Physics Exm # April, 7 Nme Plese red nd follow these instructions crefully: Red ll problems crefully before ttempting to solve them. Your work must be legible, nd the orgniztion cler. You must show ll work, including correct vector nottion. You will not receive full credit for correct nswers without dequte explntions. You will not receive full credit if incorrect work or explntions re mixed in with correct work. So erse or cross out nything you don t wnt grded. Mke explntions complete but brief. Do not write lot of prose. Include digrms. Show wht goes into clcultion, not just the finl number. For exmple! p m v " = 5kg ( m s ) = kg m s Give stndrd SI units with your results unless specificlly sked for certin unit. Unless specificlly sked to derive result, you my strt with the formuls given on the formul sheet including equtions corresponding to the fundmentl concepts. Go for prtil credit. If you cnnot do some portion of problem, invent symbol nd/or vlue for the quntity you cn t clculte (explin tht you re doing this), nd use it to do the rest of the problem. Ech free-response prt is worth points Problem # /4 Problem # /4 Totl /8 I ffirm tht I hve crried out my cdemic endevors with full cdemic honesty.
. Consider the system show below tht is described by the potentil fucntion x < V ( x) = x V x >. Wht re the llowed solutions to the time independent Schrödinger wve eqution inside the well nd inside of the brrier? Express your nswers to the different regions in terms of the overll constnt from the solutions inside of the well. In the region x <, there is no wve function. In the region x, we hve! d ψ m dx = Eψ d ψ dx = me ψ = k ψ with solutions! ψ = Asin kx + Bcos kx. In the region x >, we hve! d ψ m dx + V ψ = Eψ d ψ dx = m ( E V ) ψ = m( V E) ψ = k '!! solutions ψ = Ce k 'x + De k 'x. ψ with To keep finite ψ, we require tht s x, ψ, so C =. Therefore ψ = De k 'x. Thus the wve functions re: Asin kx + Bcos kx x ψ ( x) = De k 'x x > Applying the boundry conditions tht the wve function nd its first derivtive re continuous t x = nd the wve function vnishes t x = we hve: ψ x= : Asin + Bcos = B = ψ x= : Asin k = De k ' D = Ae k ' sin k dψ dx x= : Ak cos k = k 'De k ' Thus the wve functions re: Asin kx x ψ ( x) = Ae k ' sin( k)e k 'x x >
b. Wht is the normliztion constnt in terms of α nd β, where we defineα = k nd β = k '. To normlize we pply the normliztion condition: P = ψ * ψ dx = A sin α x dx + A sin ( α )e β e β x dx = sin α x x = A 4α = A 4α sinα + β sin α + A sin ( α )e β e β x β A = sinα α + sin α β This could lso hve been done on Mthemtic. The code is below nd if you use the definitions of α nd β provided the result is identicl to A bove.
c. In terms of V, wht is the energy of the single bound stte for r =? Hint define α = k nd β = k '. If you cnnot determine vlue for lph, use α =. From the boundry conditions t x = we get: ψ x= : Asin k = De k ' dψ dx x= : Ak cos k = k 'De k ' Dividing these two expressions produces the eqution for determining the bound stte energies. Asin k Ak cos k = De k ' k ' = k cot k k ' = kcot k β = α cotα. k ' k 'De Following the procedure from clss, we define α + β = r, so tht me m( V E)! + = mv = r, nd thus β = r α. To!! solve this trnscendentl eqution for lph we plot the two functions nd look for intersections. From the grph below (generted on Mthemtic) nd the FindRoot commnd, α =.8955. Froph we cn determine the energy of the single odd bound stte. α = me! = r E V E = α r V =.8955 V =.898V
d. Wht is the expecttion vlue of the position? Wht does the result tell you? x = ψ * xψ dx = A xsin α x dx + A sin ( α )e β xe β x dx Evluting the normliztion coefficient: A = sin.8955 (.8955) And, + sin.8955 4.8955 =.64 x = ψ * xψ dx = A xsin α x dx + A sin ( α )e β xe β x dx =.8 I chose to use Mthemtic to evlute the numbers in the problem nd the integrls. The code nd solution re below. Wht this result implies is tht if you mde huge ensemble of identicl prticles nd mesured their positions you d find on verge tht the prticle is most likely to be found outside of the well rther thn inside, even though its energy is less thn the brrier height. In cse you re interested, the grph of the wve function is below.
. Suppose tht the stte of prticle of mss m is given by the normlized wve function Ψ x,t = mω 4 mω mω π!! xe! x e i E h t.. Determine the expression for the expecttion vlue of the kinetic energy, T. Hint: To mke your clcultions esier in prts, b nd c, define α = mω!. = mω Ψ x,t 4 mω mω π!! xe! x e i E! t = α π 4 α xe α x e i E! t T = p m =! m Ψ* x,t dx d Ψ x,t dx = 3α! 4m = 3! 4m mω! = 3 4!ω
b. Determine the expression for the expecttion vlue of the potentil energy, V = mω x. V = mω x = mω Ψ * ( x,t) x Ψ( x,t) dx = 3α! 4m = 3! 4m mω! = 3 4!ω
c. Wht is the expecttion vlue for the energy of the stte? How does your nswer compre to the sum of T + V? H = Ψ * m! d + mω x Ψdx = dx m! Ψ * d Ψ dx + mω dx Ψ * x Ψdx = T + V H = T + V = 3 4!ω + 3 4!ω = 3!ω d. Wht is the probbility of finding the prticle in the rnge x? Let the α = 5. P = Ψ * Ψ dx = P = α α π P = α π α π 4 α xe α x e i E! t α π x e α x dx = α π xe α x e α + Erf α 4 + Erf α x Erf [ ] = P =.85 + [.9875 ] =.485 ~ 49% α xe α x e i E! t dx 5 π e 5 + Erf 5 This cn be evluted on Mthemtic. The code is below. Erf [ ]
Physics Equtions Useful Integrls: x n dx = xn+ n + sin x dx = cos x cos x dx = sin x cos ( qx)dx = x + sin[qx] 4q sin ( qx)dx = x sin[qx] 4q cos 3 ( qx)dx = 3sin[qx] + sin[3qx] 4q q sin 3 ( qx)dx = 3cos[qx] + cos[3qx] 4q q x cos qx dx = xsin qx dx = sin( qx)cos(qx) dx = e x dx = ex e x dx = π xe x dx = x e x dx = π 3 α α π x e α x dx = α π x e α x + Erf α x πhc λ 5 hc dλ = 97 W m λkt e 7nm 4nm ; Erf[ ] Constnts: g = 9.8 m s G = 6.67 Nm c = 3 8 m s σ = 5.67 8 k B =.38 3 J K kg ev =.6 9 J e =.6 9 C h = 6.63 34 Js; m e = 9. 3 kg =.5 MeV c m p =.67 7 kg = 938 MeV c m n =.69 7 kg = 939 MeV c m E = 6 4 kg R E = 6.4 6 m =.5 m Formuls : c = υλ E = hυ = hc λ ds dλ = πhc λ 5 hc e λkt ds dυ = πhυ 3 c hυ e ds dλ = πckt λ 4 kt λ mx =.9 3 m K T S = σt 4 ev stop = hf φ λ ' = λ + h cosφ mc! = h π ; k = π λ ; ω = π f! m ψ + Vψ = i! ψ t Ê = i! t ˆp = i! dx T ˆ =! d m dx Ĥ =! d m dx + V ˆx = x O = ψ * Ôψ dr P = ψ * ψ dx π! E n = n m = Eψ Ψ n (x,t) = sin k n x ( E n )e i! t
T = k ' k F A T + R = T = ; E < V V + 8m 4E(V E) sinh (V! E) T = + k ; E ~ V T = ; E > V V + 8m 4E(E V ) sin (E V! ) H n (q) = Y l P l P l ( θ,φ) = ε n e d q dq n e q ; q = ( l +) ( l )! 4π ( l + )! eimφ P l ( cosθ ) = ( cos θ ) ml ( cosθ ) = d l l! d cosθ l+ r L n l n L n+l d cosθ d l mω! x ( cos θ ) l ( cosθ ); ε = P l cosθ = ( ) l+ n ( ) l+ d ( dr ) l+ r L n+l ( n ) r ( n ) = e n r n ( ) n+l d ( dr ) n+l e n r r n+l n ψ nlml = n! r 3 n ( n + l)! e n 3 n l ± = ip + mω x m!ω H = ( ± ± )!ω L ± = L x ± il y r ( n ) l l+ r L n l n L =! sinθ θ sinθ θ + sin θ φ L z = i! φ P = ψ * ψ d 3 r = π π ψ * ψ r dr sinθ dθdφ Y l θ,φ ( )