CA Solution Set 2[1]

1

PS301 Quantum Physics 2 Solutions to Take Home Problem Assignment #2

Due Date Tuesday 16th December Problem 1 An electron is in a finite potential well with a width of 0.5nm and depth 2eV. Determine the energy of all the bound states in this potential using an EXCEL spreadsheet method shown in lectures.

The solution to Schrodinger’s equation will be of the form

sin( )II A kxψ = or sin( )B kx where 2k πλ

= 2

2mE=

h in region II. Classically regions I

and III are forbidden since E < U0 but applying quantum mechanics the wavefunctions

are of the form x xI Ce Deα αψ −= + and x x

III Fe Geα αψ −= + where 02

2 ( )m U Eα = −h

.We

give α in this form so that it is always real. To normalize the wavefunction D = F = 0 so that the wave does not expand indefinitely towards infinity. Next we look at the boundaries. At x=a and x=-a both the wave and it’s derivative must be continuous. We will look at even functions for the wave in region II. We get at x=-a :

cos( ) cosaI II Ce A ka A kaαψ ψ −= → = − =

And sin( ) sinaI IId d Ce kA ka kA kadx dx

αψ ψ α −= → − = − = − .

We combine these to find the logarithmic derivative to get tan( )kakα= for even solutions.

For odd solutions we do similar calculations to get cot ( )an kakα

− = for odd solutions.

We could do the same for region II and region III but we already have enough information in the above equations to find the quantised values of E. We substitute our earlier equations for α and k to get

U0 = 2eV

L=0.5nm

I II III

-a a

2

02

2cotU E mEan a

E⎛ ⎞−

− = ⎜ ⎟⎜ ⎟⎝ ⎠h

for odd solutions and

02

2tanU E mE a

E⎛ ⎞−

= ⎜ ⎟⎜ ⎟⎝ ⎠h

for even solutions.

We can evaluate the exponent in the tan and cotan functions

( )E562.2m105.0

s.J100545.1

eV.J10609.1Ekg1011.92amE2 9234

11931

2 =×××

×××××= −

−

−−−

h

where the energy is in units of eV. We substitute our values for a, U0 and E into these equations and use excel as below:

E EEU −

( )E*562.2Tan ( )E*562.2TanE

EU−

−

5.00000E-02 6.24500E+00 6.44724E-01 5.60027E+00 1.00000E-01 4.35890E+00 1.05016E+00 3.30874E+00 1.50000E-01 3.51188E+00 1.52991E+00 1.98198E+00 2.00000E-01 3.00000E+00 2.20670E+00 7.93295E-01 2.50000E-01 2.64575E+00 3.34744E+00 -7.01691E-01 3.00000E-01 2.38048E+00 5.89346E+00 -3.51298E+00 3.50000E-01 2.17124E+00 1.79385E+01 -1.57673E+01 4.00000E-01 2.00000E+00 -2.04243E+01 2.24243E+01 4.50000E-01 1.85592E+00 -6.74552E+00 8.60144E+00 5.00000E-01 1.73205E+00 -4.08452E+00 5.81657E+00 5.50000E-01 1.62369E+00 -2.93392E+00 4.55761E+00 6.00000E-01 1.52753E+00 -2.28239E+00 3.80991E+00 6.50000E-01 1.44115E+00 -1.85711E+00 3.29827E+00 7.00000E-01 1.36277E+00 -1.55367E+00 2.91644E+00 7.50000E-01 1.29099E+00 -1.32340E+00 2.61439E+00 8.00000E-01 1.22474E+00 -1.14055E+00 2.36529E+00 8.50000E-01 1.16316E+00 -9.90187E-01 2.15335E+00 9.00000E-01 1.10554E+00 -8.63038E-01 1.96858E+00 9.50000E-01 1.05131E+00 -7.53032E-01 1.80435E+00 1.00000E+00 1.00000E+00 -6.56016E-01 1.65602E+00 1.05000E+00 9.51190E-01 -5.69047E-01 1.52024E+00 1.10000E+00 9.04534E-01 -4.89972E-01 1.39451E+00 1.15000E+00 8.59727E-01 -4.17177E-01 1.27690E+00 1.20000E+00 8.16497E-01 -3.49420E-01 1.16592E+00 1.25000E+00 7.74597E-01 -2.85723E-01 1.06032E+00 1.30000E+00 7.33799E-01 -2.25304E-01 9.59103E-01 1.35000E+00 6.93889E-01 -1.67521E-01 8.61409E-01 1.40000E+00 6.54654E-01 -1.11839E-01 7.66493E-01 1.45000E+00 6.15882E-01 -5.78045E-02 6.73686E-01 1.50000E+00 5.77350E-01 -5.02108E-03 5.82371E-01 1.55000E+00 5.38816E-01 4.68609E-02 4.91955E-01 1.60000E+00 5.00000E-01 9.81580E-02 4.01842E-01

3

1.65000E+00 4.60566E-01 1.49163E-01 3.11403E-01 1.70000E+00 4.20084E-01 2.00153E-01 2.19931E-01 1.75000E+00 3.77964E-01 2.51396E-01 1.26569E-01 1.80000E+00 3.33333E-01 3.03156E-01 3.01776E-02 1.85000E+00 2.84747E-01 3.55700E-01 -7.09529E-02 1.90000E+00 2.29416E-01 4.09305E-01 -1.79889E-01 1.95000E+00 1.60128E-01 4.64260E-01 -3.04132E-01 2.00000E+00 0.00000E+00 5.20876E-01 -5.20876E-01

From this table we see that there is a change in sign between energies of 0.20 and 0.25 eV and again between 1.80 and 1.87 eV. So we zoom in on these energy ranges to obtain

E EEU −

( )E*562.2Tan ( )E*562.2TanE

EU−

−

2.00000E-01 3.00000E+00 2.20670E+00 7.93295E-01 2.05000E-01 2.95907E+00 2.29293E+00 6.66140E-01 2.10000E-01 2.91956E+00 2.38382E+00 5.35738E-01 2.15000E-01 2.88138E+00 2.47982E+00 4.01557E-01 2.20000E-01 2.84445E+00 2.58144E+00 2.63015E-01 2.25000E-01 2.80872E+00 2.68925E+00 1.19470E-01 2.30000E-01 2.77410E+00 2.80389E+00 -2.97911E-02 2.35000E-01 2.74055E+00 2.92612E+00 -1.85569E-01 2.40000E-01 2.70801E+00 3.05678E+00 -3.48768E-01 2.45000E-01 2.67643E+00 3.19684E+00 -5.20415E-01 2.50000E-01 2.64575E+00 3.34744E+00 -7.01691E-01

and

E EEU −

( )E*562.2Tan ( )E*562.2TanE

EU−

−

1.80000E+00 3.33333E-01 3.03156E-01 3.01776E-02 1.80500E+00 3.28684E-01 3.08370E-01 2.03137E-02 1.81000E+00 3.23994E-01 3.13593E-01 1.04012E-02 1.81500E+00 3.19262E-01 3.18824E-01 4.37897E-04 1.82000E+00 3.14485E-01 3.24064E-01 -9.57871E-03 1.82500E+00 3.09662E-01 3.29313E-01 -1.96512E-02 1.83000E+00 3.04789E-01 3.34571E-01 -2.97821E-02 1.83500E+00 2.99864E-01 3.39838E-01 -3.99746E-02 1.84000E+00 2.94884E-01 3.45116E-01 -5.02316E-02 1.84500E+00 2.89846E-01 3.50403E-01 -6.05565E-02 1.85000E+00 2.84747E-01 3.55700E-01 -7.09529E-02

From this we see that the even solutions are at energies of E1 = 0.2275 ± 0.0025 eV and E3 = 1.8175 ± 0.0025 eV

4

Next we calculate the solutions for the odd functions given by

02

2cotU E mEan a

E⎛ ⎞−

− = ⎜ ⎟⎜ ⎟⎝ ⎠h

E EEU −

− ( )E*562.2ancot ( )E*562.2ancotE

EU−

−

5.00000E-02 -6.24500E+00 1.55105E+00 -7.79605E+00 1.00000E-01 -4.35890E+00 9.52237E-01 -5.31114E+00 1.50000E-01 -3.51188E+00 6.53634E-01 -4.16552E+00 2.00000E-01 -3.00000E+00 4.53164E-01 -3.45316E+00 2.50000E-01 -2.64575E+00 2.98736E-01 -2.94449E+00 3.00000E-01 -2.38048E+00 1.69680E-01 -2.55016E+00 3.50000E-01 -2.17124E+00 5.57459E-02 -2.22699E+00 4.00000E-01 -2.00000E+00 -4.89614E-02 -1.95104E+00 4.50000E-01 -1.85592E+00 -1.48247E-01 -1.70767E+00 5.00000E-01 -1.73205E+00 -2.44827E-01 -1.48722E+00 5.50000E-01 -1.62369E+00 -3.40841E-01 -1.28285E+00 6.00000E-01 -1.52753E+00 -4.38138E-01 -1.08939E+00 6.50000E-01 -1.44115E+00 -5.38470E-01 -9.02683E-01 7.00000E-01 -1.36277E+00 -6.43638E-01 -7.19132E-01 7.50000E-01 -1.29099E+00 -7.55630E-01 -5.35364E-01 8.00000E-01 -1.22474E+00 -8.76770E-01 -3.47975E-01 8.50000E-01 -1.16316E+00 -1.00991E+00 -1.53249E-01 9.00000E-01 -1.10554E+00 -1.15870E+00 5.31560E-02 9.50000E-01 -1.05131E+00 -1.32797E+00 2.76650E-01 1.00000E+00 -1.00000E+00 -1.52435E+00 5.24353E-01 1.05000E+00 -9.51190E-01 -1.75733E+00 8.06136E-01 1.10000E+00 -9.04534E-01 -2.04093E+00 1.13640E+00 1.15000E+00 -8.59727E-01 -2.39706E+00 1.53734E+00 1.20000E+00 -8.16497E-01 -2.86189E+00 2.04539E+00 1.25000E+00 -7.74597E-01 -3.49989E+00 2.72530E+00 1.30000E+00 -7.33799E-01 -4.43846E+00 3.70466E+00 1.35000E+00 -6.93889E-01 -5.96941E+00 5.27552E+00 1.40000E+00 -6.54654E-01 -8.94139E+00 8.28674E+00 1.45000E+00 -6.15882E-01 -1.72997E+01 1.66838E+01 1.50000E+00 -5.77350E-01 -1.99160E+02 1.98583E+02 1.55000E+00 -5.38816E-01 2.13398E+01 -2.18786E+01 1.60000E+00 -5.00000E-01 1.01877E+01 -1.06877E+01 1.65000E+00 -4.60566E-01 6.70407E+00 -7.16463E+00 1.70000E+00 -4.20084E-01 4.99617E+00 -5.41626E+00 1.75000E+00 -3.77964E-01 3.97779E+00 -4.35576E+00 1.80000E+00 -3.33333E-01 3.29864E+00 -3.63197E+00 1.85000E+00 -2.84747E-01 2.81136E+00 -3.09610E+00 1.90000E+00 -2.29416E-01 2.44316E+00 -2.67258E+00 1.95000E+00 -1.60128E-01 2.15396E+00 -2.31409E+00 2.00000E+00 0.00000E+00 1.91984E+00 -1.91984E+00

5

From this table we see that there is a change in sign between energies of 0.85 and 0.90 eV So we zoom in on this energy ranges to obtain

E EEU −

− ( )E*562.2ancot ( )E*562.2ancotE

EU−

−

8.50000E-01 -1.16316E+00 -1.00991E+00 -1.53249E-01 8.55000E-01 -1.15723E+00 -1.02402E+00 -1.33214E-01 8.60000E-01 -1.15134E+00 -1.03828E+00 -1.13057E-01 8.65000E-01 -1.14549E+00 -1.05271E+00 -9.27750E-02 8.70000E-01 -1.13967E+00 -1.06731E+00 -7.23622E-02 8.75000E-01 -1.13389E+00 -1.08208E+00 -5.18136E-02 8.80000E-01 -1.12815E+00 -1.09703E+00 -3.11238E-02 8.85000E-01 -1.12245E+00 -1.11216E+00 -1.02876E-02 8.90000E-01 -1.11678E+00 -1.12748E+00 1.07006E-02 8.95000E-01 -1.11114E+00 -1.14299E+00 3.18466E-02 9.00000E-01 -1.10554E+00 -1.15870E+00 5.31560E-02

From this we see that the even solutions is at E2 = 0.8875 ± 0.0025 eV There is only one even solution. So the energy levels for this potential well are E1 = 0.2275 ± 0.0025 eV E2 = 0.8875 ± 0.0025 eV E3 = 1.8175 ± 0.0025 eV This problem is solved graphically in Problem 2, so it is good to check the answer. Problem 2 An electron is in a finite potential well with a width of 0.5nm and depth 2eV. Determine the energy of all the bound states in this potential using the graphical method described in lectures. From our lecture notes we see that the energies of a particle of mass m in a finite potential well of width L are given by

( )kaancotkα=− for even solutions and

( )katankα= for odd solutions and

where ( )EUm2α 02 −=h

and 2mE2kh

=

For the values given we find

( ) ( )EU10123.5EUm2α 09

02 −×=−= −

h

6

E123.5mE2k 2 ==h

and

EEU

kα 0 −=

We now plot these functions with E as the independent variable and look for the points of intersection First look at the cotan solutions

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-10

-8

-6

-4

-2

0

2

4

6

8

10

Energy

Cot

an(k

x) a

nd s

qrt(U

/E-1

)

There is a solution (intersection) between 0.8 and 1.0 eV. If we zoom in to the region around 0.8 and 1.0 we get

0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 10.8

0.9

1

1.1

1.2

1.3

So the energy of this state is E = 0.888 eV Now look at the odd (tan) solutions

7

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-10

-8

-6

-4

-2

0

2

4

6

8

10

Energy

Tan(

ka)

There is a solution between 0.1 and 0.3 eV and a second solution close to 1.8 eV. If we zoom in to the region around 0.0 to 0.2 we get

0.2 0.205 0.21 0.215 0.22 0.225 0.23 0.235 0.24 0.245 0.252

2.5

3

3.5

Which gives an energy of 0.228 eV for E1. If we zoom in to the region around 1.8 eV we get

1.75 1.76 1.77 1.78 1.79 1.8 1.81 1.82 1.83 1.84 1.850.25

0.3

0.35

0.4

8

So there is a solution at E = 1.815 eV The three eigenstates are then E1 = 0.228 eV, E2 = 0.888 eV and E3 = 1.815 eV In agreement with the values found in Problem 1 Problem 3 An electron is in a finite potential well with a width of 1.0 nm and depth 3 eV. Determine the energy of all the bound states in this potential using an EXCEL spreadsheet method shown in lectures. This problem is solved in exactly the same manner as problem 1. We look for even and odd solutions. The odd solutions are found from

02

2cotU E mEan a

E⎛ ⎞−

− = ⎜ ⎟⎜ ⎟⎝ ⎠h

while the even solutions are obtained from

02

2tanU E mE a

E⎛ ⎞−

= ⎜ ⎟⎜ ⎟⎝ ⎠h

for even solutions.

We can evaluate the exponent in the tan and cotan functions

( )E124.5m100.1

s.J100545.1

eV.J10609.1Ekg1011.92amE2 9234

11931

2 =×××

×××××= −

−

−−−

h

where the energy is in units of eV. We substitute our values for a, U0 and E into these equations and use excel as below: Odd (tan) solutions

E EEU −

( )E*562.2Tan ( )E*562.2TanE

EU−

−

5.00000E-02 7.68115E+00 2.20933E+00 5.47181E+001.00000E-01 5.38516E+00 -2.01632E+01 2.55483E+011.50000E-01 4.35890E+00 -2.27759E+00 6.63648E+002.00000E-01 3.74166E+00 -1.13849E+00 4.88015E+002.50000E-01 3.31662E+00 -6.54586E-01 3.97121E+003.00000E-01 3.00000E+00 -3.48191E-01 3.34819E+003.50000E-01 2.75162E+00 -1.10642E-01 2.86226E+004.00000E-01 2.54951E+00 9.94353E-02 2.45007E+004.50000E-01 2.38048E+00 3.04621E-01 2.07585E+005.00000E-01 2.23607E+00 5.22675E-01 1.71339E+005.50000E-01 2.11058E+00 7.73652E-01 1.33693E+006.00000E-01 2.00000E+00 1.08783E+00 9.12171E-016.50000E-01 1.90142E+00 1.52204E+00 3.79372E-017.00000E-01 1.81265E+00 2.20752E+00 -3.94869E-017.50000E-01 1.73205E+00 3.54593E+00 -1.81388E+008.00000E-01 1.65831E+00 7.68816E+00 -6.02984E+008.50000E-01 1.59041E+00 -8.54253E+01 8.70157E+019.00000E-01 1.52753E+00 -6.67694E+00 8.20446E+009.50000E-01 1.46898E+00 -3.45330E+00 4.92227E+00

9

1.00000E+00 1.41421E+00 -2.29070E+00 3.70491E+001.05000E+00 1.36277E+00 -1.67528E+00 3.03805E+001.10000E+00 1.31426E+00 -1.28395E+00 2.59821E+001.15000E+00 1.26834E+00 -1.00584E+00 2.27418E+001.20000E+00 1.22474E+00 -7.92457E-01 2.01720E+001.25000E+00 1.18322E+00 -6.19147E-01 1.80236E+001.30000E+00 1.14354E+00 -4.71913E-01 1.61546E+001.35000E+00 1.10554E+00 -3.42117E-01 1.44766E+001.40000E+00 1.06904E+00 -2.24026E-01 1.29307E+001.45000E+00 1.03391E+00 -1.13557E-01 1.14746E+001.50000E+00 1.00000E+00 -7.59273E-03 1.00759E+001.55000E+00 9.67204E-01 9.64406E-02 8.70764E-011.60000E+00 9.35414E-01 2.00857E-01 7.34558E-011.65000E+00 9.04534E-01 3.07925E-01 5.96609E-011.70000E+00 8.74475E-01 4.20077E-01 4.54398E-011.75000E+00 8.45154E-01 5.40124E-01 3.05030E-011.80000E+00 8.16497E-01 6.71559E-01 1.44937E-011.85000E+00 7.88430E-01 8.18981E-01 -3.05514E-021.90000E+00 7.60886E-01 9.88789E-01 -2.27903E-011.95000E+00 7.33799E-01 1.19037E+00 -4.56568E-012.00000E+00 7.07107E-01 1.43827E+00 -7.31162E-012.05000E+00 6.80746E-01 1.75658E+00 -1.07584E+002.10000E+00 6.54654E-01 2.18853E+00 -1.53388E+002.15000E+00 6.28768E-01 2.82047E+00 -2.19171E+002.20000E+00 6.03023E-01 3.85417E+00 -3.25115E+002.25000E+00 5.77350E-01 5.89693E+00 -5.31958E+002.30000E+00 5.51677E-01 1.20131E+01 -1.14614E+012.35000E+00 5.25924E-01 -1.03991E+03 1.04043E+032.40000E+00 5.00000E-01 -1.18647E+01 1.23647E+012.45000E+00 4.73804E-01 -5.95598E+00 6.42978E+002.50000E+00 4.47214E-01 -3.95301E+00 4.40022E+002.55000E+00 4.20084E-01 -2.93490E+00 3.35498E+002.60000E+00 3.92232E-01 -2.31205E+00 2.70428E+002.65000E+00 3.63422E-01 -1.88712E+00 2.25055E+002.70000E+00 3.33333E-01 -1.57532E+00 1.90865E+002.75000E+00 3.01511E-01 -1.33414E+00 1.63565E+002.80000E+00 2.67261E-01 -1.13991E+00 1.40717E+002.85000E+00 2.29416E-01 -9.78380E-01 1.20780E+002.90000E+00 1.85695E-01 -8.40445E-01 1.02614E+002.95000E+00 1.30189E-01 -7.19994E-01 8.50183E-013.00000E+00 0.00000E+00 -6.12761E-01 6.12761E-01

Even (cotan) solutions

E EEU −

− ( )E*562.2ancot ( )E*562.2ancotE

EU−

−

5.00000E-02 -7.68115E+00 4.52625E-01 -8.13377E+00 1.00000E-01 -5.38516E+00 -4.95953E-02 -5.33557E+00 1.50000E-01 -4.35890E+00 -4.39062E-01 -3.91984E+00 2.00000E-01 -3.74166E+00 -8.78353E-01 -2.86330E+00

10

2.50000E-01 -3.31662E+00 -1.52768E+00 -1.78894E+00 3.00000E-01 -3.00000E+00 -2.87199E+00 -1.28013E-01 3.50000E-01 -2.75162E+00 -9.03820E+00 6.28657E+00 4.00000E-01 -2.54951E+00 1.00568E+01 -1.26063E+01 4.50000E-01 -2.38048E+00 3.28277E+00 -5.66324E+00 5.00000E-01 -2.23607E+00 1.91324E+00 -4.14930E+00 5.50000E-01 -2.11058E+00 1.29257E+00 -3.40315E+00 6.00000E-01 -2.00000E+00 9.19262E-01 -2.91926E+00 6.50000E-01 -1.90142E+00 6.57011E-01 -2.55843E+00 7.00000E-01 -1.81265E+00 4.52996E-01 -2.26565E+00 7.50000E-01 -1.73205E+00 2.82013E-01 -2.01406E+00 8.00000E-01 -1.65831E+00 1.30070E-01 -1.78838E+00 8.50000E-01 -1.59041E+00 -1.17061E-02 -1.57871E+00 9.00000E-01 -1.52753E+00 -1.49769E-01 -1.37776E+00 9.50000E-01 -1.46898E+00 -2.89578E-01 -1.17940E+00 1.00000E+00 -1.41421E+00 -4.36548E-01 -9.77666E-01 1.05000E+00 -1.36277E+00 -5.96916E-01 -7.65855E-01 1.10000E+00 -1.31426E+00 -7.78845E-01 -5.35413E-01 1.15000E+00 -1.26834E+00 -9.94198E-01 -2.74145E-01 1.20000E+00 -1.22474E+00 -1.26190E+00 3.71534E-02 1.25000E+00 -1.18322E+00 -1.61513E+00 4.31910E-01 1.30000E+00 -1.14354E+00 -2.11904E+00 9.75492E-01 1.35000E+00 -1.10554E+00 -2.92297E+00 1.81743E+00 1.40000E+00 -1.06904E+00 -4.46378E+00 3.39473E+00 1.45000E+00 -1.03391E+00 -8.80618E+00 7.77227E+00 1.50000E+00 -1.00000E+00 -1.31705E+02 1.30705E+02 1.55000E+00 -9.67204E-01 1.03691E+01 -1.13363E+01 1.60000E+00 -9.35414E-01 4.97868E+00 -5.91409E+00 1.65000E+00 -9.04534E-01 3.24754E+00 -4.15207E+00 1.70000E+00 -8.74475E-01 2.38052E+00 -3.25499E+00 1.75000E+00 -8.45154E-01 1.85143E+00 -2.69658E+00 1.80000E+00 -8.16497E-01 1.48907E+00 -2.30557E+00 1.85000E+00 -7.88430E-01 1.22103E+00 -2.00946E+00 1.90000E+00 -7.60886E-01 1.01134E+00 -1.77222E+00 1.95000E+00 -7.33799E-01 8.40077E-01 -1.57388E+00 2.00000E+00 -7.07107E-01 6.95280E-01 -1.40239E+00 2.05000E+00 -6.80746E-01 5.69287E-01 -1.25003E+00 2.10000E+00 -6.54654E-01 4.56928E-01 -1.11158E+00 2.15000E+00 -6.28768E-01 3.54550E-01 -9.83318E-01 2.20000E+00 -6.03023E-01 2.59459E-01 -8.62482E-01 2.25000E+00 -5.77350E-01 1.69580E-01 -7.46930E-01 2.30000E+00 -5.51677E-01 8.32424E-02 -6.34920E-01 2.35000E+00 -5.25924E-01 -9.61625E-04 -5.24962E-01 2.40000E+00 -5.00000E-01 -8.42838E-02 -4.15716E-01 2.45000E+00 -4.73804E-01 -1.67899E-01 -3.05905E-01 2.50000E+00 -4.47214E-01 -2.52972E-01 -1.94242E-01 2.55000E+00 -4.20084E-01 -3.40727E-01 -7.93568E-02 2.60000E+00 -3.92232E-01 -4.32517E-01 4.02844E-02 2.65000E+00 -3.63422E-01 -5.29907E-01 1.66485E-01

11

2.70000E+00 -3.33333E-01 -6.34792E-01 3.01458E-01 2.75000E+00 -3.01511E-01 -7.49547E-01 4.48036E-01 2.80000E+00 -2.67261E-01 -8.77263E-01 6.10002E-01 2.85000E+00 -2.29416E-01 -1.02210E+00 7.92682E-01 2.90000E+00 -1.85695E-01 -1.18985E+00 1.00415E+00 2.95000E+00 -1.30189E-01 -1.38890E+00 1.25871E+00 3.00000E+00 0.00000E+00 -1.63196E+00 1.63196E+00

From EXCEL Tables we see that there are solutions at E1 = 0.675 ± 0.025 eV E2 = 1.175 ± 0.025 eV E3 = 1.825 ± 0.025 eV E4 = 2.575 ± 0.025 eV These solutions can be refined (as in Problem 1) to obtain more significant figures if required.

12

Problem 4 A particle of mass m is moving in the following potential

⎪⎩

⎪⎨

⎧

>≤≤−

<=

Lfor x 0L x 0for U

0for x 0)x(V 0

a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0 d) Write down the form of the continuity conditions at x = 0 and at x = L e) Write down the form of the logarithmic derivative at x = 0 and x = L

b) Schrödinger Equation

Region I: ( ) 0UEm2dxdEU

dxd

m2 o22

2

o2

22=ψ−+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒

where ( )o21 UEm2k −=h

Region II: 0Em2dxdE0

dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

22

2=ψ+

ψ⇒

where Em2k 22h

=

Region III: ( ) 0UEm2dxdEU

dxd

m2 o22

2

o2

22=ψ−+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒

c) Form of the solutions for E > U0: Region I: xikxik 11 BeAe −+=ψ Region II: xikxik 22 DeCe −+=ψ Region III: xikxik 11 GeFe −+=ψ d) Boundary condition at x = 0:

Uo

0 L

I II III

13

( ) ( ) DCBA0x0x III +=+⇒=ψ==ψ ( ) ( ) DikCikBikAik

dx0xd

dx0xd

2211III −=−⇒

=ψ=

=ψ

Boundary condition at x = L: ( ) ( ) LikLikLikLik

IIIII1122 GeFeDeCeLxLx −− +=+⇒=ψ==ψ

( ) ( ) Lik2

Lik2

Lik1

Lik1

IIIII 1111 GeikFeikBeikCeikdx

Lxddx

Lxd −− −=−⇒=ψ

==ψ

e) Logarithmic derivative

The logarithmic derivative is defined as dxd1 ψ

ψ. Therefore at x = 0 the log derivative

becomes ( ) ( )

DCDikCik

BABikAik

dx0xd1

dx0xd1 2211II

II

I

I +−

=+−

⇒=ψ

ψ=

=ψψ

while at x = L the log derivative is ( ) ( )

LikLik

Lik2

Lik2

LikLik

Lik1

Lik1III

III

II

II11

11

22

22

GeFeGeikFeik

DeCeBeikCeik

dxLxd1

dxLxd1

−

−

−

−

+−

=+−

⇒=ψ

ψ=

=ψψ


⎪⎩

⎪⎨

⎧

>≤≤

<=

Lfor x 0L x 0for U

0for x 0)x(V 0

a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0 d) Write down the form of the continuity conditions at x = 0 and at x = L e) Write down the form of the logarithmic derivative at x = 0 and x = L

b) Schrödinger Equation (for E > U0)

Region I: 0Em2dxdE0

dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒

Uo

0 L

I II III

14

where Em2k 21h

=

Region II: ( ) 0UEm2dxdEU

dxd

m2 o22

2

o2

22=ψ−+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

22

2=ψ+

ψ⇒


is real and positive

Region III: 0Em2dxdE0

dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒

c) Form of the solutions for E > U0: Region I: xikxik 11 BeAe −+=ψ Region II: xikxik 22 DeCe −+=ψ Region III: xikxik 11 GeFe −+=ψ d) Boundary condition at x = 0:

( ) ( ) DCBA0x0x III +=+⇒=ψ==ψ ( ) ( ) DikCikBikAik

dx0xd

dx0xd

2211III −=−⇒

=ψ=

=ψ

Boundary condition at x = L: ( ) ( ) LikLikLikLik

IIIII1122 GeFeDeCeLxLx −− +=+⇒=ψ==ψ

( ) ( ) Lik2

Lik2

Lik1

Lik1

IIIII 1111 GeikFeikBeikCeikdx

Lxddx

Lxd −− −=−⇒=ψ

==ψ

e) Logarithmic derivative

The logarithmic derivative is defined as dxd1 ψ

ψ. Therefore at x = 0 the log derivative

becomes ( ) ( )

DCDikCik

BABikAik

dx0xd1

dx0xd1 2211II

II

I

I +−

=+−

⇒=ψ

ψ=

=ψψ

while at x = L the log derivative is ( ) ( )

LikLik

Lik2

Lik2

LikLik

Lik1

Lik1III

III

II

II11

11

22

22

GeFeGeikFeik

DeCeBeikCeik

dxLxd1

dxLxd1

−

−

−

−

+−

=+−

⇒=ψ

ψ=

=ψψ


⎪⎩

⎪⎨

⎧

>≤≤

<∞=

Lfor x 0L x 0for U

0for x )x(V 0

a) Sketch this potential

15

b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0 d) Write down the form of the continuity conditions at x = 0 and at x = L e) Write down the form of the logarithmic derivative at x = L

b) Schrödinger Equation (for E > U0) Region I: Here the potential is infinite.


dxd

m2 o22

2

o2

22=ψ−+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒


is real and positive


dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

22

2=ψ+

ψ⇒

where Em2k 22h

=

c) Solutions to the Schrodinger equation: Region I: ψI(x) = 0 Region II: ( ) xikxik

II11 BeAex −+=ψ

Region III: xikxikIII

22 DeCe −+=ψ d) Continuity conditions: at x = 0 we find

( ) ( ) BABA00x0x III −=⇒+=⇒=ψ==ψ so the wavefunction in region II becomes ( ) xksiniA2AeAex 1

xikxik 11 =−=ψ − At x = L we find

( ) ( )( ) ( ) Lik

2Lik

2Lik

1Lik

1IIIII

LikLikLikLikIIIII

2211

2211

DeikCeikBeikAeikdx

Lxddx

LxdDeCeBeAeLxLx

−−

−−

−=−⇒=ψ

==ψ

+=+⇒=ψ==ψ

e) Logarithmic derivative at x = 0

Uo

0 L

I II III∞

16

( ) ( )BA

BkAk0dx

0xd1dx

0xd1 11III

II

II

I +−

=⇒=ψ

ψ=

=ψψ

at x = L ( ) ( )

LikLik

Lik2

Lik2

LikLik

Lik1

Lik1III

III

II

II22

22

11

11

DeCeDekCek

BeAeBekAek

dxLxd1

dxLxd1

−

−

−

−

+−

=+−

⇒=ψ

ψ=

=ψψ


⎪⎩

⎪⎨

⎧

>≤≤

<∞=

Lfor x 0L x 0for U

0for x )x(V 0

a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E < U0 d) Write down the form of the continuity conditions at x = 0 and at x = L e) Write down the form of the logarithmic derivative at x = L

b) Schrödinger Equation (for E < U0) Region I: Here the potential is infinite.

Region II: ( ) 0EUm2dxdEU

dxd

m2 o22

2

o2

22=ψ−−

ψ⇒ψ=ψ+

ψ−

h

h 0dxd 2

2

2=ψα−

ψ⇒

where ( )EUm2o2 −=α

his real and positive since E < U0.


dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

22

2=ψ+

ψ⇒

where Em2k 22h

=

c) Solutions to the Schrodinger equation: Region I: ψI(x) = 0 Region II: ( ) xx

II BeAex α−α +=ψ

Uo

0 L

I II III∞

17



( ) ( ) BABA00x0x III −=⇒+=⇒=ψ==ψ At x = L we find

( ) ( )( ) ( ) Lik

2Lik

2LLIIIII

LikLikLLIIIII

22

22

DeikCeikBeAedx

Lxddx

LxdDeCeBeAeLxLx

−α−α

−α−α

−=α−α⇒=ψ

==ψ

+=+⇒=ψ==ψ


( ) ( )BA

BA0dx

0xd1dx

0xd1 III

II

II

I +α−α

=⇒=ψ

ψ=

=ψψ

at x = L ( ) ( )

LikLik

Lik2

Lik2

LL

LLIII

III

II

II22

22

DeCeDeikCeik

BeAeBeAe

dxLxd1

dxLxd1

−

−

α−α

α−α

+−

=+α−α

⇒=ψ

ψ=

=ψψ


⎪⎩

⎪⎨

⎧

>≤≤−

<∞=

Lfor x 0L x 0for U

0for x )x(V 0

a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0 d) Write down the form of the continuity conditions at x = 0 and at x = L e) Write down the form of the logarithmic derivative at x = L

b) Schrödinger Equation (for E < U0) Region I: Here the potential is infinite.

-Uo

0 L

I II III∞

18


dxd

m2 o22

2

o2

22=ψ++

ψ⇒ψ=ψ−

ψ−

h

h 0kdxd 2

12

2=ψ+

ψ⇒

where ( )o21 UEm2k +=h

is real and positive.


dxd

m2 22

2

2

22=ψ+

ψ⇒ψ=ψ+

ψ−

h

h 0kdxd 2

22

2=ψ+

ψ⇒

where Em2k 22h

=

c) Solutions to the Schrodinger equation: Region I: ψI(x) = 0 Region II: ( ) xikxik

II11 BeAex −+=ψ



( ) ( ) BABA00x0x III −=⇒+=⇒=ψ==ψ At x = L we find

( ) ( )( ) ( ) Lik

2Lik

2Lik

1Lik

1IIIII

LikLikLikLikIIIII

2211

2211

DeikCeikBeikAeikdx

Lxddx

LxdDeCeBeAeLxLx

−−

−−

−=−⇒=ψ

==ψ

+=+⇒=ψ==ψ


( ) ( )BA

BA0dx

0xd1dx

0xd1 III

II

II

I +α−α

=⇒=ψ

ψ=

=ψψ

at x = L Problem 9 Consider reflection of a particle from a step potential of height U0 at x = 0 but now with an infinitely high potential wall added at a distance +L from the step. If the particle is incident from the left with an energy E > U0 then a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0. d) Show that the reflection coefficient at x = 0 is R = 1. (a)

19

(b)

Region 1: 121

22ψE

dxψd

m2=−

h

Region 2: 22022

22EU

dxd

m2ψ=ψ+

ψ−h

Region 3: ψ3 =0 (c) Solutions Region 1: xikxik

111 BeAeψ −+=

Region 2: xikxik2

22 DeCeψ −+= Region 3: ψ3 =0 Match boundary conditions (i) at x = 0

( ) ( ) DCBA0ψ0ψ 21 +=+⇒= (1) ( ) ( ) ( ) ( )DCikBAik

dx0ψd

dx0ψd

2121 −=−⇒= (2)

(ii) at x = L

( ) ( ) Lik2Lik

LikLikLik

322

2

222 Ce

eeCD0DeCeLψLψ −=−=⇒=+⇒=−

− (3)

substituting for D in equations (1) and (2)

( ) ( )Lik2Lik2 22 e1CCeCBA1 −− −=−=+⇒ (4)

0 Lx

U(x)

E

20

( ) ( )Lik22

Lik2211

22 e1CkCeCkBkAk)2( −− +=+=−⇒ (5) let ( )Lik2 2e1α −−= and ( )Lik2 2e1β −+= then equations (4) and (5) become

( )α

BACCαBA +=⇒=+ (6)

and ( )

βkBAk

CβCkBkAk2

1211

−=⇒=− (7)

equating the values of C from (6) and (7) gives ( ) ( )

( ) ( )( )( )

( ) ( )( )( ) ( )( )Lik2

2Lik2

1

Lik22

Lik21

21

21

2121

11222

1

22

22

e1ke1ke1ke1k

βkαkβkαk

AB

βkαkBβkαkA

BαkAαkBβkAβkβk

BAkα

BA

++−+−−

=+−

=⇒

+=−⇒

−=+⇒=

=+

we can reorganize this equation by multiplying above and below by Lik2e− this gives ( ) ( )

( ) ( )( ) ( )( )( ) ( )( )

( )( )

( )( )

( )( )Lktanikk

LktanikkLksinikLkcoskLksinikLkcosk

Lkcosk2Lksinik2Lkcosk2Lksinik2

eekeekeekeek

AB

βkαkBβkαkA

BαkAαkBβkAβkβk

BAkα

BA

212

212

2122

2122

2221

2221LikLik

2LikLik

1

LikLik2

LikLik1

2121

11222

1

2222

2222

−+

−=−+

−=

+−−−

=++−+−−

=⇒

+=−⇒

−=+⇒=

=+

−−

−−

The reflection coefficient R is then

( )( )

( )( )

( )( )

( )( ) 1

LktankkLktankk

LktanikkLktanikk

LktanikkLktanikk

LktanikkLktanikk

AB

*A*B

ABR

222

122

222

122

212

212

212

2122

212

2122

=++

=

+−

−+

=−+

−===

Note that you have to take the complex conjugate and not the square of B/A. Problem 10 Consider reflection of a particle from a step potential of height U0 at x = 0 but now with an infinitely high potential wall added at a distance +L from the step. If the particle is incident from the left with an energy E < U0 then a) Sketch this potential b) Write down the form of the Schrödinger equation in each region of this potential c) Write down the form of the solutions in each region of this potential for E > U0. d) What is the probability that the electron will be found at x = +L/2? (a) This is very similar to problem 9 above

21

(b)

Region 1: 0ψEkdxψd

0ψEm2dxψd

ψEdxψd

m2 1121

2

1221

2

121

22=+⇒=+⇒=−

h

h

Region 2: ( )

0ψαdxψd

0ψEUm2dxψd

ψEψUdxψd

m2

22

22

2

20222

2

22022

22

=−⇒

=−−⇒=+−h

h

Region 3: ψ3 =0 (c) Solutions Region 1: xikxik

111 BeAeψ −+=

Region 2: xαxα2 DeCeψ −+=

Note that we have to keep both exponential solutions, since region 2 does not extend to infinity. Region 3: ψ3 =0 Match boundary conditions (i) at x = 0

( ) ( ) DCBA0ψ0ψ 21 +=+⇒= (1) ( ) ( ) ( ) ( )DCαBAik

dx0ψd

dx0ψd

121 −=−⇒= (2)

(ii) at x = L

( ) ( ) Lα2Lα

LαLαLα

32 CeeeCD0DeCeLψLψ −=−=⇒=+⇒=−

− (3)

0 Lx

U(x)

E

22

(d) In region (2) the solution is ( )xα2Lα2xαxαLα2xαxαxα

2 ee1CeeCeCeDeCeψ −−− −=−=+= probability that the particle is in region II is

( ) ( ) ( )xα4Lα4xα2Lα2xα22xα2Lα2xα22 eeee21Ceee1Cexψ −−− +−=−=

therefore the probability that the electron is at L/2 is

( )

( ) ( )Lα3Lα2LαLα2LαLα

Lα2Lα4LαLα2Lα2Lα4Lα42

Lα2Lα22Lα2

2

2

ee2eCee21Ce

eeee21Ceeeee21Ce2Lψ

+−=+−=

+−=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=⎟

⎠⎞

⎜⎝⎛ −−−−

Problem 11 A particle of mass m and energy E moving in a region where the potential energy zero encounters a potential dip of width L and depth U = -U0.

⎪⎩

⎪⎨

⎧

>≤≤−

<=

Lfor x 0L x 0for U

0for x 0)x(U 0

Find the reflection and Transmission coefficients for this particle.

First write down the Schrödinger equation in each region Region I & III:

0)x(ψk

dx)x(ψd

0)x(ψmE2dx

)x(ψd)x(ψE)x(ψ0dx

)x(ψdm2

212

2

22

2

2

22

=+⇒

=+⇒=+−h

h

Region II:

0 L

0

-U0

E

I II III

(a) Sketch of the Potential given:

U(x)

23

( )

0)x(ψkdx

)x(ψd

0)x(ψEUm2dx

)x(ψd)x(ψE)x(ψUdx

)x(ψdm2

222

2

022

2

02

22

=+

=++⇒=−−h

h

where

( )

UEm2

k and mE2k 202

2221

hh

+==

The solutions to the Schrödinger equation in regions I and III are

II region from aveIncident w Fe

Wave Reflected WaveIncident BeAexik

III

xikxikI

1

11

==ψ

+=+=ψ −

The solution in region II is L x from WaveReflected Iregion from WavedTransmitte DeCeψ xikxik

II22 =+=+= −

Now use the continuity equations at x = 0 and x = L Continuity condition at x = 0

( ) ( )DCikBAik 0 at x dxψd

dxψd (ii)

DCBA 0 at x ψψ )i(

21III

III

−=−⇒==

+=+⇒==

continuity condition at x = L

( ) Lik1

Lik-Lik2

IIIII

LikL-ikLikIIIII

122

122

FeikDeCeik Lx at dx

ddx

d (iv)

FeDeCe Lx at )iii(

=−⇒=ψ

=ψ

=+⇒=ψ=ψ

To find the Transmission coefficient we eliminate B, C and D from equations (i) to (iv) above Multiply eqn (iii) by ik2 and add to equation (iv) this gives

( ) Lik21

Lik2

12 FeikikCeik2 += can now solve for C in terms of F

( ) Lik-Lik

2

21 21 eFeik2

ikikC +=

Multiply eqn (iii) by -ik2 and add to equation (iv) this gives ( ) Lik

12L-ik

212 FeikikDeik2 −=

can now solve for D in terms of F ( ) LikLik

2

12 21 eFeik2

ikikD −=

Next, multiply eqn (i) by ik1 and add to equation (ii)

( ) ( )DikikCikikAik2DikCikBikAikDikCikBikAik

21211

2211

1111

−++=⇒−=−+=+

24

Now substitute for C and D in the last equation

( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )[ ]( )( ) ( )( )[ ] LikLik2

12Lik-2

2121

LikLik1221

Lik-212121

LikLik1221

Lik-Lik212121

LikLik

2

1221

Lik-Lik

2

21211

122

122

2121

2121

FeeikikeikikAkk4

FeeikikikikeikikikikAkk4

eFeikikikikeFeikikikikAkk4

eFeik2

ikikikikeFeik2

ikikikikAik2

−−+=−⇒

−−+++=−⇒

−−+++=−⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

( )( ) ( )( )[ ]Lik212

Lik-221

Lik21

22

1

eikikeikikeAkk4F

−−+=⇒

The transmission coefficient T, is given by 1

32

kk

AFT = , however since k3 = k1 we get

( )( ) ( )( )[ ] ( )( ) ( )( )[ ]Lik-212

Lik221

L-ik21

Lik212

Lik-221

Lik21

2

22

1

22

1

eikikeikikekk4

eikikeikikekk4

AFT

+−−−−−−+==

+

The reflection coefficient R, is found in the same y eliminating C, D and F and finding B

in terms of A then 2

ABR = . Alternatively R = 1-T.

25

Problem 12 An electron of energy E is incident on a potential barrier described by the potential function

( )⎪⎩

⎪⎨

⎧

>≤≤−

<=

B/A xfor 0B/Ax 0for BxA

0 xfor 0xU

derive an expression for the tunneling probability of an electron with energy E < A through this barrier. Calculate the transmission probability of a 5 eV electron tunneling through this barrier for A = 10 eV and B = 25 eV.nm-1. First make a sketch of the potential

The particle encounters the barrier at x = 0 and exits the barrier at x = L = (A-E)/B. Following the lecture notes, the tunneling probability is given by

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛∫

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∫−≈

b

a2

b

adxE)x(Vm22expdxxα2expP

h

First evaluate the integral ( ) ( ) ( ) dxx

BEABm2dxEBxAm2dxE)x(Vm2 L

02

L

02

b

a2 ∫∫ ⎟

⎠⎞

⎜⎝⎛ −

−=−−=∫

−hhh

( )dxxLBm2 L

02 ∫ −=h

Let u = L-x then du = -dx so the integral becomes

( ) ( )2

3

22

323

2

L

0

232

L

02 B

EAm232mBL2

32LBm2

32uBm2

32duuBm2 −

===−=−= ∫ hhhhh

The tunneling probability is then

( ) ( ) ⎟⎠⎞

⎜⎝⎛ −−=⎟

⎠⎞

⎜⎝⎛

∫−≈ 3b

aEAm2

B34expdxxα2expPh

Now plug in numbers given

E

U(x)

x L

A

26

( )

( ) 2

31931

9

1934

1065.407.3exp

)10602.15(kg1011.92

m1010602.125s.J1005.13

4expP

−

−−

−

−−

×=−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

×××××

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××

−≈

So about 4.6% of the electrons tunnel through the barrier, the rest are reflected. Problem 13 An electron of energy E is incident on a potential barrier described by the potential function

( )⎪⎩

⎪⎨

⎧

>≤≤

<=

L xfor 0Lx 0for Ax

0 xfor 0xU

derive an expression for the tunneling probability of an electron with energy E < A through this barrier. Calculate the transmission probability of a 5 eV electron tunneling through this barrier for A = 10 eV and L = 2 nm. First Sketch the potential

The particle encounters the potential at the point EL/A and exits at x = L. Following the lecture notes, the tunneling probability is given by

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛∫

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∫−≈

b

a2

b


h

First evaluate the integral ( ) ( ) dx

AExAm2dxEAxm2dxE)x(Vm2 L

AEL2

L

AEL2

b

a2 ∫∫ ⎟

⎠⎞

⎜⎝⎛ −=−=∫

−hhh

let u = x-(E/A) then du = dx and the integral becomes

E

U(x)

x L

AL

27

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−==∫

23

2

32323

2

2323

2

L

AEL23

2

L

AEL

212

AE1AmL2

32

AE1LAm2

32

AELLAm2

32u

32Am2duuAm2

hh

hhh


( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−−=⎟

⎠⎞

⎜⎝⎛

∫−≈23

2

3b

a AE1AmL2

34expdxxα2expP

hh

Now plug in numbers given

( ) ( ) ( )

99.0)1044.4exp(

1051102kg1011.910602.1102

1005.134expP

4

23393119

34

=×−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−×××××××

××−≈

−

−−−−

Problem 14 An electron of energy E is incident on a potential barrier described by the potential function

( )

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

>

≤≤−

<

=

BA xfor 0

BAx

BA-for xBA

BA- xfor 0

xU

derive an expression for the tunneling probability of an electron with energy E < A through this barrier. Calculate the transmission probability of a 5 eV electron tunneling through this barrier for A = 10 eV and B = 25 eV.nm-1. First Sketch the potential

The particle encounters the potential at the point –(A-E)/B and exits at +(A-E)/B. Following the lecture notes, the tunneling probability is given by

E

U(x)

x 0

A

-A/B A/B

28

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛∫

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∫−≈

b

a2

b


h

Let L = (A-E)/B, then the limits of integration are –L to L. Evaluate the integral ( ) ( ) ( ) dxx

BEABm2dxEBxAm2dxE)x(Vm2 L

L2

L

L2

b

a2 ∫∫ ⎟

⎠⎞

⎜⎝⎛ −

−=−−=∫

−

−hhh

( )dxxLBm2 L

L2 ∫

−

−=h

Because the integral is symmetric about the origin we can write

( ) ( )dxxLBm22dxxLBm2 L

02

L

L2 ∫∫ −=−=

−hh

Let u = L-x then du = -dx so the integral becomes

( ) ( )2

3

22

323

2

L

0

232

L

02 B

EAm234mBL2

34LBm2

34uBm2

34duuBm22 −

===−=−= ∫ hhhhh


( ) ( ) ⎟⎠⎞

⎜⎝⎛ −−=⎟

⎠⎞

⎜⎝⎛

∫−≈ 3b

aEAm2

B38expdxxα2expPh

Plug in numerical values

( )

( ) 3

31931

9

1934

1015.214.6exp

)10602.15(kg1011.92

m1010602.125s.J1005.13

8expP

−

−−

−

−−

×=−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

×××××

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××

−≈

Problem 15 A 2 kg block oscillates with an amplitude of 10 cm on a spring of force constant 120 N/m. (a) In which quantum state is the block? (b)The block has a slight electrical charge and may therefore drop to a lower energy level by emitting a photon. What is the minimum energy decrease possible, and what would the corresponding fractional change in energy be?

A quantum oscillator in the nth state has an energy ω⎟⎠⎞

⎜⎝⎛ += h

21nEn , where ω is the

angular frequency of the oscillator given by mk

=ω .

In this problem 11

s75.7kg2

m.N120mk −

−

===ω .

29

The total energy of the oscillator is given by E = kA2/2, where A is the amplitude of the oscillation, (see notes on SHO). So in this problem

( ) J6.0m10.0m.N12021kA

21E 212 === −

Equating this energy to the quantum energy of the system we find 32

134n 1029.721

s75.7s.J1062.62J6.0

216.0nJ6.0

21nE ×=−

××π××

=−ω

=⇒=ω⎟⎠⎞

⎜⎝⎛ += −−h

h

so we see that the system is in a state with a VERY high quantum number, i.e. it is in a classical state. When the system makes a transition between two states it emits a photon with energy Eph =hω, which in this case is

J101.5s75.7s.J1062.6 33134 −−− ×=××=ωh . A tiny amount of energy. This is the minimum energy decrease possible, i.e. the oscillator emits just one photon. The fractional change in energy of the system is

323333

10105.8J6.0

J101.5EE −−

−

≈×=×

=Δ

The fractional change is one part in 1032 again, too small to measure. Problem 16 What is the most probable location to find a particle that is in the first excited state (n = 1 state) of a harmonic oscillator potential? The wavefunction for the first excited state of the quantum SHO is

( ) ( )2sexps2Ns 211 −=ψ

where N1 is the normalization constant and xkms41

2 ⎟⎠⎞

⎜⎝⎛=h

The probability of finding the particle at some position s is then ( ) ( )222

12

1 sexps4Ns)s(P −=ψ= To find the most probable place we differentiate w.r.s. x and set the result = 0 for a maximum

( )[ ] [ ] ( ) 0dxdssexps8s8N

dxdssexps4N

dsd

dxds

ds)s(dP

dx)s(dP 232

1222

1 =−−=−==

[ ] ( ) 0dxdssexp.s.s88N 222

1 =−−⇒

this expression is zero for s = 0 (a minimum) and 8(1 – s2) = 0 ⇒ s = ±1 (a maximum) Therefore, the probability is at a maximum when

s = ±1 41241

2 kmx1xkms ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⇒±=⎟

⎠⎞

⎜⎝⎛=⇒

h

h

Problem 17 Assuming that the vibrations of a 35Cl2 diatomic molecule are equivalent to those of a harmonic oscillator with force constant k = 329 Nm-1, what is the zero-point energy of

30

vibration of this molecule? Estimate the energy difference between the ground state and the first excited state of this molecule. The mass of a 35Cl atom is 34.9688 amu We must first work out the reduced mass of the 35Cl2 molecule

21

21mm

mmμ

+=

in this case m1 = m2 = m so amu2m

m2mμ

2== ,

To convert from atomic mass units (amu) to kg: 1 amu = 1.6605×10-27kg, so m(35Cl) = 34.9688×1.6605×10-27kg = 5.807×10-26 kg Then

( ) eV1050.3J1061.5kg10807.5

m.N329221

mk2

21

μk

21ω

21E 221

26

1

0−−

−

−×=×=

×

×==== hhhh .

The separation between the energy levels is ΔE = E1 – E0 and since ω21nE h⎟⎠⎞

⎜⎝⎛ += we

see that ωω210ω

211EEEΔ 21 hhh =⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ +=−= .

We have already found above that eV1050.3J1061.5ω21 221 −− ×=×=h , so

eV1000.7J1012.1J1061.52EΔ 22021 −−− ×=×=××= Problem 18 Calculate the wavelength of a photon required to excite a transition between neighboring energy levels of a harmonic oscillator of mass equal to that of the oxygen atom (15.999 atomic mass units) and force constant 544 Nm-1. Again, following the approach outlined in problem 2 we find that

( )

eV1071.4J1054.7

amu/kg1066054.1amu99.15m.N544

21

mk

21ω

21E

221

27

1

0

−−

−

−

×=×=

××=== hhh

then

eV1042.9J1050.1J1054.72λhcωEEEΔ 22021

01−−− ×=×=××===−= h

solve for λ

m1032.1J1050.1

hcλ 520

−−

×=×

=

Problem 19 The wavefunction of the ground state of a harmonic oscillator of force constant k and mass m is

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛= 2

41

0 x2αexp

πα)x(ψ where ,

ωmα 0

h= and

mkω0 =

31

Obtain an expression for the probability of finding the particle outside the classically allowed region. The particle is said to be outside of the classical region if E < V(x), i.e. the total energy is less than the potential energy. In the ground state E = hω0/2 and the non-classical region is therefore,

2200 xm

21

21

ω<ωh i.e. α

=ω

>1

mx

0

2 h or

⎪⎪⎩

⎪⎪⎨

⎧

α−<

α>

1x

1x

The probability of finding the particle in the nonclassical region is therefore

( ) ( )dxxexpdxxexpdxdxP1

221

1

21

2∫∫∫∫∞

α

α−

∞−

∞

α

α−

∞−

α−πα

+α−πα

=ψ+ψ=

( ) ( ) 16.0dttexp12dxxexp21

2

1

2 ≈−π

=α−πα

∫∫∞∞

α

where we have let tx =α so α= dtdx . The value of the integral is approximated numerically. Problem 20 One possible solution for the wavefunction ψn for the simple harmonic oscillator is

( ) ⎟⎠⎞

⎜⎝⎛−−= 22

n x2αexpxα21Aψ

where A is the normalization constant. (a) What is the energy of the oscillator when it is in this state? (b) find <x2> for this state. (a) By comparing this wavefunction with those of the harmonic oscillator we see that the particle is in the n = 2 state and therefore the energy of the particle is

ω25ω

212E2 hh =⎟⎠⎞

⎜⎝⎛ +=

(b) The expectation value is found from

( ) ( )

( ) ( ) ( ) ( )

( ) ( )dxxαexpxα4xα4xA

dxxαexpxα21xAdxxαexpxα21xA

dxx2αexpxα21xx

2αexpxα21Aψxψx

262422

2222222222

2222222

2*2

2

−+−=

−−=−−=

⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛−−==

∫

∫∫

∫∫

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∞

∞−

We now evaluate the three integral using integral tables. (see the following website http://integrals.wolfram.com/index.jsp)

32

( )

( )

( ) 2726

2524

2322

α16π15dxxαexpx

α8π3dxxαexpx

α4πdxxαexpx

=−

=−

=−

∫

∫

∫

∞

∞−

∞

∞−

∞

∞−

so collecting terms gives

( ) ( )

22323

2

2323232

272

25232

2624222

Aα2π5

415

46

41

απA

α415

α23

α41πA

α16π15α4

α8π3α4

α4πA

dxxαexpxα4xα4xAx

=⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦

⎤⎢⎣

⎡ +−=⎥⎦

⎤⎢⎣

⎡+−=

−+−= ∫∞

∞−

Finally, substituting the normalization constant for the n = 2 state gives 2

32

3

2323

2

2232

232

kω

165

ωm165

α165

π81

α2π5

!22π1

α2π5A

α2π5x ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛===⎟⎟

⎠

⎞⎜⎜⎝

⎛==

hh

Problem 21 Using the fact that the commutator [ ] hip,x = , evaluate the following commutators:

(i) [ ]x,p , (ii) [ ]p,x 2 , (iii) [ ]2p,x , (iv) ⎥⎦⎤

⎢⎣⎡ 2p,

x1 .

(i) [ ] ψ−=⎥⎦⎤

⎢⎣⎡

∂ψ∂

−∂ψ∂

+ψ−=ψ⎟⎠⎞

⎜⎝⎛

∂∂

−−ψ∂∂

−=ψ−ψ=ψ hhhh ix

xx

xix

ixxx

ipxxpx,p

So [ ] hix,p −= Alternatively we could use the fact that [A,B] = AB – BA = -BA-AB = -[B,A]. So [ ] [ ] hip,xx,p −=−= (ii)

[ ] ( ) ( )[ ] ( )

[ ] xi2xiixxip,xxxixppxxxixpxpxxxixpxpxxp,xxpxpx

xxppxxpxpxxxpxpxxpxpxxxppxxppxp,x22

222222

hhhh

hhh

=+=+=+−=+−=+−=+−=

−+−=−+−=−=−=

(iii) Similarly, [ ] ( ) [ ][ ] [ ] ( ) [ ] [ ] pi2pipip,xppp,xxppxppp,xxpppxppp,x

xppxppp,xxppxppxppxxppxppxpppxxppxp,x 222222

hhh =+=+=−+=−+=−+=−+−=−+−=−=

33

(iv) This one is a little harder so we will expand the commutator brackets

now 2

222

xxi

xip

∂∂

−=⎟⎠⎞

⎜⎝⎛

∂∂

⎟⎠⎞

⎜⎝⎛

∂∂

= hhh using the definition of the linear momentum

operator, Then

ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡

∂∂

−∂∂

∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−=

ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦⎤

⎢⎣⎡

∂∂

+−⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−=ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦⎤

⎢⎣⎡

∂∂

+∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−=

ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−=ψ⎟⎠⎞

⎜⎝⎛ −=ψ⎥⎦

⎤⎢⎣⎡

2

2

222

22

22

22

2

22

2

2

2

22222

xx1

xx1

xxx1

x1

xxx1

xx1

x1

xxx1

xx1

x1

xxxx1

x1

xxx1

x1pp

x1p,

x1

h

hh

h

{ }ixpx

2pix12

x2

xx12

x12 32

23

2

232 −=⎟

⎠⎞

⎜⎝⎛−=ψ

⎭⎬⎫

⎩⎨⎧

∂∂

+−−= hh

hh

hh

Problem 22 Determine the expectation value of the position and momentum of a harmonic oscillator in the first excited (n = 1) state. The wavefunction for the first excited state of the quantum SHO is

( ) ( )2sexps2Ns 211 −=ψ


2 ⎟⎠⎞

⎜⎝⎛=h

then dskm

dxskm

x412412

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=

hh

The expectation value of x is given by

( )∫∫∞

∞−

∞

∞−

−⎟⎟⎠

⎞⎜⎜⎝

⎛=ψψ= dssexps4

kmNdxxx 23

212211

*11 h

where we have changed the variable of integration from x to s

( ) ( ) 01se21

km

2Ndssexps4

kmNx 2s

21

21

23212

211

2 =+−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

∞

∞−⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ −∞

∞−∫

hh , since e-∞ =0

The wavefunction for the first excited state of the quantum SHO is ( ) ( )2sexps2Ns 2

11 −=ψ


2 ⎟⎠⎞

⎜⎝⎛=h

then dskm

dxskm

x412412

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=

hh

34

( ) ( )

( )( ) ( )

( ) ( )

( )[ ] 01se2e2kmkm

Ni

dssexps4s4kmkm

Ni

dsdxds2sexps222sexps2

kmNi

ds2sexps2dxds

dsdi2sexps2

kmNdxpp

2ss41

2

21221

2341

2

21221

222212

21

22212

211

*11

22

=++−⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−−⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−⎟⎠⎞

⎜⎝⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛=ψψ=

∞

∞−−−

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∫

∫

∫∫

h

hh

h

hh

hh

hh

since e-∞ =0 Problem 23 Determine the expectation value of the kinetic energy of a harmonic oscillator in the first excited (n = 1) state. The wavefunction for the first excited state of the quantum SHO is

( ) ( )2sexps2Ns 211 −=ψ


2 ⎟⎠⎞

⎜⎝⎛=h

then dskm

dxskm

x412412

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=

hh

The expectation value of the kinetic energy is given by

( ) ( )

( ) ( )

( ) ( ) ( )

( )( ) ( )ds2sexps2s62sexps2kmm2

N

ds2sexps22dsd2sexps2km

m2N

ds2sexps2dsd

m22sexps2kmN

ds2sexps2dxds

dsd

dsd

m22sexps2NdxKK

23221

2

221

22221

2

221

22

222

21

221

222

2211

*11

−+−−⎟⎠⎞

⎜⎝⎛−=

−−⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−=

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟

⎠⎞

⎜⎝⎛=

−⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−=ψψ=

∫

∫

∫

∫∫

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∞

∞−

h

h

h

h

h

h

h

[ ] ω=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛π=π−π⎟

⎠⎞

⎜⎝⎛−= h

h

h

h

h

h

43

mk

43km

m4N363km

m2N

2121

2

221

21

2

221

Where we have evaluated the normalization constant from

!n2

1Nn41n

π=

35

Problem 24 Consider a system whose state ψ is given in terms of three eigenfunctions of the system,ϕ1, ϕ2, ϕ3, as

321 φ32φ

32φ

33ψ ++=

a) verify that ψ is normalised b) Calculate the probability of finding the system in any one of the states ϕ1, ϕ2 and ϕ3. c) Verify that the total probability is one d) Consider now a collection of 810 identical systems, all in the same state ψ. If

measurements are made on each system, how many will be found in each of the states ϕ1, ϕ2 and ϕ3.

a) We use the orthonormality condition ijji δφφ = , where i and j = 1, 2 and 3 we can verify that

192

94

93φφ

32φφ

32φφ

33ψψ 33

2

22

2

11

2

=++=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

b) Since ψ is normalised the probability of finding the system in the state iφ is 2

ii ψφP = using this expression we find

310

320

321

33φφ

32φφ

32φφ

33ψφP

22

3121112

11 =×+×+×=++==

940

321

320

33φφ

32φφ

32φφ

33ψφP

22

3222122

22 =×+×+×=++==

921

320

320

33φφ

32φφ

32φφ

33ψφP

22

3323132

23 =×+×+×=++==

c) As expected the total probability is

192

94

31PPPP 321 =++=++=

d) the number of systems that will be found in the state 1φ is

27031810P810N 11 =×=×=

similarly,

36094810P810N 22 =×=×= and 180

92810P810N 33 =×=×= .

Problem 25 A particle of mass m is confined in an infinite potential well of width a. At time t = 0 the wavefunction of the particle is

( ) ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

ax5πsin

a51

ax3πsin

a53

axπsin

aA0,xψ

36

a) Find A so that the wavefunction ψ(x,0) is normalised. b) If measurements of the energy are carried out, what are the values that will be found

and what are their corresponding probabilities. c) Calculate the average energy of the system at t = 0. d) Find the wavefunction ψ(x,t) at any later time t. Since the particle is in an infinite potential well, the eigenfunctions of this system are

( ) ⎟⎠⎞

⎜⎝⎛=

axπnsinxφ a

2n

and these eigenfunctions are orthonormal, that is, ijji δφφ = , so we can write the wavefunction at time t = 0 as

( ) 531 φ101φ

103φ

2A

ax5πsin

a51

ax3πsin

a53

axπsin

aA0,xψ ++=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

the normalisation condition then gives

a) 56A 1

101

103

2Aφφ

101φφ

103φφ

2Aψψ

2

553311

2=⇒=++=++=

so the correctly normalised wavefunction is

( ) 531 φ101φ

103φ

530,xψ ++=

b) The expectation value of the Hamiltonian (energy operator) for the particle when it is

in the nth state is:

dxφdxd

m2φφHφE

a

0n2

22*nnnn ∫ ⎟

⎟⎠

⎞⎜⎜⎝

⎛−==h

and we know that

n2

222

2

222

2

22

n2

22φ

ma2πn

axπnsin

a2

ma2πn

axπnsin

a2

dxd

m2φ

dxd

m2hhhh

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−

then

2

222a

0n

*n2

222a

0n2

22*nnnn

ma2πndxφφ

ma2πndxφ

dxd

m2φφHφE hhh

=∫=∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−==

since the eigenfunctions are orthonormal. If a measurement is carried out on the system we would obtain an energy

2

222

nma2πnE h

= with a corresponding probability of ( ) 2nnn ψφEP = . Since the initial

wavefunction contains only three eigenstates of H, the results of energy measurements with their corresponding probabilities are:

( )53ψφEP ;

ma2πφHφE 1112

22

111 ====h

( )103ψφEP ;

ma2π9φHφE 3332

22

333 ====h

37

( )101ψφEP ;

ma2π25φHφE 5552

22

555 ====h

note that the probability of finding the particle in any other state is zero, so the probability of measuring any other value for the energy is zero. The average energy is

2

22

2

22

531n

nnma10π29

ma2π25

1019.

103

106E

101E

103E

53EPE hh

=⎟⎠⎞

⎜⎝⎛ ++=++=∑=

c) the wavefnction at any time t later than t = 0 is found by putting in the appropriate

time dependence factors:

( ) ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛−=

hhh

tiEexpxφ

101tiE

expxφ103tiE

expxφ53t,xψ 5

53

31

1

where the values of En are given above. Problem 26 A particle of mass m is confined in an infinite potential well of width a. At time t = 0 the wavefunction of the particle is ψ = A where A is a constant. a) Find A so that the wavefunction ψ(x,0) is normalised. b) Find an expression for the wavefunction at any later time t c) If measurements are carried out on this system at time t = 0, what is the probability that the particle will be found in the (i) n = 1, (ii) n = 2 and (iii) n = 5 state. d) Calculate the average energy of the system at t = 0. (a) Normalisation

( )L1A1L2A1

L

0

dx2A1dx2x,0ψ =⇒=⇒=⇒=∞

∞−∫∫

(b) We can write the wavefunction at any time t as a linear expansion of the eigenfunctions:

( ) ∑ ⎟⎠⎞

⎜⎝⎛−=

n

nnn

tiEexp)x(φct,xψh

where the eigenfunctions and eigenvalues for a particle in a box are given by

2

222

nnmL2πnE and

Lxπnsin

L2)x(φ h

=⎟⎠⎞

⎜⎝⎛=

the coefficients cn, are given by

dxL

xnsinAL2Adx

Lxnsin

L2c

L

0

L

0

nn ∫∫ ⎟⎠⎞

⎜⎝⎛ π

=⎟⎠⎞

⎜⎝⎛ π

=ψφ=

The result of the integral is

38

( )[ ] ( )[ ] [ ]n

L

0

L

0

n

)1(1nLA

L2ncos1

nLA

L21ncos

nLA

L2

Lxncos

nLA

L2dx

LxnsinA

L2c

−−⎟⎠⎞

⎜⎝⎛π

=π−⎟⎠⎞

⎜⎝⎛π

=−π⎟⎠⎞

⎜⎝⎛π

−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ π

⎟⎠⎞

⎜⎝⎛π

−=⎟⎠⎞

⎜⎝⎛ π

= ∫

and substituting the value for A determined in part a gives

( )⎪⎩

⎪⎨

⎧

π=−−⎟⎠⎞

⎜⎝⎛π

=even nfor 0

odd nfor n

22)1(1

nL

L2c n2n

So the wavefunction at any later time t is given by

( ) ∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ π

π−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−φ=ψ

odd nn

tniEexpx

Lnsin

L2

n22tniE

exp)x(nnct,xhh

Note that the dimensions of ψ(x) are correct. (c) The probability of finding the particle at position x at tine t = 0 is given by

⎪⎩

⎪⎨

⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛π==

even nfor 0

odd nfor n

22cP

2

2nn

therefore

032.05

22cP

even is n since 0cP

811.022cP

2255

222

2211

=⎟⎟⎠

⎞⎜⎜⎝

⎛π

==

==

=⎟⎟⎠

⎞⎜⎜⎝

⎛π

==

(d) The average energy is found from

∑ ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ π⎟⎟⎠

⎞⎜⎜⎝

⎛π

=

∑=∑=++++=

odd n

nn

2n

nnn44332211avg

2mL2

222n 2

n22

EcEP...EPEPEPEPE

h

Problem 27 A particle of mass m is confined in an infinite potential well of width L. At time t = 0 the wavefunction of the particle is

( ) Axx,0ψ = where A is a constant. a) Find A so that the wavefunction ψ(x,0) is normalised. b) Find an expression for the wavefunction at any later time t

39

c) If measurements are carried out on this system at time t = 0, what is the probability that the particle will be found in the (i) n = 1, (ii) n = 2 and (iii) n = 5 state.

d) Calculate the average energy of the system at t = 0. (a) Normalisation

( ) 3L3A1

3

3L2A1L

0

dx2x2A1dx2x,0ψ =⇒=⇒=⇒=∞

∞−∫∫


( ) ∑ ⎟⎠⎞

⎜⎝⎛−=

n

nnn

tiEexp)x(φct,xψ

h


2

222

nnmL2πnE and

Lxπnsin

L2)x(φ h

=⎟⎠⎞

⎜⎝⎛=


dxL

xπnsinxAL2Axdx

Lxπnsin

L2ψφc

L

0

L

0

nn ∫∫ ⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛==

Either integrate pay parts or look up integral tables to evaluate the integral. The result is

( )( ) ( )n2222

L

022

L

0

n

)1(Lπnπn

LAL2πncosLπn

πnLA

L2

Lxπncosxπn

LxπnsinL

πnLA

L2dx

LxπnsinxA

L2c

−−=⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛= ∫

and substituting the value for A determined in part A gives

( ) ( ) ( )nn224

n22n )1(

πn6)1(Lπn

πnL

L6)1(Lπn

πnLA

L2c −−=−−=−−=


( ) ( )∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

n

n

n

tniEexpx

Lπnsin

L12

πn)1(tniE

exp)x(nφnct,xψhh

Note that the dimensions of ψ(x) are correct. (c) The probability of finding the particle at position x at tine t = 0 is given by

( ) 22

2n2

nnπn6)1(

πn6cP =⎥

⎦

⎤⎢⎣

⎡−−==

therefore

40

024.0π256cP

152.0π46cP

608.0π6cP

2255

2222

2211

===

===

===


∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

∑=∑=++++=

∞=

=

n

0n 2

222

22

nn

2n

nnn44332211avg

mL2πn

πn6

EcEP...EPEPEPEPE

h

Problem 28 A particle of mass m is confined in an infinite potential well of width L. At time t = 0 the wavefunction of the particle is

( ) x)-Ax(Lx,0ψ = where A is a constant. a) Find A so that the wavefunction ψ(x,0) is normalised. b) Find an expression for the wavefunction at any later time t c) If measurements are carried out on this system at time t = 0, what is the probability that the particle will be found in the (i) n = 1, (ii) n = 2 and (iii) n = 5 state. d) Calculate the average energy of the system at t = 0. (a) Normalisation

( ) ( )

5L

30A130

5L2A

130

5L630

5L1530

5L102A15

5L2

5L3

5L2A1L

05

5x4

4xL22L3

3x2A

1L

0

dx4x3Lx22L2x2A1L

0

dx2xL2x2A1dx2x,0ψ

=⇒=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⇒

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−⇒=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛+−⇒=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛+−⇒

=⎟⎠⎞⎜

⎝⎛ +−⇒=−⇒=

∞

∞−∫∫∫


( ) ∑ ⎟⎠⎞

⎜⎝⎛−=

n

nnn

tiEexp)x(φct,xψ

h


41

2

222

nnmL2πnE and

Lxπnsin

L2)x(φ h

=⎟⎠⎞

⎜⎝⎛=


( ) ( )

( ) dxL

xπnsinxL

xπnsinxLAL2dx

LxπnsinxLxA

L2

dxL

xπnsinxLxAL2dxxLAx

Lxπnsin

L2ψφc

L

0

2L

0

L

0

L

0

nn

∫∫

∫∫

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛−=−⎟

⎠⎞

⎜⎝⎛==

Look up integrals in integral tables.

( ) ( )

( )L

0

222233

L

0

2

n22L

022

L

0

LxπnsinxπLn2

LxπncosxπnL2

πnLdx

Lxπnsinx

1πn

Lπncosπn

LL

xπncosxπnL

xπnsinLπn

LdxL

xπnsinx

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛

−−=−=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

∫

∫

( ) ( )[ ] ( )[ ]⎥⎥⎦

⎤

⎢⎢⎣

⎡−−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−= 2)1(πn2

πnL2πncosπn2

πnL n22

33

322

33

3

Collect terms together

( ) ( )[ ]

( ) ( )[ ]

[ ] [ ] [ ]⎭⎬⎫

⎩⎨⎧

−−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−−+−−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−−+−−=

n33

n33

3

6n

33

3

n2233

3n

3

n2233

3n

3

n

)1(1πn154)1(1

πnL2

L60)1(1

πnL2A

L2

2)1(πn2πn

L1πn

LAL2

2)1(πn2πn

L1πn

LAL2c

So we see that

⎪⎩

⎪⎨

⎧=

oddn for πn158

evenn for 0c

33n


( ) ∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

odd n33

n

tniEexpx

Lπnsin

L30

πn8tniE

exp)x(nφnct,xψhh

(c) The probability that the particle will be found in the nth state is

⎪⎩

⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡==oddn for

πn158

evenn for 0

cP 2

33

2nn

42

so

52

3255

222

2

3211

1045.6π125158cP

even isn since 0cP

998.0π158cP

−×=⎥⎦

⎤⎢⎣

⎡==

==

=⎥⎦

⎤⎢⎣

⎡==


∑ ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

∑=∑=++++=

odd n33

nn

2n

nnn44332211avg

2mL2

22π2nπn158

EcEP...EPEPEPEPE

h

Problem 29 Calculate the commutators [ ] [ ] [ ]z,L ,y,L ,x,L zzz , where Lz is the z-componet of the angular momentum and px, py and pz are the x, y, and z components of the linear momentum. Recall the definition of Lz: Lz = xpy - ypx then [ ] [ ] ( ) ( )

[ ] yip,xy

yxpxypyxpxxpxypxxpxypxxpxypxxp

ypxpxxypxpx,ypxpx,L

x

xxxyxyxyxy

xyxyxyz

h==

+−=+−−=+−−=

−−−=−=

similarly [ ] [ ] ( ) ( )

[ ] xiy,px

xypyxpyypxypyypyxpyypyxpyypyxp

ypxpyyypxpy,ypxpy,L

y

yyxyxyxyxy

xyxyxyz

h−==

−=+−−=+−−=

−−−=−=

and finally [ ] [ ] ( ) ( )

0zypzxpzypzxp

ypxpzzypxpz,ypxpz,L

xyxy

xyxyxyz

=+−−=

−−−=−=

Problem 30 Suppose that an electron in the hydrogen atom is in a state characterised by the angular momentum quantum numbers l and ml. Show that, although in such a state neither Lx nor Ly is well defined, Lx

2 + Ly2 is well defined. Express the value of the latter quantity in

terms of l and ml. We know that the wavefunction ψn,l,m, of the hydrogen atom is an eigenfunction of both the L2 operator and the Lz operator, that is,

( ) m,,n2

m,,n2 ψ1ψL ll hll += and m,,nm,,nz ψmψL lll h=

We can write 2z

22y

2x

2z

2y

2x

2 LLLLLLLL −=+⇒++=

43

Thus ( ) ( ) ( ) ( )( ) m,,n

22m,,n

22m,,n

2m,,n

2z

2m,,n

2y

2x ψm1ψmψ1ψLLψLL lllllll hllhhll −+=−+=−=+

Thus, although ψn,l,m, is not an eigenfunction of the Lx and Ly operators, it is an eigenfunction of the ( )2

y2x LL + operator and the corresponding eigenvalue is

( )( ) 22m1 hll l−+ .

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CA Solution Set 2[1]