H PHY 326 3rd homework Optional complement to the second ... · H* PHY 326 - 3rd homework - Optional complement to the second exercise - H. Panagopoulos, December 2014 *L H* We shall

H* PHY 326 - 3rd homework - Optional complement to the second exercise -

H. Panagopoulos, December 2014 *L

H* We shall use the quantity u = Π2Ñ2�I2ma2M as our unit of energy.

In this unit: E = z * u, where z is a dimensionless unknown.

Thus, e.g., when z = 1 the energy assumes its unperturbed value.

Denoting also: ¶ = V0�u, the quantization condition takes the form:

z-¶ tanIHΠ�4L z-¶ M = z cotIHΠ�4L z MFor a given value of ¶ this condition can be solved numerically for z.

Let us plot the left hand side HblueLand the right hand side HredL of the condition.

The abscissa of the leftmost intersection point is

the smallest value of z Hthat is, of the energyLwhich satisfies the quantization condition. *L

H* Let us give a particular value to the perturbation V0: ¶ = 0.3 *L

In[1]:= Plot@8Sqrt@z - 0.3D Tan@Pi � 4 Sqrt@z - 0.3DD, Sqrt@zD Cot@Pi � 4 Sqrt@zDD<, 8z, 0, 30<D

Out[1]=

5 10 15 20 25 30

-20

-10

10

20

H* Let us compute the energy perturbatively IEH0L+E

H1LMfor various values of V0: ¶ = 0, 0.1, 0.2, ..., 0.9, 1.0 *L

In[2]:= Table@8e, 1 + e H1 � 2 + 1 � PiL<, 8e, 0, 1, 0.1<DOut[2]= 880., 1.<, 80.1, 1.08183<, 80.2, 1.16366<,

80.3, 1.24549<, 80.4, 1.32732<, 80.5, 1.40915<, 80.6, 1.49099<,80.7, 1.57282<, 80.8, 1.65465<, 80.9, 1.73648<, 81., 1.81831<<

H* Now let us compute the energy variationally for the same values of V0: ¶ = 0,

0.1, 0.2, ..., 0.9, 1.0 *L

In[3]:= c1^2 + 9 c3^2 + 25 c5^2 +

e Integrate@H2 � aL Hc1 Sin@Pi Hx + a � 2L � aD + c3 Sin@3 Pi Hx + a � 2L � aD +

c5 Sin@5 Pi Hx + a � 2L � aDL^2, 8x, -a � 4, a � 4<DOut[3]= c1

2+ 9 c3

2+ 25 c5

2+

e H-60 c3 c5 - 20 c1 H3 c3 + c5L + 15 c12 H2 + ΠL + 5 c3

2 H-2 + 3 ΠL + 3 c52 H2 + 5 ΠLL

30 Π

In[4]:= Table@8e, Minimize@8%, c1^2 + c3^2 + c5^2 � 1<, 8c1, c3, c5<D@@1DD<, 8e, 0, 1, 0.1<DOut[4]= 880., 1.<, 80.1, 1.0817<, 80.2, 1.16313<,

80.3, 1.24429<, 80.4, 1.32517<, 80.5, 1.40577<, 80.6, 1.48608<,80.7, 1.5661<, 80.8, 1.64583<, 80.9, 1.72525<, 81., 1.80436<<

H* We note that the perturbative and variational estimates of the energy

are very close to each other, especially for small values of V0 *L

H* Now let us also find the exact values of the ground state

energy by numerically solving the quantization condition. *L

In[5]:= Table@8e, FindRoot@Sqrt@zD Cot@Pi Sqrt@zD � 4D � Sqrt@z - eD Tan@Pi � 4 Sqrt@z - eDD,8z, 1.2<D@@1, 2DD<, 8e, 0, 1, 0.1<D

Out[5]= 880., 1.<, 80.1, 1.0817<, 80.2, 1.16312<,80.3, 1.24426<, 80.4, 1.32511<, 80.5, 1.40568<, 80.6, 1.48595<,80.7, 1.56592<, 80.8, 1.64559<, 80.9, 1.72494<, 81., 1.80398<<

H* We note that the exact values of the energy are extremely close

to the variational estimates, and always smaller, as they must. *L

H* Finally, let us plot on the same graph the perturbative HredL,variational HblueL and exact HgreenL values of

the ground state energy for different values of V0 *L

2 PHY325_hw3_2014sol.nb

In[6]:= ListPlot@8%2, %4, %5<, Joined ® True, PlotStyle ® 8Red, Blue, Green<D

Out[6]=

0.2 0.4 0.6 0.8 1.0

1.2

1.4

1.6

1.8

In[7]:= Quit

PHY325_hw3_2014sol.nb 3