For questions 1 and 2 you may wish to note the following Gaussian integrals: 71

Question

For questions 1 and 2 you may wish to note the following Gaussian integrals: 71 where n is a non-negative integer and a is a real constant. When applying these formulae be careful to note the limits of integration.! 1. We have seen in the lectures that the ground-state wave function of the harmonic oscillator with potential V(x) m2x2 is: mo o)T 1 (x), deter (a) By using the properties of the raising opertor, (a-Yh (x) = Vn+1 y, mine the wave functions of the first two excited states, (x) and V2 (x) (b) Check the orthogonality of 0(x), | (x) and V2 (x) by explicit integration. Hint There is actually only one integration which you have to perform.) (c) Calculate the expectation value of the kinetic energy in the ground state 10 marks

Explanation / Answer

1. given ground state wave function

phio(x) = (m*w/pi*h')^1/4 e^[(-mw/2h')x^2]

wherte h' is reduced planks constant

a. a+phin(x) = sqroot(n + 1)phi(n+1)(x)

here a+ = sqroot(mw/2h')(x - i*p/mw) [ where p is momentum vector, i is iota]

hence

phi1(x) = sqroot(mw/2h')(x - i*p/mw)(m*w/pi*h')^1/4 e^[(-mw/2h')x^2]/sqroot(2)

phi2(x) = (mw/2h')(x - i*p/mw)^2*(m*w/pi*h')^1/4 e^[(-mw/2h')x^2]/sqroot(6)

b. lets find orthogonality of phio(x) and phi1(x)

<phi1(x)|phio(x)> = [integrate from -infinity to +infinity][phi1*(x)phio(x)dx]

phi1*(x) = sqroot(mw/2h')(x + i*p/mw)(m*w/pi*h')^1/4 e^[(-mw/2h')x^2]/sqroot(2)

<phi1(x)|phio(x)> = [integrate from -infinity to +infinity][sqroot(mw/2h')(x + i*p/mw)(m*w/pi*h')^1/4 e^[(-mw/2h')x^2]/sqroot(2) * (m*w/pi*h')^1/4 e^[(-mw/2h')x^2]dx]

for orthogonality this should be 0

[integrate from -infinity to +infinity][sqroot(mw/2h')(x + i*p/mw)(m*w/pi*h')^1/4 e^[(-mw/2h')x^2]/sqroot(2) * (m*w/pi*h')^1/4 e^[(-mw/2h')x^2]dx] = 0

[integrate from -infinity to +infinity][(x + i*p/mw) e^[(-mw/2h')x^2] * e^[(-mw/2h')x^2]dx] = 0

[integrate from -infinity to +infinity][x e^[(-mw/h')x^2 ]dx] = -[integrate from -infinity to +infinity][i*p/mw*e^[(-mw/h')x^2 ]]

now RHS is imaginary part and can be ignored

[integrate from -infinity to +infinity][x e^[(-mw/h')x^2 ]dx] = 0

form the given identity fopr n = 0

integration from 0 to infinity xe^(-x^2/a^2)dx = a^2/2

[integrate from -infinity to +infinity][x e^[(-mw/h')x^2 ]dx] = -[integrate from -infinity to +infinity][i*p/mw*e^[(-mw/h')x^2 ]]

the LHS is an odd function of x hence [integration from 0 to infinity] + [integration from -infinty to 0] = 0

hence phio(x) and phi1(x) are orthogonal

similiarly we can find

phi1(x) and phi2(x) are also orthogonal

hence

phi0(x) and phi2(x) are also orhtogonal

c. the expectation of kinetic energy of a harmonic osscilator in nth state is given by

En = (n + 0.5)h'w

hence

Eo = h'w/2

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