called agen values of the particle ie, energy is quantized En E3 n=3 E2 n=2 E1 n=1 Zero point energy: For from the above topic it us clear that to minimum energy of a particle is not zero, it is called zero pant energy . The zero point energy of particle in a box of length 'L' is, E, = Thb = 2 8mL2 2mL2 Particle in 3-dimensional potential box: For 3-D box, the Schrodinger equation is 22 22 22 81'm [E-v] 4=0 + dx2 dy2 232 a h For free peaticle, V=0, the schrodinger eq: becomes, 224 + 234 + 23 + 8ITm [E]4 =0 dx2 dy2 23' 2 h2 Warle length will be, 4(x,y,z) = (2/2) 3/2 Sen (nx nTox) Sin (nyty) Sin (73 T[3) Dx,Py, 3 positive integer and cannot be zero a The Eigen

E = h2 ( n x + n22 + 22) The energy eigen value, 8mL2 Cx=1,2,3 An The Zero point energy E, = h2 (1+1+1) 8mL2 E, = 3h2 8mL2 Degeneracy An energy eigen value is said to be degenerate if 2. different combination of Dx, ny, nz yelds the same energy, ie, different wave function having same energy, Ans it is said to be degenerate If corresponding to an energy value there is only one wave function / then it is said to be non-degenerate The zero-point energy is non - degenerate because only nx ny=nz =1 leads to this energy value Energy levels and degeneraties in a box 3. cubical is : Energy combination of Degeneracy 3E, (1,1,1) I Ans 6E, (1,1,2) (12,1) (2,1,1) 3 9E, (1,2,2) (21,2) (2,2,1) 3 allE, (1,1,3) (123;1) (3,1,1) 3 12E, (2,2,02) I 14E, (3,1,2) (2,1,3) (3,12) 1 6 (133,2) (1,2,3), (3,211)

PYQ. Quantim mechanical operator corresponding to linear momentum (May 20-21) P=ihD Aos. , Px = -ih Fy = -iba S3 dy = -iba 23 Poshon operator : x energy operator = iha ^ Hamiltonian operator H = -h2 D'+v at ( D2 22 + 2m dx2 dy2 dz2 2. Calculate the permitted energy level of an electron in a box I A6 wide (May21) ergy, AnS Energy of particle in 1-D box, En = n2hh 8mL2 L= 1A° = 10th , h= 6626x10 -34 Js, m = 9.1x10 kg :- En = n°(6.626x1034)2 rate 8x 9.1x10-3x (16-10)2 J = n superscript(2) (6.031x16-88) 1.6 x 10-19 ev En = 37.7n2ev = { E1 37.7ev E2 150.7ev 3. The life time of excited stable of atom is about 10-s, Calculate the minimum uncertainity in determination of the energy of the exuited State (May -21) Ans By energy - hme unuetainity relation, DE x At=h = h 2 4TT -8 At=10's, A E = 6.626x10 -34 1-6x10-92 = 5.272x1027 J = 3.29x10 -8ey

4. for electron in 1-D infruite potential well/box of width an IA° Calculate the seperation between the two (May21) lowest energy level. Ans: En = n'h2 = n°(6-626x10"84)" -20 = n2x37.Ter 8mL2 E1= 37.7ev , E2 = 150.7ev, DE = 113ev DE = 113 x 1 - 6 x 10 - 9 = 1.808x10 5 find normalization constant A for the wavelength Ans 4 = Aeimp for $ = o to 2 IT ( July 2019). , Ans: The normalization condchon for the imp wavefunction, 2TT $4*4 do =1 4=Ae 4* Ae = -inc o 2TJ A. e.e dp=1 imp -imp 0 A = 2TL // .: 4 = e inp 2TT 6. The wave function for a particle in a box is, 4 = A Sin [3TTX] OLASL. Evaluate A so that wave function is normalized Ans: for normalised wave function, 14*4dz= ( : 4 = = Asin [ ) a A [1-cos dx = 2

A2 2 [x- Sin (6TTX) sin (6TT) - 2 6 6 TI/2 0 A2L = I or A= 2/L 2 4 = V 2/1 Sin (3TTX) , OLXLL 7 Calculate the minimum values of energy and momenhum of an electron confined on 1-D box of length IA° ( June -16) En= = n2/22 , Emin min is when, n 1 8mL2 : E u h2 = (6.626x103) min 8m L2 = 6-03 x10 - 8 = 37.69ev We know, E = p2 +V (here V=0) 2m P= 2ME = V2 x 9.1x10 x 6.03x10'8 -24 = 3.313 X10 kgms-'