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4604_Solutions_Set2

# 4604_Solutions_Set2 - PHY4604 Problem Set 2 Solutions PHY...

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PHY4604 Problem Set 2 Solutions Department of Physics Page 1 of 14 PHY 4604 Problem Set #2 Solutions Problem 1 (10 points): The time-dependent Schrödinger’s equation is ) , ( ) ( ) , ( 2 ) , ( 2 2 2 t x x V x t x m t t x i Ψ + Ψ = Ψ h h , where we take the potential energy, V, to be only a function of x (not of time). We look for solutions of the form ) ( ) ( ) , ( t x t x Φ = Ψ ψ . The time-dependent equation separates into two equations (separation constant E). ) ( ) ( ) ( ) ( 2 2 2 2 x E x x V x d x d m ψ ψ ψ = + h and ) ( ) ( t E dt t d i Φ = Φ h . Thus, h / ) ( iEt e t = Φ . (a) (3 points) For normalizable solutions, show that the separation constant E must be real. Solution: Following Griffith’s problem 2.1 we set Γ + = i E E r , where E r and Γ are real. Thus, h h h / / / ) ( ) ( ) , ( t iE t iEt r e e x e x t x Γ = = Ψ ψ ψ and +∞ Γ +∞ = = Ψ Ψ dx x x e dx t x t x t ) ( ) ( 1 ) , ( ) , ( / 2 ψ ψ h . The second term is independent of time and hence Γ = 0 and E is real. One can prove E is real by noting that if V(x) is real then ( ) +∞ +∞ = dx x x H dx x H x op op ) ( ) ( ) ( ) ( ψ ψ ψ ψ , where ) ( 2 2 ) ( 2 2 2 2 x V x d d m V m p H op op x op + = + = h . The time-independent Schrödinger equation is ) ( ) ( x E x H op ψ ψ = which yields +∞ +∞ = dx x x E dx x H x op ) ( ) ( ) ( ) ( ψ ψ ψ ψ ( ) ( ) +∞ +∞ +∞ = = dx x x E dx x x E dx x x H op ) ( ) ( ) ( ) ( ) ( ) ( ψ ψ ψ ψ ψ ψ Thus, E = E * which means that E is real. (b) (3 points) Show that E corresponds to the total energy and that ) ( ) ( ) , ( t x t x Φ = Ψ ψ corresponds to a state with definite energy ( i.e. E = 0). Solution: E dx x x E dx x H x H E op op = = >= >=< < +∞ +∞ ) ( ) ( ) ( ) ( ψ ψ ψ ψ

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PHY4604 Problem Set 2 Solutions Department of Physics Page 2 of 14 2 2 2 2 2 ) ( ) ( ) ( ) )( )( ( ) ( ) )( ( E dx x x E dx x H H x dx x H x H E op op op op = = = >= >=< < + +∞ +∞ ψ ψ ψ ψ ψ ψ Thus, ( E) 2 = <E 2 > - <E> 2 = 0. Also note that since h / ) ( ) , ( iEt e x t x = Ψ ψ we have E dx x x E dx t x t i t x dx t x E t x E E op op = = Ψ Ψ = Ψ Ψ >= >=< < +∞ +∞ +∞ ) ( ) ( ) , ( ) , ( ) , ( ) , ( ψ ψ h . (c) (4 points) Show that the time-independent wave function, ) ( x ψ , can always be taken to be a real function. (Hint: take V to be real and see Griffith’s problem 2.1) Solution: If ) ( x ψ satisfies ) ( ) ( x E x H op ψ ψ = we can take the complex conjugate of both sides and get ) ( ) ( x E x H op = ψ ψ , but H op * = H op and E * = E and hence ) ( ) ( x E x H op = ψ ψ . Thus, ( ) ( ) ) ( ) ( ) ( ) ( x x E x x H op + = + ψ ψ ψ ψ and we see that ( ) ) ( ) ( x x + ψ ψ is a solution to the time- independent equation and ( ) ) ( ) ( x x + ψ ψ is real. Problem 2 (50 points): Consider an infinite square well defined by V(x) = 0 for 0 < x < a , and V(x) = otherwise. (a) (4 points) Show that the stationary states are given by h / ) ( ) , ( t iE n n n e x t x = Ψ ψ with ) / sin( 2 ) ( a x n a x n π ψ = and 2 2 2 2 2 ma n E n h π = , and n is a positive integer.
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4604_Solutions_Set2 - PHY4604 Problem Set 2 Solutions PHY...

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