Solutions to Problem Set 4, Problem 1 - 3
Physics 480 /Fall 1999
Professor Klaus Schulten /Prepared by Guochun Shi

Problem 1    Wave Function of One - Dim, 
      Harmonic Oscillator

\(\(\(a\)\()\)\)\ First, \ 
    let' s\ define\ some\ \(constants : 
			\(w : \ omega\)\), 
    the\ frequency\ of\ the\ oscillator\)
\(			m : \ mass\ of\ the\ oscillator\)
\(			hbar : \ Plank\ constant\ divided\ by\ 2  π\)

w = 1.0;
m = 1;
hbar = 1;
\(Phi[y_, \ n_] := 
      If[n == 0, 
        Sqrt[Sqrt[m\ \(w/Pi\)\ /hbar]]\ Exp[\(-y^2\)/2], 
		Simplify[
          Sqrt[1/2] \((y\ \ Phi[y, \ n - 1] - 
                D[Phi[y, \ n - 1], y])\)]]\)

Now we can get the exact forms of these wave functions:

Phi[x, 0]
Phi[x, 1]
Phi[x, 2]
Phi[x, 3]
Phi[x, 4]
Phi[x, 5]

\(
g = {}\)
\(\(g = 
        Append[g, \ 
          Plot[0.751125544464942507`\ E\^\(-\(x\^2\/2\)\), \ {x, \ \(-5\), \ 
              5}, DisplayFunction -> Identity, 
			\ PlotLabel -> 
              StringForm["grould state"]]];\)\)
\(\(g = 
        Append[g, \ 
          Plot[1.06225193202719681`\ E\^\(-\(x\^2\/2\)\)\ x, \ {x, \ \(-5\), \ 5}, DisplayFunction -> Identity, 
			PlotLabel -> 
              StringForm["1st excited state"]]];\)\)
\(\(g = 
        Append[g, \ 
          Plot[1.50225108892988501`\ E\^\(-\(x\^2\/2\)\)\ \((\(-0.5`\) + 
                  1.`\ x\^2)\), \ {x, \ \(-5\), \ 5}, 
            DisplayFunction -> Identity, 
			PlotLabel -> 
              StringForm["2nd excited state"]]];\)\)
\(\(g = 
        Append[g, \ 
          Plot[2.12450386405439406`\ E\^\(-\(x\^2\/2\)\)\ \((\(-1.5`\)\ x + 
                  1.`\ x\^3)\), \ {x, \ \(-5\), \ 5}, 
            DisplayFunction -> Identity, 
			PlotLabel -> 
              StringForm["3rd excited state"]]];\)\)
\(\(g = 
        Append[g, \ 
          Plot[3.00450217785977047`\ E\^\(-\(x\^2\/2\)\)\ \((\(\(0.750000000000000088`\)\(\)\) - 
                  3.00000000000000044`\ x\^2 + 
                  1.`\ x\^4)\), \ {x, \ \(-5\), \ 5}, 
            DisplayFunction -> Identity, 
            PlotLabel -> StringForm["4th excited state"]]];\)\)
\(\(g = 
        Append[g, \ 
          Plot[4.24900772810878901`\ E\^\(-\(x\^2\/2\)\)\ \((3.75000000000000133`\ x - 5.00000000000000088`\ x\^3 + 
                  1.`\ x\^5)\), \ {x, \ \(-5\), \ 5}, 
            DisplayFunction -> Identity, 
            PlotLabel -> StringForm["5th excited state"]]];\)\)
Show[GraphicsArray[Partition[g,  3]],  
      DisplayFunction -> $DisplayFunction]

b)

H[y_,  n_] := 
        If[n == 0,  1,  
          If[n == 1,  2  y,  
            Simplify[
              2  y H[y,  n - 1]  -  
                2 (n  - 1) H[y,  n - 2]]]];
H[y, 0]
H[y, 1]
H[y, 2]
H[y, 3]
H[y, 4]
H[y, 5]

(Subscript[ϕ, n]) (x) = (1/(Sqrt[(2^n) n! (Sqrt[π])])) 
          exp {-((x^2)/2)} (Subscript[H, n]) (x) = (1/(Sqrt[(2^n) n! (Sqrt[π])])) exp {-((x^2)/2)} H (x,  n)
\(Substituing\ H \((x, 
        n)\)\ in\ the\ above\ equation\ would\ yiled\ the\ same\ wavefunctions\ as\ in\ part\ \((a)\)
\)

c)

K[y_,  n_] = 
        Simplify[
          Sum[(-1)^
                i  n!/i!/(n - 
                      2 i)!  (2 y)^(n - 2 i), {i,  0,  
              IntegerPart[n/2]}]];
K[y, 0]
K[y, 1]
K[y, 2]
K[y, 3]
K[y, 4]
\(K[y, 5]
\)

These polynomials are just the same as H (y,  
        n) in part b

Problem 2    Displaced Harmonic Oscillator

a)   At t > 0,  
      we can rewrite the potential as

		V (x)=(1/2) m(ω^2) (x^2) - (Subscript[F, 0]) x =(1/2) m(ω^2) (((x - ((Subscript[F, 0])/m)(ω^2)))^2) - (((Subscript[F, 0])^2)/(2  m(ω^2)))
By letting y=x - ((Subscript[F, 0])/m)(ω^2), \(we\ can\ transform\ the\ original\ Schrodinger\ Equation\ from
		[\(-\(h\^2\/\(2  m\)\)\) d\^2\/dx\^2 + \(1\/2\) \(mω\^2\) 
                x\^2 - \(F\_0\) 
                x] \(ϕ\_n\^F\) \((x)\) = \(E\_n\) \(ϕ\_n\^F\) \((x)\)\)
\(to
		[\(-\(h\^2\/\(2  m\)\)\) 
              d\^2\/dy\^2 + \(1\/2\) \(mω\^2\) 
              y\^2] \(ϕ\_n\^F\) \((y)\) = \(E\_n\) \(ϕ\_n\^F\) \((y)\)\)
with
\(\(E\^'\)\_n = E\_n + F\_0\^2\/\(2  mω\^2\)\)
\(The\ solutions\ to\ this\ equation\ is\ well\ \(known : 
		\(ϕ\_n\^F\) \((y)\)\)\)=(1/(Sqrt[(2^n) n! (Sqrt[π])])) 
        exp {-((y^2)/2)} (Subscript[H, n]) (y)
\(So\ the\ original\ equation' 
        s\ solutions\ \(are : 
		\(ϕ\_n\^F\) \((x)\)\)\)=(1/(Sqrt[(2^n) n! (Sqrt[π])])) 
        exp {-((((x + a))^2)/2)} (Subscript[H, n]) (x + a)
	where  a  = ((Subscript[F, 0])/m)(ω^2)

b)       (∫_(-∞)^(+∞))((Subscript[ϕ, n])^(F*)) (x, 0) ψ (x, 
              0) \[DifferentialD]x
			=(∫_(-∞)^(+∞))(1/(Sqrt[(2^n) n! π]))exp{-((y^2)/2)}(Subscript[H, n])(y)exp{-((((y + a))^2)/2)}\[DifferentialD]y
			=(∫_(-∞)^(+∞))(1/(Sqrt[(2^n) n! π]))exp{-(y^2)}(Subscript[H, n])(y)exp{-ay - ((a^2)/2)}\[DifferentialD]y
			=(∫_(-∞)^(+∞))(1/(Sqrt[(2^n) n! π]))exp{-(y^2)}(Subscript[H, n])(y)(∑_(i = 0)^∞) (1/(n!)) ((((a/2)))^i) (Subscript[H, i]) (y) 
                    exp {-((a^2)/4)} \[DifferentialD]y
			=\(\(1\/\@\(\(2\^n\) \(n!\) π\)\) \(∑\+\(i = 0\)\%∞\( 1\/\(n!\)\) \(\((a\/2)\)\^i\) \(δ\_\(n, 
                        i\)\) \(\@π\) \(2\^n\) \(n!\) 
                    exp {\(-\(a\^2\/4\)\)}\)
			 = \(\@2\^n\) \(1\/\(n!\)\) \(\((a\/2)\)\^n\) exp {\(-\(a\^2\/4\)\)}\)
Thus
	\(P\_n = \(\(|\)\(∫\_\(-∞\)\%\(+∞\)\(ϕ\_n\^\(\(F\)\(*\)\)\) \((x, 0)\) ψ \((x, 
                0)\) \[DifferentialD]x\)\( | \^2\)\(
\)\(			\)\(\(=\)\(\(1\/\(n!\)\) \(\((a\^2\/2)\)\^n\) 
              exp {\(-\(a\^2\/2\)\)}\)\)\)\)
and

\(\(\(		\)\(∑\+\(i = 0\)\%∞ P\_n = \(∑\+\(i = 0\)\%∞\( 1\/\(n!\)\) \(\((a\^2\/2)\)\^n\) 
            exp {\(-\(a\^2\/2\)\)} = \(exp {a\^2\/2} exp {\(-\(a\^2\/2\)\)} = 
          1\)\)\)\)\)

c)

P[n_,  a_] := (a a/2)^n Exp[-a a/2] /n!;
g = {}
\(\(For[i = 0, \ i <= 5, \ \(i++\), \ 
        g = Append[g, 
            Plot[P[i, \ a], \ {a, \ 0, \ 5}, \ 
              DisplayFunction -> Identity, 
\ \ \ \ \ \ \ \ \ PlotLabel -> 
                StringForm["n=" <> ToString[i]]]]];\)\)
\(Show[GraphicsArray[Partition[g, \ 3]], \ 
      DisplayFunction -> $DisplayFunction]
\)

\(
\(\(d\)\()\)\)\)
(Subscript[P, 2]) is larger than any other (Subscript[P, i]) (i ≠ 
          2) implies that
(Subscript[P, 2]) > (Subscript[P, 3]) and (Subscript[P, 2]) > (Subscript[P, 1])
(Subscript[P, 2]) = (Subscript[P, 3])  \[Implies] a = 2
(Subscript[P, 2]) = (Subscript[P, 1])  \[Implies] a = (Sqrt[6])
Thus (Subscript[F, 0])' s range is
2  m(ω^2) < (Subscript[F, 0]) < (Sqrt[6]) m(ω^2)

Problem 3    Two - Dimensional Harmonic Oscillator

\(\(\(a\)\()\)\)
		[\(-\(h\^2\/\(2  m\)\)\) \((d\^2\/dx\_1\^2 + 
                  d\^2\/dx\_1\^2)\) + \(1\/2\) \(mω\^2\) \((x\_1\^2 + 
                  x\_2\^2)\)] \(ϕ\_E\) \((x\_1, 
          x\_2)\)
 = \(\([\(-\(h\^2\/\(2  m\)\)\) \((d\^2\/dx\_1\^2 + 
                    d\^2\/dx\_1\^2)\) + \(1\/2\) \(mω\^2\) \((x\_1\^2 + x\_2\^2)\)]\) \(ϕ\_\(E\_1\)\) \((x\_1)\) \(ϕ\_\(E\_2\)\) \((x\_2)\)
 = \(\([\(-\(h\^2\/\(2  m\)\)\) 
                    d\^2\/dx\_1\^2 + \(1\/2\) \(mω\^2\) 
                    x\_1\^2]\) \(ϕ\_\(E\_1\)\) \((x\_1)\) \(ϕ\_\(E\_2\)\) \((x\_2)\) + \([\(-\(h\^2\/\(2  m\)\)\) 
                    d\^2\/dx\_2\^2 + \(1\/2\) \(mω\^2\) 
                    x\_2\^2]\) \(ϕ\_\(E\_1\)\) \((x\_1)\) \(ϕ\_\(E\_2\)\) \((x\_2)\)
 = \(\(E\_1\) \(ϕ\_\(E\_1\)\) \((x\_1)\) \(ϕ\_\(E\_2\)\) \((x\_2)\) + \(E\_2\) \(ϕ\_\(E\_1\)\) \((x\_1)\) \(ϕ\_\(E\_2\)\) \((x\_2)\)
 = \(Eϕ\_E\) \((x\_1, 
                x\_2)\)\)\)\)\)

b)
       Since (Subscript[E, 1]) = ((Subscript[n, 1]) + (1/2)) hω,  
    (Subscript[E, 2]) = ((Subscript[n, 2]) + (1/2)) hω
E = (Subscript[E, 1]) + (Subscript[E, 2]) = ((Subscript[n, 1]) + (Subscript[n, 2]) + 1) hω <= 
          4  hω implies  (Subscript[n, 1]) + (Subscript[n, 2]) <= 3,  
    so the valid values for ((Subscript[n, 1]),  (Subscript[n, 2])) are
(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 
      1), (1, 2), (2, 0), (2, 1), (3, 0)
and their plots are:

g = {};
Phi[x_,  0] := Sqrt[Sqrt[Pi]] Exp[-x x/2];
Phi[x_,  1] := Sqrt[Sqrt[Pi]/2] x Exp[-x x/2];
Phi[x_,  2] := 
        Sqrt[Sqrt[Pi]/2] (2 x x  - 
              1)  Exp[-x x/2];
\(\(Phi[x_, \ 3] := 
        Sqrt[Sqrt[Pi]/3]\ \ \((2\ x^3\  - \ 3  x)\) 
          Exp[\(-x\)\ x/2];\)
	\)
\(\(For[i = 0, \ i <= 3, \ \(i++\), \ 
        For[j = 0, \ j <= 3 - i, \ \(j++\), 
          g = Append[
              g, 
		Plot3D[
                Phi[x, \ i]\ Phi[y, \ j], \ {x, \ \(-4\), \ 
                  4}, \ {y, \ \(-4\), \ 4}, \ 
				DisplayFunction -> 
                  Identity, 
			\ \ PlotLabel -> 
                  StringForm["n1=" <> ToString[i] <> ",n2=" <> 
                      ToString[j]]]]]];\)\)
Show[GraphicsArray[Partition[g,  3]],  
      DisplayFunction -> $DisplayFunction]

c)

g = {};
\(\(For[i = 0, \ i <= 3, \ \(i++\), \ 
        For[j = 0, \ 
          j <= 3 - i, \ \(j++\), 
		g = 
            Append[g, 
              ContourPlot[
                Phi[x, \ i]\ Phi[y, \ j], \ {x, \ \(-4\), \ 
                  4}, \ {y, \ \(-4\), \ 4}, \ 
					DisplayFunction -> 
                  Identity, Contours -> {0}, 
                ContourShading -> False, 
					PlotLabel -> 
                  StringForm["n1=" <> ToString[i] <> ", n2=" <> 
                      ToString[j]]]]]];\)\)
Show[GraphicsArray[Partition[g,  3]],  
      DisplayFunction -> $DisplayFunction]