Physics 2203, Fall 2012 Modern Physics

Transcription

1 Physics 03, Fall 01 Modern Physics. Wednesday, Sept. 6 th, 01: Ch. 7: Schrödinger s equadon. Harmonic Oscillator Time dependence In class exercise drawing wave funcdons Announcements Friday, Review for first midterm Turn in Dtle and modvadon for you paper. Monday, First midterm.

2 For a standing wave we know the variables are separated. Ψ(x,t) = ψ(x)φ(t) In the Standing Wave example Ψ(x,t) = ψ(x)e iωt In general we will separate variabes: i Ψ t Ψ(x,t) = ψ(x)φ(t) i φ t ψ = m = m ψ x φ Ψ x Which gives i φ ' φ = m ψ" ψ φ is a function of only t ψ is a function of only x So each side must =constant The constant is the Energy E i φ ' φ = i de iωt dt 1 e iωt = ω Eψ = m ψ x

3 Schrödinger s EquaDon is really an equadon about operators on the wavefuncdon. ( K + U )Ψ = E Ψ K = p m = m p = i d dx d dx : because U (x) = U(x) not an operator Check with ψ (x)=ae i( kx ωt) K ψ = m p ψ = i dψ dx d ψ dx = k m ψ = Kψ = kψ = pψ Check with ψ (x)=ae i( kx ωt) E = i d dt E ψ = i dψ dt = ωψ = Eψ

4 E ψ = i dψ dt = ωψ = Eψ In QM this process where an operator acdng on the wave funcdon produces a constant Dmes the wave funcdon has a name. E is an Eigenvalue Ψ is an Eigenfunc-on Other Operators Hamiltonian H H = m d dx + U E is an Eigenvalue Ψ Is an Eigenfunc-on Most General Case H Ψ(x,t) = m Time independent H Ψ(x) = d dx +U(x) Ψ(x,t) = i dψ(x,t) dt m d dx + U(x) ψ (x) = Eψ

5 Separate the variables again ψ (x,t) = ψ (x)φ(t) m ψ x + U(x)ψ (x) = Eψ (x) ψ x = m ( U(x) E)ψ (x) i dφ(t) dt = Eφ(t) Lets do the harmonic oscillator! F= kx: U=kx /

6 Many problems in physics can be represented by a Quantum Mechanical Harmonic Oscillator. The best known examples are the vibradonal modes of molecules. For a Harmonic Potential F=-Kx U= K x where K is spring constant For a classic HO ω= K m ψ x ψ x = m ( U(x) E)ψ (x) = m mω x E ψ (x) Ground State ψ ψ should be symmetric about 0 ψ should have no nodes: ----Remember this

7 ψ x = m mω x E ψ (x) Trial Function α x ψ (x)=c 0 e Correct symmetry α x C 0 e x α x C 0 e x = m mω x E C 0e α x = ( 4α x α )C 0 e α x = m mω x E C 0 e α x ( 4α x α ) = m mω x mω E = x α m + α m E So what do we do now? Why are we solving for x? The answer should be independent of x! E is an Eigenvalue, it can t depend upon x!

8 mω E = x α m + α m E can t be a funcdon of x, so the terms muldplying x must be zero. mω α m E = α m = 0 m ω 4 = α E 0 = α m = m m ω = ω 0 4 ψ(x)=c 0 e mω 0 x E 0 = ω 0

9 ψ(x)=c 0 e αx E 0 = ω 0 mω = α ψ (x)=c 0 e mω 0 x ψ * (x)ψ (x)dx=c 0 e α x dx = π α α >0 e mω 0 x dx π ψ * (x)ψ (x)dx= C 0 = 1 mω 0 C 0 = mω ψ (x) = mω 1/4 π π 1/4 e mω 0 x

10 The allowed solutions of the Schrödinger Equation are ψ n (x) = C n e mω 0 x H n (x) : H n are Hermite polynomials Zero nodes one node Classical Turning points C n = 1 mω n n! π 1/4 How do you interpret the n=0 state? H 0 = 1: y= mω x n nodes H 1 = y H = 4y H 3 = 1y 48y + 16y 4 E n = ω n + 1 n=0, 1,,...

11 E n = ω n + 1 n=0, 1,,... ΔE=E n E m TransiDons are dipole acdvated, so x is the appropriate measure. A Harmonic oscillator is very different from a H atom or the infinite potendal well. There are selecdon rules on the possible transidons! A property that you can prove is ψ n * xψ m dx = 0 unless n=m±1 Allowed Transitions Δn=±1

12 i Ψ(x,t) t = m x +U(x) Ψ(x,t) E Ψ(x,t) = ( K +U ) Ψ(x,t) Fortunately, almost every thing we do with QM involves stadonary states. Ψ(x,t) = ψ(x)e iωt Prove that the wave func-on for this sta-onary state sa-sfies the -medependent Schrödinger Equa-on! Lets show that a free pardcle has a wave funcdon that sadsfies the Dmedependent Schrödinger EquaDon. Ψ(x,t) = Ae i(kx ωt ) i Ψ t = ωae i(kx ωt ) = EΨ n K = E QED Ψ x = k m Aei(kx ωt ) = KΨ

13 Simple example: Consider the first two states of the infinite well. Ψ 1 (x,t) = ψ 1 (x)e iω 1t Ψ (x,t) = ψ (x)e iω t = L = L sin πx L e iω 1t sin πx L e iω t If these two funcdons sadsfy the Schrödinger Eqn. then any linear sum will also be a soludon! Ψ(x,t) = αψ 1 (x,t) + βψ (x,t) Ψ 1 (x,t) Ψ (x,t) Ψ(x,t) stationary state independent of t stationary state independent of t is not time independent!!!

14 Ψ(x,t) = αψ 1 (x,t) + βψ (x,t) = αψ 1 (x) + βψ (x)e ω 1t Let α=β= 1 with ω 1 = ω ω 1 With a little work book ( ) Ψ(x,t) = 1 ψ 1 (x) + ψ (x) + ψ 1 ψ cos ω 1 t period τ = π ω 1 t = 0: x<l/ add t = τ : x<l/ subtract L

15 Can't go to infinity as x ± In the regions were E>U ψ ''=(negative numer)ψ The wave function curves toward the axis and oscillates In the regions were U>E ψ ''=(positive number)ψ The wave function is concave away for the axis

16 n= n= n= n=1 n=1 n=1 L L L L L L Rules are: the first state has no nodes and the second state has one.

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20 HW #5, QuesDon 7. The figure to the right shows three different potendal wells. The potendal energy goes to infinity at the boundaries. In each case draw in the wave funcdon for the ground state and the first excited state. In your sketches, pay abendon to the condnuity condidons and to changes in the wavelength and amplitude. Use a verdcal arrow to indicate the posidon where the pardcle is most likely to be found <x>. E E 1 E E 1 E 1