Particle in a box, Harmonic oscillator, Rotational motion, Hydrogen atom
Particle in a box, Harmonic oscillator, Rotational motion, Hydrogen atom
April 7, 2014
0.1
Particle in a box,
0.1.1 Normalize the wavefunction. The eigenfunctions ψn (x) are orthonormal that means : i)orthogonal and ii)normalized. Since the eigenfunctions are orthogonal Z +∞ Z +∞ Z +∞ Z +∞ dx|ψ(x)|2 = |A|2 dx|ψ1 (x)|2 + 2 1 = |A|2 dx|ψ2 (x)|2 + 3 dx|ψ3 (x)|2 (1) −∞
−∞
−∞
and since they are normalized and defined in [0, L] Z +∞ Z +∞ Z +∞ dx|ψ1 (x)|2 + 2 dx|ψ2 (x)|2 + 3 dx|ψ3 (x)|2 = 6 −∞
−∞
−∞
(2)
−∞
so that after normalization: i √ √ 1 h ψ(x) = √ ψ1 (x) + 2ψ2 (x) + 3ψ3 (x) 6
(3)
0.1.2 Using r ψn (x) =
Figure 1: Wavefunction ψ(x) =
nπx 2 sin L L
√1 6
(4)
√ √ ψ1 (x) + 2ψ2 (x) + 3ψ3 (x)
0.1.3 r ψn (x) =
2 nπx sin L L 1
(5)
also we have Z
+∞
dx|ψn (x)|2 =
−∞
L
Z
dx|ψn (x)|2 = 0
(6)
0
and L
Z
dxψn (x)ψm (x) = 0
(7)
0
m 6= n
(8)
however notice Z
L/2
dxψn (x)ψm (x) 6= 0
(9)
0
(10)
√ √ 2 Figure 2: Wavefunction |ψ(x)|2 = √16 ψ1 (x) + 2ψ2 (x) + 3ψ3 (x) Z Z 1 L/2 1 L/2 dx|ψ1 (x)|2 + dx|ψ2 (x)|2 + dx|ψ3 (x)|2 + 3 0 2 0 0 0 √ Z L/2 √ Z L/2 √ Z L/2 2 3 6 + dxψ1 (x)ψ2 (x) + dxψ1 (x)ψ3 (x) + dxψ2 (x)ψ3 (x) 3 0 3 0 3 0 Z
L/2
dx|ψ(x)|2 =
1 6
Z
L/2
notice that out of symmetry of the eigenfunctions Z L/2 Z L/2 Z L/2 1 dx|ψ1 (x)|2 = dx|ψ2 (x)|2 = dx|ψ3 (x)|2 = 2 0 0 0
(11)
the rest are just trigonometric integrals and once you complete the calculation you should find the probability is about 0.9. 2
0.2
Harmonic oscillator
0.2.1 We are given ∆E = hν
(12)
∆E h
(13)
so that ν=
now to find the force constant k we need to equate it with s k 1 ν= 2π µ
(14)
where the reduced mass is µ=
mH mBr mH + mBr
(15)
0.2.2 The vibrational states are just given by the HO eigenfunctions below in Fig3
Figure 3: The first harmonic oscillator eigenstates
3
0.2.3 Proceed like in section 0.2.1 but notice we have a different reduced mass µ0 =
m0H mBr m0H + mBr
(16)
so that changes all the other equations ν=
∆E h
now to find the force constant k we need to equate it with s 1 k ν= 2π µ
0.3
(17)
(18)
Rotational motion
0.3.1 Again to compute the moment of inertia we juts need the reduced mass µ=
mC mO mC + mO
(19)
and hence I = µR2
(20)
where R is the bond length. The rotational eigenvalues are just El =
l(l + 1)¯h2 2I
(21)
so that the first three rotational states are E0 , E1 and E2 .
0.3.2 Consider that for each l we have ml = −l, −l+1, −l+2.....−1, 0, 1, ..., l−2, l−1, l degenerate states. So how many states are those?
0.3.3 If ml = 0 the rotating motion about the z-axis is taking place on the xy plane at the origin as at the center of fig.4 the center of rotation is the center of mass.
4
Figure 4: For ml = 0 we only have the components Lx and Ly .
0.4
Hydrogen atom
0.4.1 In fig5 you have a representation of the 2p probability density you can describe
Figure 5: 2p state the maximum in terms of Bohr radius, nodes and symmetry of the probability 5
distribution in connection to its angular momentum for this energy state for l = 1. The same should be done for the n = 4 and l = 2 state.
0.4.2 Remember the selection rules for rotational transitions require that for the atom to have a permanent dipole moment we must have ∆l
= ±1
(22) (23)
0.4.3 Using the Rydberg formula hν = En2 − En1 = −13.6eV (
1 1 − 2) 2 n2 n1
(24)
positive for a photon emitted. It also works if the n1, n2 restriction is relaxed. In that case the negative energy means a photon (of positive energy) is absorbed.
6
April 7, 2014
0.1
Particle in a box,
0.1.1 Normalize the wavefunction. The eigenfunctions ψn (x) are orthonormal that means : i)orthogonal and ii)normalized. Since the eigenfunctions are orthogonal Z +∞ Z +∞ Z +∞ Z +∞ dx|ψ(x)|2 = |A|2 dx|ψ1 (x)|2 + 2 1 = |A|2 dx|ψ2 (x)|2 + 3 dx|ψ3 (x)|2 (1) −∞
−∞
−∞
and since they are normalized and defined in [0, L] Z +∞ Z +∞ Z +∞ dx|ψ1 (x)|2 + 2 dx|ψ2 (x)|2 + 3 dx|ψ3 (x)|2 = 6 −∞
−∞
−∞
(2)
−∞
so that after normalization: i √ √ 1 h ψ(x) = √ ψ1 (x) + 2ψ2 (x) + 3ψ3 (x) 6
(3)
0.1.2 Using r ψn (x) =
Figure 1: Wavefunction ψ(x) =
nπx 2 sin L L
√1 6
(4)
√ √ ψ1 (x) + 2ψ2 (x) + 3ψ3 (x)
0.1.3 r ψn (x) =
2 nπx sin L L 1
(5)
also we have Z
+∞
dx|ψn (x)|2 =
−∞
L
Z
dx|ψn (x)|2 = 0
(6)
0
and L
Z
dxψn (x)ψm (x) = 0
(7)
0
m 6= n
(8)
however notice Z
L/2
dxψn (x)ψm (x) 6= 0
(9)
0
(10)
√ √ 2 Figure 2: Wavefunction |ψ(x)|2 = √16 ψ1 (x) + 2ψ2 (x) + 3ψ3 (x) Z Z 1 L/2 1 L/2 dx|ψ1 (x)|2 + dx|ψ2 (x)|2 + dx|ψ3 (x)|2 + 3 0 2 0 0 0 √ Z L/2 √ Z L/2 √ Z L/2 2 3 6 + dxψ1 (x)ψ2 (x) + dxψ1 (x)ψ3 (x) + dxψ2 (x)ψ3 (x) 3 0 3 0 3 0 Z
L/2
dx|ψ(x)|2 =
1 6
Z
L/2
notice that out of symmetry of the eigenfunctions Z L/2 Z L/2 Z L/2 1 dx|ψ1 (x)|2 = dx|ψ2 (x)|2 = dx|ψ3 (x)|2 = 2 0 0 0
(11)
the rest are just trigonometric integrals and once you complete the calculation you should find the probability is about 0.9. 2
0.2
Harmonic oscillator
0.2.1 We are given ∆E = hν
(12)
∆E h
(13)
so that ν=
now to find the force constant k we need to equate it with s k 1 ν= 2π µ
(14)
where the reduced mass is µ=
mH mBr mH + mBr
(15)
0.2.2 The vibrational states are just given by the HO eigenfunctions below in Fig3
Figure 3: The first harmonic oscillator eigenstates
3
0.2.3 Proceed like in section 0.2.1 but notice we have a different reduced mass µ0 =
m0H mBr m0H + mBr
(16)
so that changes all the other equations ν=
∆E h
now to find the force constant k we need to equate it with s 1 k ν= 2π µ
0.3
(17)
(18)
Rotational motion
0.3.1 Again to compute the moment of inertia we juts need the reduced mass µ=
mC mO mC + mO
(19)
and hence I = µR2
(20)
where R is the bond length. The rotational eigenvalues are just El =
l(l + 1)¯h2 2I
(21)
so that the first three rotational states are E0 , E1 and E2 .
0.3.2 Consider that for each l we have ml = −l, −l+1, −l+2.....−1, 0, 1, ..., l−2, l−1, l degenerate states. So how many states are those?
0.3.3 If ml = 0 the rotating motion about the z-axis is taking place on the xy plane at the origin as at the center of fig.4 the center of rotation is the center of mass.
4
Figure 4: For ml = 0 we only have the components Lx and Ly .
0.4
Hydrogen atom
0.4.1 In fig5 you have a representation of the 2p probability density you can describe
Figure 5: 2p state the maximum in terms of Bohr radius, nodes and symmetry of the probability 5
distribution in connection to its angular momentum for this energy state for l = 1. The same should be done for the n = 4 and l = 2 state.
0.4.2 Remember the selection rules for rotational transitions require that for the atom to have a permanent dipole moment we must have ∆l
= ±1
(22) (23)
0.4.3 Using the Rydberg formula hν = En2 − En1 = −13.6eV (
1 1 − 2) 2 n2 n1
(24)
positive for a photon emitted. It also works if the n1, n2 restriction is relaxed. In that case the negative energy means a photon (of positive energy) is absorbed.
6
