Modern Physics. Unit 3: Operators, Tunneling and Wave Packets Lecture 3.3: The Momentum Operator

Transcription

1 Modern Physics Unit 3: Operators, Tunneling and Wave Packets Lecture 3.3: The Momentum Operator Ron Reifenberger Professor of Physics Purdue University 1

2 There are many operators in QM H Ψ= EΨ, or ˆop Ψ= Ψ ˆ op = = Ψ ( ) Ψ( ) d Ψ ( ) Ψ( ) d What is the momentum operator? Needed to answer the question what is the epectation value of the momentum? dt dt d d p = m v = m = m ΨΨ d = Ψ i Ψd The Answer!

3 d d p = m v = m = m d dt dt Ψ Ψ now d Ψ Ψ m d m d dt Ψ Ψ = Ψ+Ψ t t is like a dummy variable; does not operate on! t Ψ from Sch. Eq : + U ( ) Ψ= i m t Ψ Ψ = i t m Working it out requires some effort i Ψ Ψ i U( ) Ψ and = i + U ( ) Ψ t m Ψ Ψ Ψ i m Ψ+Ψ d = m i U( ) Ψ d + t t + Ψ m + Ψ i m Ψ i U( ) d Ψ m

4 simplifying gives : + Ψ i Ψ i = m i + U( ) Ψ Ψ d + m Ψ i U( ) d Ψ m m + + Ψ im Ψ im = i Ψ + U U Ψ + Ψ Ψ d ( ( ) Ψ ) d i d ( ( ) Ψ) + Ψ Ψ Ψ Ψ = i d d i d Ψ Ψ = Ψ Ψ A useful identity : Ψ Ψ Ψ Ψ + Ψ Ψ Ψ Ψ Ψ Ψ Ψ = Ψ+ Ψ+ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ + Ψ Ψ = Ψ Ψ Ψ They are the same! d

5 Now Ψ Ψ i Ψ Ψ p = m Ψ+Ψ d d t t = Ψ Ψ i Ψ Ψ = + Ψ Ψ Ψ = i Ψ Ψ i Ψ Ψ Ψ Ψ Ψ = Ψ Ψ = Ψ i d i p op p = i Ψ Ψ Ψ d Ψ Ψ d d The term in red brackets must vanish when evaluated at = ± because Ψ for large must be zero. This of course assumes the normalization integral is finite. 5

6 Eample 1: Epectation values of momentum? Epectation values can be calculated as before. For eample, what is the epectation value for a) the momentum (p), and b) the square of the momentum (p ) of an electron in the n= state of an infinite square well? π a) Ψ ( ) = sin L L + p = Ψ i d n= Ψ + L/ π π π = i sin cos d = 0 L L L L L/ odd even Recall standing wave has two equal but opposite components; one moving to left, the other moving to right. Therefore, the total momentum should be zero. 6

7 π π Ψ ( ) = sin Ψ ( ) = sin L L L L + b) p = i i d n= Ψ Ψ + π π = sin i i sin d L L L + L/ π π π = ( i) sin cos d L L L L L/ L/ π + ( ) π π π = sin sin d L L L L L L/ + L/ π ( ) π = sin d L L L L/ 7

8 p + L/ π n= π = ( ) sin d L L L L/ 4π sin π L = ( ) L L 8π L π L = ( ) = L L 4π L + L/ L/ Js N m s kg m kg m = = = 4 m m s s Units: ( ) s Sanity check: p n= 1 4π < En= >= < KE >+< PE >= +< U >= + 0 m m L Recall ( h ) 1 4π = π = m E L n= ( ) h h ml nh h = = = from solving Schrodinger ' s Equation 8mL 8mL ml ( see L.05) 8

9 Not all wavefunctions are eigenfunctions of an operator. Consider the n= wavefunction of a particle in an infinite one-dimensional barrier ( ) π sin L L Does i Ψ ( ) = [ something] Ψ( ) Ψ = π π π i sin i cos = L L L L L Conclude that wavefunctions of infinite one-dimensional barrier are NOT eigenfunctions of momentum. This means that it is not possible to predict a definite value for momentum in a measurement of MOMENTUM. However, it is possible to predict a definite value for energy in a single measurement of ENERGY. (Why?) 9

10 What are the eigenfunctions of momentum? p ˆ op = p i i Ψ ( ) = pψ( ) eigenvalue problem what is Ψ( ) that solves this equation? dψ ( ) p p = d = i d Ψ( ) i p n( Ψ ( ) ) = i + C p p p i + C i i C Ψ ( ) = e = e e = Ae = Ae where p k ( debroglie) ik A is some constant Free particle wavefunction No quantization on p! 10

11 Eample : Prove that the wavefunction for a free particle is an eigenfunction of momentum? p ˆ op = p i ik If Ψ ( ) = Ae ( free particle), then we have ˆ ik ( )( ik pψ= i = ) Ae i ik Ae = [ k] Ψ conclude that pˆ Ψ= [ something] Ψ This implies that the free particle Ψ is an eigenfunction of momentum. 11

12 Eample 3: What is the epectation value of p for a free particle? p ˆ op = p i ik If Ψ ( ) = Ae ( free particle), then we have p = + Ψ Ψ i i d + ΨΨ d 1

13 Focus on numerator : p + ik = ( ) = ( k ) + + ik + + ΨΨ p k < E >= = m m ik A e i i Ae d + ik ik i A e Ae d ik [ ] [ ] ( ) = A e ik ik Ae d = k ΨΨd ΨΨ = = d d ( k) 13

14 Appendi: Transforming Wavefunctions In general, if the momentum for a particle in a quantum system has a definite value p, then we require pˆ Φ ( k) = p Φ( k) This means Φ( k) is an eigenfunction of momentum. If you know Ψ( ), how do find Φ( k)? You need the Fourier transform integral! 14

15 The comple form of the Fourier Integral From the theory of Fourier analysis, any smooth function Ψ() may be represented as follows + 1 ik Ψ ( ) = Φ( k) e dk π Note that both +k and k are eplicitly included. If Ψ() is known, Φ(k) can ALWAYS be found from + 1 ik Φ ( k) = Ψ( ) e d π It is common to call and k Fourier transform pairs. More about this topic in Week 4 lectures. 15

16 Calculating <p>: Do you use Φ(k) or Ψ()? If you use Φ eigenvalue operator If you use Ψ + p = k Φ k Φ k dk ( ) ( ) ik = Ψ Φ k π ( ) e d + + d 1 i k = Ψ ( ) e Φ( k) dk + d ( ) Ψ( ) = d ( k) dk ik = Ψ ( ) k e Φ( k) dkd π d ik = Ψ ( ) i e Φ( k) dkd π d = Ψ i d d π i d k Probability density integrated over k 16