to the potential V to get V + V 0 0Ψ. Let Ψ ( x,t ) =ψ x dx 2

Transcription

1 Physics 0 Homework # Spring 017 Due Wednesday, 4/1/17 1. Griffith s 1.8 We start with by adding V 0 to the potential V to get V + V 0. The Schrödinger equation reads: i! dψ dt =! d Ψ m dx + VΨ + V 0Ψ. Let Ψ ( x,t ) =ψ x ( )P t ( ) =ψ P and inserting this into the SWE we get: i!ψ dp dt =! P d ψ m dx + VPψ + V 0Pψ. Now divide by ψ P and notice that both sides equal a constant. We call this constant the energy, E. Thus we have i! 1 P side we get, dp dt V 0 =! m 1 ψ d ψ dx + V = E. From the left hand i! 1 dp P dt V 0 = E dp P = i E + V 0 E + V! dt ln P = i 0! t P(t) = e i t. The right hand side gives the solutions to the time independent Schrödinger wave equation. And we have that the wave function E+V 0! Ψ( x,t) =ψ ( x)p(t) =ψ e i t = ψ e i E! t e i V0! t, which picks up a phase factor e iv 0 Since this is in the time component, the expectation values are unaffected by the phase factor.. Griffith s 1.18 Quantum mechanics is important when the debroglie wavelength of the particle λ is greater than the characteristic size of the system d. Thus λ > d, where h λ = 3m T. Or, d < λ = h. We can relate this to the temperature of the 3m T system, d < h 3m T T < h. Thus when the temperature of the system is 3m d h less than, quantum mechanics is relevant. 3m d a. For electrons: ( 34 Js) h T < 3m d = kg E+V 0! ( ) J K ( m) = K and thus free electrons in a solid are always quantum mechanical.! t.

2 b. For sodium nuclei: ( 34 Js) h T < 3m Na d = kg ( ) J K ( m) = 3.1K and thus nuclei are most always not quantum mechanical. c. For gasses we use the ideal gas law, where for one molecule ( N = 1 ) we assume that the molecule takes up a volume of approximately d 3. Thus PV = N T P d 3 T < h 3m Na d = h 3m Na 3 5 ( ) = T d = T P T T < 1 h 3m Na P 5 For helium at atmospheric pressure: T < 1 h 3m 3 5 P 1 3 and 3 T 5 3 < h 3m Na ( ) 5 3 ( ) ( ) P 5 = Js J K kg P 3. Thus we have and helium nuclei at atmospheric pressure is quantum mechanical. d. For hydrogen in outer space: ( 34 Js) h T < 3m Na d = kg ( ) J K 0.01m and hydrogen in outer space is not quantum mechanical. 3. Griffith s.3 There are two ways to answer this problem. d ψ Method I: The SWE reads: dx = me ψ!. The two cases we have are: Case 1: E = 0 d ψ = 0 ψ = A + Bx dx BC 's : ψ 0 ( ) = 0 = A + B( 0) A = 0 ( ) = 0 = B( a) B = 0 ψ a ψ = ( ) N m ( ) = K 5 =.93K

3 Case : E < 0 d ψ dx BC 's : ψ 0 ψ = 0 = me! ψ = k ψ ψ = Ae kx + Be kx ( ) = 0 = A + B A = B ψ = B e kx e kx ψ ( a) = 0 = B( e ka e ka ) ( ) B = 0 e ka = e ka 1 = e ka ln(1) = 0 = ka k = 0 Method II: The lowest energy in the infinite square well is E 1 = π! ma. Why cannot the energy be zero or negative? For the case of E = p < 0 implies that m the momentum is imaginary and a measurement would not yield a real value. For the case of E = 0, the uncertainty principle, the uncertainty in the position of the particle is Δx ~ a and therefore ΔxΔp! Δp!. But E = 0 if then a E = 0 = p Δp = 0, which violates the uncertainty principle. m 4. Griffith s.9 First we normalize the wave function over the given region, a P = 1 = A x (a x) dx. On Mathematica we have that A = 30. The 0 Mathematica code for the problem is below. To determine the expectation value of the Hamiltonian, we perform a a ( ) dψ a [ ] *! H = ψ * Hψ dx = ψ * m! dx = dx Ax(a x) m d Ax(a x) dx dx = 5! ma, where the integral was done on Mathematica. See the code below. a 5 ( )

4 5. Consider a particle bound in a 1D potential with wave function given by Ae 5ikx cos( ψ ( x) 3π a x) a x a = 0 x > a a. What is the normalization constant A? To normalize the wave function we compute P = 1 = a ψ * ψ dx = Ae 5ikx cos( 3π a x) Ae 5ikx cos( 3π a x) dx = A cos ( 3π a x)dx a a a a a and we find that the normalization constant is A = a. The normalization was done on Mathematica. The code follows. b. What is the probability of finding the particle between 0 x a 4? From the normalized wavefunction we can calculate the probability and a a P = 1 = ψ * ψ dx = a e 5ikx cos( 3π a x) a e5ikx cos( 3π a x) dx = 1 a 4 1 6π = a The integral was done on Mathematica and the code follows. c. What are the expectation values of x, p, x, & p?

5 a a x = ψ * xψ dx = a e 5ikx cos( 3π a x)x a e5ikx cos( 3π a x) dx = 0 a a a a x = ψ * x ψ dx = a e 5ikx cos( 3π a x)x a e5ikx cos( 3π a x) dx = 3π 36π a a a p = ψ * i! d a dx ψ dx = i! a e 5ikx cos( 3π a x) d dx a e5ikx cos( 3π a x) a dx = 5!k a a p = ψ * i! d dx i! d a dx ψ dx =! a e 5ikx cos( 3π a x) d dx a e5ikx cos( 3π a x) a dx 9π =! k k a + 5 a a

6 6. Determine the odd solutions to the finite square well. Determine the energy of the single bound state with E < V 0. Normalize your solutions in each region to determine the unknown coefficient A in each region. Plot your solution for ψ (x). The mathematica code and plots are given in the attached file. 7. Determine the normalization coefficients for the second energy state of the even solutions to the finite square well. That is, renormalize the solutions and determine B in each region for E 3. Plot your solution for ψ 3 (x), along with the solutions for ψ (x) from above and ψ 1 (x) from class. The mathematica code and plots are given in the attached file.

7 (* # # # # # *) [*] (* # *) (* *) = * [ * / ] * [ * / ] * [ * ] { - - / } = * [ * ] { - / / } = - * [ * / ] * [ * / ] * [- * ] { / } [ ] - [] > + [ ] [ ] - + [ ] -

8 111 FiniteWell_HW_S17.nb (* *) [ ] [{- * [][ - ]} { }] [- * [] == [ - ] {}] { } (* = *) [ ] (* *) = [ / ] * [] + - [ * ] ( * )(- / ) (* -/ / *) (* *) = = = [ - ] = (* *) = - * [] * [] * [ * * / ]

9 FiniteWell_HW_S17.nb 1113 = * [ * * / ] = * [] * [] * [- * * / ] (* *) = [ { - - / } ] = [ { - / / } ] = [ { / } ] (* -/ +/ *) [ {{{- / {}} { / {}}}} { ψ}] ψ x

10 4 111 FiniteWell_HW_S17.nb (* # *) (* *) [ ] = * [ * / ] * [ * / ] * [ * ] { - - / } = [( * [ * ]) { - / / }] = * [ * / ] * [ * / ] * [- * ] { / } [ ] [] > + + [ ] - + [ ] + + [ ] + (* *)

11 FiniteWell_HW_S17.nb 1115 (* *) [ ] [{ * [][ - ]} { }] [ * [] == [ - ] {}] { }

12 6 111 FiniteWell_HW_S17.nb (* = *) [ ] (* *) = [ / ] * [] + - [ * ] ( * )(- / ) (* -/ / *) (* *) = = = [ - ] = (* *) = * [] * [] * [ * * / ] = * [ * * / ] = * [] * [] * [- * * / ] (* *) = [ { - - / } ] = [ { - / / } ] = [ { / } ]

13 FiniteWell_HW_S17.nb 1117 (* -/ +/ *) [ {{{- / {}} { / {}}}} { ψ}] 1.5 ψ x