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Quantum orbital angular momentum operators (spherical coordinates)

The quantum orbital angular momentum operators in spherical coordinates are derived using the following diagram:

where

x=rsin\theta cos\phi\; \; \; \; y=rsin\theta sin\phi\; \; \; \; z=rcos\theta\; \; \; \; r=\sqrt{x^{2}+y^{2}+z^{2}}\; \; \; \; \; \; \; \; 77

Therefore,

\frac{\partial r}{\partial x}=\frac{\partial \sqrt{x^{2}+y^{2}+z^{2}}}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}=\frac{x}{r}=sin\theta cos\phi\; \; \; \; \; \; \; \; 78

Similarly,

\frac{\partial r}{\partial y}=sin\theta sin\phi\; \; \; \; \; \; \; \; 79

\frac{\partial r}{\partial z}= cos\theta\; \; \; \; \; \; \; \; 80

Furthermore, by differentiating cos\theta=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} implicitly with respect to x and separately with respect to y, and rearranging, we have

\frac{\partial\theta}{\partial x}=\frac{cos\theta cos\phi}{r}\; \; \; \; \; \; \; \; 81

\frac{\partial\theta}{\partial y}=\frac{cos\theta sin\phi}{r}\; \; \; \; \; \; \; \; 82

 

Question

Show that sin\phi=\frac{y}{\sqrt{x^{2}+y^{2}}}, cos\phi=\frac{x}{\sqrt{x^{2}+y^{2}}} and sin\theta=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}}.

Answer

To find expressions for sin\phi and cos\phi, we let \theta=\frac{\pi}{2} for the first three equations of eq77, which gives us x=rcos\phi, y=rsin\phi and z=0. So,

cos\phi=\frac{x}{\sqrt{x^{2}+y^{2}}}\; \; \; \; \; \;sin\phi=\frac{y}{\sqrt{x^{2}+y^{2}}}

Substituting these two expressions back into either the first or second equation of eq77, we have

sin\theta=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}}

 

Implicit differentiation of sin\theta=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}},  sin\phi=\frac{y}{\sqrt{x^{2}+y^{2}}} and cos\phi=\frac{x}{\sqrt{x^{2}+y^{2}}} with respect to z, x and y respectively gives

\frac{\partial\theta}{\partial z}=-\frac{sin\theta}{r}\; \; \; \; \; \; \; \; 83

\frac{\partial\phi}{\partial x}=-\frac{sin\phi}{rsin\theta}\; \; \; \; \; \; \; \; 84

\frac{\partial\phi}{\partial y}=\frac{cos\phi}{rsin\theta}\; \; \; \; \; \; \; \; 85

Since \phi is independent of z

\frac{\partial\phi}{\partial z}=0\; \; \; \; \; \; \; \; 86

Applying the multivariable chain rule to f\left ( r,\theta,\phi \right ), we have:

\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial x}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; 87

\frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial y}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial y}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; 88

\frac{\partial}{\partial z}=\frac{\partial r}{\partial z}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial z}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial z}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; 89

Substitute i) eq78, eq81 and eq84 in eq87, ii) eq79, eq82 and eq85 in eq88 and iii) eq80, eq83 and eq86 in eq89, we have

\frac{\partial}{\partial x}=sin\theta cos\phi\frac{\partial}{\partial r}+\frac{cos\theta cos\phi}{r}\frac{\partial}{\partial \theta}-\frac{sin\phi }{rsin\theta}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; \; \; 90

\frac{\partial}{\partial y}=sin\theta sin\phi\frac{\partial}{\partial r}+\frac{cos\theta sin\phi}{r}\frac{\partial}{\partial \theta}+\frac{cos\phi }{rsin\theta}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; \; \; 91

\frac{\partial}{\partial z}=cos\theta \frac{\partial}{\partial r}-\frac{sin\theta}{r}\frac{\partial}{\partial \theta}\; \; \; \; \; \; \; \; \; \; 92

respectively.

Substitute eq77, eq90, eq91 and eq92 in eq72, eq73 and eq74, we have

\hat{L}_x=\frac{\hbar}{i}\left ( -sin\phi\frac{\partial}{\partial \theta}-\frac{cos\theta cos\phi}{sin\theta}\frac{\partial}{\partial \phi}\right )\; \; \; \; \; \; \; \; 93

\hat{L}_y=\frac{\hbar}{i}\left (cos\phi\frac{\partial}{\partial \theta}-\frac{cos\theta sin\phi}{sin\theta}\frac{\partial}{\partial \phi}\right )\; \; \; \; \; \; \; \; 94

\hat{L}_z=\frac{\hbar}{i}\frac{\partial}{\partial \phi}\; \; \; \; \; \; \; \; 95

respectively.

Substitute eq93, eq94 and eq95 in eq75, we have, with some algebra

\hat{L}^{2}=-\hbar^{2}\left ( \frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{cos\theta}{sin\theta}\frac{\partial}{\partial \theta}\right )

or equivalently

\hat{L}^{2}=-\hbar^{2}\left \[ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial \theta}\left ( sin\theta\frac{\partial}{\partial \theta}\right )\right ]\; \; \; \; \; \; \; \; 96

\hat{L}^{2} is the quantum orbital angular momentum operator and each of its eigenvalues is the square of the orbital angular momentum of an electron.

Question

Show that eq76 is \frac{\hbar}{i}\left ( \boldsymbol{\mathit{\phi}}\frac{\partial}{\partial \theta}- \boldsymbol{\mathit{\theta}}\frac{1}{sin \theta}\frac{\partial}{\partial \phi}\right ) in spherical coordinates.

Answer

Substitute eq93, eq94, eq95 and unit vectors in spherical coordinates \boldsymbol{\mathit{i}}= \boldsymbol{\mathit{r}}sin\theta cos\phi+ \boldsymbol{\mathit{\theta}}cos\phi cos\theta-\boldsymbol{\mathit{\phi}}sin\phi, \boldsymbol{\mathit{j}}= \boldsymbol{\mathit{r}}sin\theta sin\phi+ \boldsymbol{\mathit{\theta}}cos\theta sin\phi-\boldsymbol{\mathit{\phi}}cos\phi and \boldsymbol{\mathit{k}}= \boldsymbol{\mathit{r}}cos\theta- \boldsymbol{\mathit{\theta}}sin\theta in eq76. we have, after some algebra, we have

\boldsymbol{\mathit{\hat{L}}}=\frac{\hbar}{i}\left ( \boldsymbol{\mathit{\phi}}\frac{\partial}{\partial\theta}- \boldsymbol{\mathit{\theta}}\frac{1}{sin\theta}\frac{\partial}{\partial\phi}\right )\; \; \; \; \; \; \; \; 97

 

Question

Show that \boldsymbol{\mathit{\hat{L}}}\cdot\boldsymbol{\mathit{\hat{L}}}=\hat{L}^{2}.

Answer

Substituting eq97 in \hat{L}^{2}=\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{L}} and using \frac{\partial\boldsymbol{\mathit{\phi}}}{\partial\theta}=0, \frac{\partial\boldsymbol{\mathit{\theta}}}{\partial\theta}=-\boldsymbol{\mathit{r}}, \frac{\partial\boldsymbol{\mathit{\phi}}}{\partial\phi}=-\boldsymbol{\mathit{r}} sin\theta-\boldsymbol{\mathit{\theta}}cos\theta and \frac{\partial\boldsymbol{\mathit{\theta}}}{\partial\phi}=\boldsymbol{\mathit{\phi}}cos\theta, we have

\hat{L}^{2}=\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{L}}=-\hbar^{2}\left [\frac{\partial^{2}}{\partial\theta^{2}}+\frac{cos\theta}{sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right ]\; \; \; \; \; \; \; \; 98

 

 

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Quantum orbital angular momentum operators (Cartesian coordinates)

The quantum orbital angular momentum operators in Cartesian coordinates are derived from the classical angular momentum components

L_x=r_yp_z-r_zp_y

L_y=r_zp_x-r_xp_z

L_z=r_xp_y-r_yp_x

by replacing the position and linear momentum components with their corresponding operators. Since r_x,r_y,r_z are position components with position operators \hat{r}_x,\hat{r}_y,\hat{r}_z respectively, and p_x,p_y,p_z are linear momentum components with linear momentum operators \frac{\hbar}{i}\frac{\partial}{\partial r_x},\frac{\hbar}{i}\frac{\partial}{\partial r_y},\frac{\hbar}{i}\frac{\partial}{\partial r_z} respectively (see eq4), we have

\hat{L}_x=\frac{\hbar}{i}\left (r_y \frac{\partial}{\partial r_z}-r_z\frac{\partial}{\partial r_y}\right )\; \; \; \; or\; \; \; \;\hat{L}_x=\frac{\hbar}{i}\left (y \frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right )\; \; \; \; \; \; \; \; 72

\hat{L}_y=\frac{\hbar}{i}\left (r_z \frac{\partial}{\partial r_x}-r_x\frac{\partial}{\partial r_z}\right )\; \; \; \; or\; \; \; \;\hat{L}_y=\frac{\hbar}{i}\left (z \frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right )\; \; \; \; \; \; \; \; 73

\hat{L}_z=\frac{\hbar}{i}\left (r_x \frac{\partial}{\partial r_y}-r_y\frac{\partial}{\partial r_x}\right )\; \; \; \; or\; \; \; \;\hat{L}_z=\frac{\hbar}{i}\left (x \frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right )\; \; \; \; \; \; \; \; 74

From eq70, we have

\hat{L}^{2}=\hat{L}_x^{\, \, 2}+\hat{L}_y^{\, \, 2}+\hat{L}_z^{\, \, 2}\; \; \; \; \; \; \; \; 75

Since \hat{L}^{ 2} is defined as the operator for the square of the magnitude of , each of its eigenvalues is the square of the magnitude of the orbital angular momentum of an electron. 

 

Question

Can we construct an angular momentum operator using \boldsymbol{\mathit{L}}=\boldsymbol{\mathit{i}}L_x+\boldsymbol{\mathit{j}}L_y+\boldsymbol{\mathit{k}}L_z, such that

\hat{\boldsymbol{\mathit{L}}}=\boldsymbol{\mathit{i}}\hat{L}_x+\boldsymbol{\mathit{j}}\hat{L}_y+\boldsymbol{\mathit{k}}\hat{L}_z\; \; \; \; \; \; \; \; 76

which is then used to generate eigenvalues?

Answer

\hat{L}^{2}, a scalar operator, is preferred over \hat{\boldsymbol{\mathit{L}}}, a vector operator, because it easier to manipulate in quantum computations. \hat{L}^{2} commutes with its component operators, allowing us to simultaneously determine the eigenvalues of \hat{L}^{2} and say, \hat{L}_z (which is useful, for example in the verification of singlet and triplet eigenstates). It also commutes with the time-independent Hamiltonian \hat{H}, also a scalar operator, implying that we can select a common complete set of eigenfunctions for \hat{L}^{2} and \hat{H}. Note that \hat{L}^{2} has a form that is consistent with the angular portion of \hat{H} in spherical coordinates (compare eq49 with eq96). An eigenvalue of \hat{H}_{angular} is the energy associated with the angular motion of an electron, while the square root of an eigenvalue of \hat{L}^{2} is the magnitude of the orbital angular momentum of an electron in a particular state.

 

 

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Classical orbital angular momentum

The classical definition of angular momentum is a pseudo-vector, which can be separated into its 3 components in Cartesian coordinates as follows:

If \boldsymbol{\mathit{A}} and \boldsymbol{\mathit{B}} are two vectors

\boldsymbol{\mathit{A}}=\boldsymbol{\mathit{i}}A_x+\boldsymbol{\mathit{j}}A_y+\boldsymbol{\mathit{k}}A_z\; \; \; \; \; \;\boldsymbol{\mathit{B}}=\boldsymbol{\mathit{i}}B_x+\boldsymbol{\mathit{j}}B_y+\boldsymbol{\mathit{k}}B_z

where A_x is the component of A in the x-direction and \boldsymbol{\mathit{i}},\boldsymbol{\mathit{j}},\boldsymbol{\mathit{k}} are unit vectors in the x,y,z directions.

The cross product of the two vectors is:

\boldsymbol{\mathit{C}}=\boldsymbol{\mathit{A}}\times\boldsymbol{\mathit{B}}=\boldsymbol{\mathit{i}}(A_yB_z-A_zB_y)+\boldsymbol{\mathit{j}}(A_zB_x-A_xB_z)+\boldsymbol{\mathit{k}}(A_xB_y-A_yB_x)\; \; \; \; \; \; \; \; 69

Since \boldsymbol{\mathit{C}}=\boldsymbol{\mathit{i}}C_x+\boldsymbol{\mathit{j}}C_y+\boldsymbol{\mathit{k}}C_z

C_x=A_yB_z-A_zB_y

C_y=A_zB_x-A_xB_z

C_z=A_xB_y-A_yB_x

Comparing eq59a and eq69,

\boldsymbol{\mathit{L}}=\boldsymbol{\mathit{r}}\times\boldsymbol{\mathit{p}}=\boldsymbol{\mathit{i}}(r_yp_z-r_zp_y)+\boldsymbol{\mathit{j}}(r_zp_x-r_xp_z)+\boldsymbol{\mathit{k}}(r_xp_y-r_yp_x)

and

L_x=r_yp_z-r_zp_y

L_y=r_zp_x-r_xp_z

L_z=r_xp_y-r_yp_x

L_x, L_y and L_z are the classical orbital angular momenta about the x-axis, y-axis and z-axis respectively. Since the magnitude of a vector \boldsymbol{\mathit{v}}=x\boldsymbol{\mathit{i}}+y\boldsymbol{\mathit{j}}+z\boldsymbol{\mathit{k}} is \vert\boldsymbol{\mathit{v}}\vert=\sqrt{x^{2}+y^{2}+z^{2}}, we have \vert\boldsymbol{\mathit{L}}\vert^{2}=L_{x}^{\, \, 2}+L_{y}^{\, \, 2}+L_{z}^{\, \, 2} or simply

L^{2}=L_{x}^{\, \, 2}+L_{y}^{\, \, 2}+L_{z}^{\, \, 2}\; \; \; \; \; \; \; \; 70

In other words, L^{2} is square of the magnitude of the vector \boldsymbol{\mathit{L}}. The significance of L^{2} will be explored in subsequent articles.

 

Question

Show that \boldsymbol{\mathit{\tau}}=\frac{d\boldsymbol{\mathit{L}}}{dt}.

Answer

From eq59a,

\frac{d\boldsymbol{\mathit{L}}}{dt}=\boldsymbol{\mathit{r}}\times\frac{d\boldsymbol{\mathit{p}}}{dt}+\boldsymbol{\mathit{p}}\times\frac{d\boldsymbol{\mathit{r}}}{dt}=\boldsymbol{\mathit{r}}\times m\frac{d\boldsymbol{\mathit{v}}}{dt}+m\boldsymbol{\mathit{v}}\times\boldsymbol{\mathit{v}}=\boldsymbol{\mathit{r}}\times\boldsymbol{\mathit{F}}=\boldsymbol{\mathit{\tau}}

Hence,

\boldsymbol{\mathit{\tau}}=\frac{d\boldsymbol{\mathit{L}}}{dt}\; \; \; \; \; \; \; \; 71

 

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Relation between magnetic dipole moment and angular momentum

In classical electrodynamics, a magnetic dipole moment \mu is associated with a current loop (see diagram below), and represents the magnitude and orientation of a magnetic dipole. It is a pseudo-vector, whose direction is perpendicular to the plane of the loop and given by the right hand thumb rule.

The magnitude of a magnetic dipole moment of a current loop is defined as

\mu=IA\; \; \; \; \; \; \; \; 60

where I is current, and A is the area of the loop.

Rewriting eq60 in terms of I=\frac{q}{t}=\frac{qv_{\perp}}{2\pi r} (where q is charge, t is time and v_{\perp} is the tangential velocity of the charged particle) and using A=\pi r^{2}, we have \mu=\frac{q}{2m}rmv_{\perp}, where m is the mass of the charged particle. Substitute eq59 in \mu=\frac{q}{2m}rmv_{\perp}, we have the relation between magnetic dipole moment and angular momentum:

\boldsymbol{\mathit{\mu}}=\gamma\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 61

where \gamma=\frac{q}{2m} is the classical gyromagnetic ratio.

When placed in an external magnetic field , the magnetic dipole moment experiences a torque, whose energy  is

U=-\boldsymbol{\mathit{\mu}}\cdot\boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 62

 

Question

How is eq62 derived?

Answer

In order to rotate a current loop, we must do work W against the torque \tau due to the magnetic field \boldsymbol{\mathit{B}}. For a rotating system (see diagram I below), W=-\int F_{\perp}ds, where according to convention, the negative sign is added so that work on the system by the field is positive.

For small \theta, ds=rd\theta. Since \boldsymbol{\mathit{\tau}}=\boldsymbol{\mathit{r}}\times\boldsymbol{\mathit{F}} and W=-\Delta U, we have

U_f-U_i=\int F_{\perp}ds=\int_{\theta_i}^{\theta_f}Fsin\theta rd\theta=\int_{\theta_i}^{\theta_f}\tau d\theta\; \; \; \; \; \; \; \; 62a

Diagram II below shows a square current loop with side 1 parallel to side 3 and side 2 parallel to side 4 (not shown in diagram). The length of each side is b.

Since \tau=r\times F=\frac{b}{2}Fsin\theta+\frac{b}{2}Fsin\theta and F=BIb (\theta=90^{\circ}),

U_f-U_i=\int_{\theta_i}^{\theta_f}IBb^{2}sin\theta\, d\theta=IA B\left (- cos\theta_f+cos\theta_i \right)\; \; \; \; \; \; \; \; 62b

where A is the area of the loop.

Substitute eq60 in the above equation,

U_f-U_i=\mu B\left (cos\theta_i-cos\theta_f \right)\; \; \; \; \; \; \; \; 63

Comparing eq62a and eq62b, torque can also be expressed as

\boldsymbol{\mathit{\tau}}=BIAsin\theta=\boldsymbol{\mathit{\mu}}\times\boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 64

The torque exerted by the magnetic field on the magnetic dipole tends to rotate the dipole towards a lower energy state. So, we let \theta_i=90^{\circ} with U_i=0 and eq63 becomes U_f=-\mu Bcos\theta_f, or simply:

U=-\boldsymbol{\mathit{\mu}}\cdot \boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 65

 

Other than a torque, the external magnetic field may exert another force on the current loop. For a magnetic field pointing in the z-direction, \boldsymbol{\mathit{B}}=B_z(x,y,z)\boldsymbol{\mathit{k}}, where B_z(x,y,z) is a scalar function. We substitute eq65 in F_z=-\frac{\partial U}{\partial z} to give

\boldsymbol{\mathit{F_z}}=\frac{\partial(\boldsymbol{\mathit{\mu}}\cdot\boldsymbol{\mathit{B}})}{\partial z}=\frac{\partial[(\mu_x\boldsymbol{\mathit{i}}+\mu_y\boldsymbol{\mathit{j}}+\mu_z\boldsymbol{\mathit{k}})\cdot B_z(x,y,z)\boldsymbol{\mathit{k}}]}{\partial z}=\mu_z\frac{\partial B_z(x,y,z)}{\partial z}\; \; \; \; \; \; \; \; 66

where \boldsymbol{\mathit{i}}, \boldsymbol{\mathit{j}} and \boldsymbol{\mathit{k}} are unit vectors, and \frac{\partial B_z(x,y,z)}{\partial z} is the gradient of the external magnetic field along the z-direction.

For a uniform field, B_z(x,y,z)=B_0 (i.e. a constant) and \boldsymbol{\mathit{F_z}}=0; while for an inhomogenous field, B_z(x,y,z)=B_0+\alpha z, where \alpha is a scalar representing the change in B_0 along the z-axis, and \boldsymbol{\mathit{F_z}}\neq 0.

Substituting eq61 in eq62,

U=-\gamma\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 67

For an inhomogenous magnetic field pointing in the z-direction, the above equation becomes

U=-\gamma(B_0+\alpha z)\boldsymbol{\mathit{k}}\cdot(L_x\boldsymbol{\mathit{i}}+L_y\boldsymbol{\mathit{j}}+L_z\boldsymbol{\mathit{k}})=-\gamma(B_0+\alpha z)L_z\; \; \; \; \; \; \; \; 68

Eq68 is used as a starting point in analysing results from the Stern-Gerlach experiment.

 

Question

Does the magnetic field \boldsymbol{\mathit{B}}=(B_0+\alpha z)\boldsymbol{\mathit{k}} violate Maxwell’s 2nd equation of \nabla\cdot\boldsymbol{\mathit{B}}=0 where \nabla=\frac{\partial}{\partial x}\boldsymbol{\mathit{i}}+\frac{\partial}{\partial y}\boldsymbol{\mathit{j}}+\frac{\partial}{\partial z}\boldsymbol{\mathit{k}}?

Answer

The field should be \boldsymbol{\mathit{B}}=-\alpha x\boldsymbol{\mathit{i}}+(B_0+\alpha z)\boldsymbol{\mathit{k}}, which satisfies \nabla\cdot\boldsymbol{\mathit{B}}. However, the precession of the spin magnetic moment of a silver atom around B_0 in the Stern-Gerlach experiment is so fast that the x-component of the spin moment averages to zero, resulting in an effective field of \boldsymbol{\mathit{B}}=(B_0+\alpha z)\boldsymbol{\mathit{k}} interacting with the atom.

 

Question

Why is \nabla\cdot\boldsymbol{\mathit{B}}=0?

Answer

A simple way of showing \nabla\cdot\boldsymbol{\mathit{B}}=0 is to consider a two-dimensional diagrammatic representation of the vector function \boldsymbol{\mathit{B}}=-y\boldsymbol{\mathit{i}}+x\boldsymbol{\mathit{j}} below, where each point in the two-dimensional space is associated with a vector.

When y=0, the successive points in the negative x direction and the positive x direction are associated with the vectors -\boldsymbol{\mathit{j}},-2\boldsymbol{\mathit{j}},\cdots and \boldsymbol{\mathit{j}},2\boldsymbol{\mathit{j}},\cdots respectively. Similarly, when x=0, the successive points in the negative y direction and the positive y direction are associated with the vectors \boldsymbol{\mathit{i}},2\boldsymbol{\mathit{i}},\cdots and -\boldsymbol{\mathit{i}},-2\boldsymbol{\mathit{i}},\cdots respectively. Clearly, \nabla\cdot\boldsymbol{\mathit{B}}=0.

 

 

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Classical angular momentum

Angular momentum is the rotational analogue of linear momentum p. For a particle rotating in a plane at radius r about O (see diagram below), its velocity component v_\perp that is perpendicular to r is

v_\perp=\frac{2\pi r}{T}=r\omega\; \; \; \; \; \; \; \; 58

where T is the period (time taken to complete a revolution) and \omega=\frac{2\pi}{T} is the angular velocity.

The rotational kinetic energy KE_{rot} of the particle is

KE_{rot}=\frac{1}{2}mv_{\perp}^{2}=\frac{1}{2}mr^{2}\omega^{2}=\frac{1}{2}I\omega^{2}

where I=mr^{2} is the moment of inertia.

 

Question

What is moment of inertia and why is it equal to mr^{2}?

Answer

The moment of inertia is the rotational equivalent of a particle’s inertia in linear motion. For a particle in linear motion, its inertia is quantified by its mass. For a particle in rotational motion, its inertia is dependent on both its mass and the distribution of that mass relative to O (i.e. dependent on m and r for a point mass). Since \frac{1}{2}mv_{\perp}^{\, \, 2}=\frac{1}{2}mr^{2}\omega^{2}, where \omega is the angular velocity of the rotating particle, we define I=mr^{2} such that there is a correspondence between \omega and v_{\perp}, and between I and m.

 

Consequently, angular momentum L, which is the rotational equivalent of linear momentum p=mv, is defined as:

L=I\omega=mr^{2}\frac{v_{\perp}}{r}=rmv_{\perp}\; \; \; \; \; \; \; \; 59

Since a particle’s orbit may be circular or non-circular (see diagram above), its angular momentum is generally expressed as:

L=rmvsin\theta

or equivalently,

\boldsymbol{\mathit{L}}=\boldsymbol{\mathit{r}}\times\boldsymbol{\mathit{p}}

Therefore, angular momentum is a pseudo-vector with a direction indicated by the right-hand rule.

 

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One-particle, time-dependent Schrodinger equation

The one-particle, time-dependent Schrodinger equation is a partial differential equation whose solutions are the one-particle, time-dependent wave functions of quantum-mechanical systems.

Even though the equation is widely regarded as a postulate, we can derive it using a general travelling wave equation \psi(x,t)=Acos\frac{2\pi}{\lambda}(x-ct). Since cosine is an even function, Acos\frac{2\pi}{\lambda}(x-ct)=Acos\frac{2\pi}{\lambda}(ct-x), which in the complex square-integrable form is: \psi(x,t)=Ae^{-i\frac{2\pi}{\lambda}(ct-x)} . Since c=v\lambda, we have \psi(x,t)=Ae^{-i2\pi( vt-\frac{x}{\lambda} )}. Substituting Planck’s relation and de Broglie’s hypothesis in the wave equation, which is a mathematical description of the properties of a quantum-mechanical particle, we have \psi(x,t)=Ae^{-\frac{i}{\hbar}(Et-xp )}, where \hbar=\frac{h}{2\pi}.

The total energy of the particle is: E=T+V=\frac{p^{2}}{2m}+V, and so

E\psi=\frac{p^{2}\psi}{2m}+V\psi\; \; \; \; \; \; \; \; 54

To develop an expression for E\psi, we find the partial derivative of \psi with respect to t:

\frac{\partial}{\partial t}Ae^{-\frac{i}{\hbar}(Et-xp)}=Ae^{\frac{i}{\hbar}xp}\frac{\partial}{\partial t}e^{-\frac{i}{\hbar}Et}=-A\frac{iE}{\hbar}e^{-\frac{i}{\hbar}(Et-xp)}=-\frac{iE}{\hbar}\psi

E\psi=i\hbar\frac{\partial\psi}{\partial t}\; \; \; \; \; \; \; \; 55

As for p^{2}\psi, we find the the 2nd-order partial derivative of \psi with respect to x:

\frac{\partial^{2}}{\partial x^{2}}Ae^{-\frac{i}{\hbar}(Et-xp)}=Ae^{-\frac{i}{\hbar}Et}\frac{\partial^{2}}{\partial x^{2}}e^{\frac{i}{\hbar}xp}=\left ( \frac{i}{\hbar}p\right )^{2}Ae^{-\frac{i}{\hbar}(Et-xp)}=-\frac{p^{2}}{\hbar^{2}}\psi

p^{2}\psi=-\hbar^{2}\frac{\partial^{2}\psi}{\partial x^{2}}\; \; \; \; \; \; \; \; 56

Substituting eq55 and eq56 in eq54, we have

i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi\; \; \; \; \; \; \; \; 57

Eq57 is the one-particle, one-dimensional, time-dependent Schrodinger equation, which has the general solution \psi(x,t)=\psi(x)e^{-\frac{i}{\hbar}Et}.

 

Question

Show that \psi(x,t)=\psi(x)e^{-\frac{i}{\hbar}Et} is a solution to eq57.

Answer

For LHS of eq57

i\hbar\frac{\partial\psi(x,t)}{\partial t}=-i\hbar\psi(x)i\frac{E}{\hbar}e^{-i\frac{Et}{\hbar}}=E\psi(x,t)

So,

\left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\right ]\psi(x,t)=E\psi(x,t)

 

 

 

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1-D classical wave equation

The 1-D classical wave equation is a mathematical description of one-dimensional waves that occur in classical physics.

To derive the equation, we consider a vibrating string of mass density \rho=\frac{dm}{dx} in a 2-dimensional plane, where \rho is the change in mass of the string dm with respect to the change in unit length of the string.

The diagram above shows that the string (blue line) has been displaced from its relaxed length. When this happens, atoms or molecules within the string are pulled apart and gain potential energy. An associated restoring force called tension T therefore arises in the string. As the string can be perceived as being composed of small interconnected segments, with each segment pulling adjacent segments and being pulled upon by adjacent segments, we have, for a particular segment (shown in red in the above diagram), 2 tension vectors of equal magnitude acting at its ends. Since the vectors are tangent to the curve and point in opposite directions, they do not add to zero. Ignoring the effects of gravity, the resultant vector is the restoring force acting on the segment. This restoring force (represented by the red vector in the diagram below) is approximately vertical if the amplitude of the wave is small:

The vertical component of T at x+dx is Tsin\theta, and the gradient of T at x+dx is \frac{dy\vert_{x+dx}}{dx}=tan\theta.

For small angles, sin\theta=tan\theta=\theta and so, Tsin\theta=T\theta and \frac{dy\vert_{x+dx}}{dx}=\theta. Combining the 2 equations, we have Tsin\theta=T\frac{dy\vert_{x+dx}}{dx}. Similarly, for T at x, we have Tsin\alpha=T\frac{dy\vert_{x}}{dx}. Therefore, the net vertical force F=(dm)a is:

(dm)a=T\left [ \frac{dy\vert_{x+dx}}{dx}-\frac{dy\vert_{x}}{dx}\right ]=Td\left ( \frac{dy}{dx} \right )

Substituting \frac{d^{2}}{dx^{2}}=\frac{d\left ( \frac{dy}{dx} \right )}{dx} in the above equation gives (dm)a=T\frac{d^{2}y}{dx^{2}}dx, which is then substituted with \rho=\frac{dm}{dx} and a=\frac{d^{2}y}{dt^{2}} to yield:

\rho\frac{d^{2}y}{dt^{2}}=T\frac{d^{2}y}{dx^{2}}

Since y is a function of position x and time t, we replace the derivatives with partial derivatives:

\rho\frac{\partial^{2}y}{\partial t^{2}}=T\frac{\partial^{2}y}{\partial x^{2}}\; \; \; \; \; \; \; \; 52

For the wave equation to be applicable to both standing and travelling waves, the wavefunction should take the form of y(x,t)=Asin\frac{2\pi}{\lambda}(x-ct) where \lambda is wavelength and c is the velocity of the travelling wave (if c=0, y(x,t) reduces to a standing wave). Substituting y(x,t) into eq52, we have \rho\frac{\partial^{2}y}{\partial t^{2}}=-\left ( \frac{2\pi c}{\lambda} \right )^{2}\rho Asin\frac{2\pi}{\lambda}(x-ct) and T\frac{\partial^{2}y}{\partial x^{2}}=-T\left ( \frac{2\pi }{\lambda} \right )^{2} Asin\frac{2\pi}{\lambda}(x-ct). Therefore, \rho=\frac{T}{c^{2}}, which when substituted in eq52, gives

\frac{\partial^{2}y}{\partial x^{2}}= \frac{1 }{c^{2}}\frac{\partial^{2}y}{\partial t^{2}}\; \; \; \; \; \; \; \; 53

Eq53 is the 1-dimensional classical wave equation.

 

 

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One-particle, time-independent Schrodinger equation in spherical coordinates

The Laplacian \nabla^{2} in spherical coordinates can be derived from eq87, eq88 and eq89. Substituting eq78 through eq86 in eq87, eq88 and eq89, adding the resulting three equations and simplifying, we have

\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial \theta^{2}}+\frac{cos\theta}{r^{2}sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{r^{2}sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\; \; \; \; \; \; \; \; 46

or

\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r} \right )+\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta}\right )\right ]\; \; \; \; \; \; \; \; 47

It can be easily shown that eq47 is equivalent to eq46 by computing the derivatives in eq47. Hence, eq45 in spherical coordinates is

\small \hat{H}=-\frac{\hbar^{2}}{2m}\left \{ \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\right \}+V(r)\; \; \; \; \; \; \; \; 48
For the case of a particle confined to a spherical surface of zero relative potential, \small r is constant and \small V(r)=0. The first term of the Laplacian operating on a wavefunction \small \psi(r,\theta,\phi) will return a result of zero. We can therefore discard it and eq48 becomes:

\hat{H}_{angular}=-\frac{\hbar^{2}}{2mr^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta} \right )\right ]\; \; \; \; \; \; \; \; 49

Substituting eq96 in the above equation gives

\hat{H}_{angular}=\frac{\hat{L}^{2}}{2I}\; \; \; \; \; \; \; \; 50

where I=mr^{2}.

Hence, an eigenvalue of \hat{H}_{angular} is the energy associated with the angular motion of a particle. This implies that the linear (or radial) part of the Hamiltonian is:

\hat{H}_{radial}=-\frac{\hbar^{2}}{2m} \frac{1}{r^{2}}\frac{\partial}{\partial r}\left ( r^{2}\frac{\partial}{\partial r}\right ) +V(r)=-\frac{\hbar^{2}}{2m} \left ( \frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^{2}}{\partial r^{2}}\right ) +V(r)\; \; \; \; \; \; \; \; 51

Similarly, an eigenvalue of \hat{H}_{radial} is the energy associated with the linear motion of a particle. We can, therefore, rewrite eq48 as:

\hat{H}=\hat{H}_{radial}+\hat{H}_{angular}

 

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One-particle, time-independent Schrodinger equation

The one-particle, time-independent Schrodinger equation is a partial differential equation whose solutions are the one-particle, time-independent wave functions of quantum-mechanical systems.

Even though it is widely regarded as a postulate, we can list the scientific findings that could have inspired Schrodinger to develop the equation.

 

Year Development Formula Scientist
1500s Equations of motion Linear rotational motion equations Many scientists including Galileo and Newton
1746 1-D classical wave equation, whose solutions are wave functions that describe the motion of waves. \frac{\partial^{2}u(x,t)}{\partial x^{2}}=\frac{1}{c^{2}}\frac{\partial^{2}u(x,t)}{\partial t^{2}} Jean d’Alembert
1924 de Broglie’s hypothesis: all matter exhibit wave-like properties p=\frac{h}{\lambda} Louis de Broglie

 

Many scientists at that time were trying to develop models of the atom, in particular, the behaviour of electrons in an atom. It is possible that when de Broglie proposed that all matter have wave—like properties, Schrodinger thought that an equation similar to the classical wave equation could be derived to describe the wave motion of an electron in an atom. Since the classical wave equation is solved by the separation of variables method, we can express u(x,t) as u(x,t)=h(x)g(t)=\psi(x)cos\omega t. Substituting u(x,t) in the classical wave equation, we have:

cos\omega t\frac{\partial^{2}}{\partial x^{2}}\psi(x)=-\frac{1}{c^{2}}\psi(x)\omega^{2}cos\omega t

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{\omega^{2}}{c^{2}}\psi(x)=0

Substitute \omega=2\pi v and c=v\lambda in the above equation

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{4\pi^{2}}{\lambda^{2}}\psi(x)=0\; \; \; \; \; \; \; \; 43

The total energy of an electron is E=KE+PE=\frac{p^{2}}{2m}+V(x) or p=\sqrt{2m[E-V(x)]}, where p  is the momentum of the electron. Substituting p in de Broglie’s relation, we have \sqrt{2m[E-V(x)]}=\frac{h}{\lambda}, which we then substitute in eq43 to give:

\frac{\partial^{2}}{\partial x^{2}}\psi(x)+\frac{2m}{\hbar^{2}}[E-V(x)]\psi(x)=0

where \hbar=\frac{h}{2\pi}, or

\hat{H}\psi=E\psi\; \; \; \; \; \; \; \; 44

where \hat{H}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V.

\hat{H} is called the Hamiltonian operator and Eq44 is the one-dimensional, one-particle time-independent Schrodinger equation. As it is an eigenvalue equation, \psi (known as a wave function) is an eigenfunction and E is the corresponding eigenvalue. For a three-dimensional, one-particle system, the Hamiltonian is:

\hat{H}=-\frac{\hbar^{2}}{2m}\nabla^{2} +V(x,y,z)\; \; \; \; \; \; \; \; 45

where \nabla^{2}=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} and is called the Laplacian.

 

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Postulates of quantum mechanics

The postulates of quantum mechanics are fundamental mathematical statements that cannot be proven. Nevertheless, they are statements that everyone agrees with.

Examples of other postulates of science and mathematics are Newton’s 2nd law, F=ma, and the Euclidean statement that a line is defined by two points respectively.

Generally, the postulates of quantum mechanics are expressed as the following 7 statements:

1) The state of a physical system at a particular time t is represented by a vector in a Hilbert space.

We call such a vector, a wavefunction \psi(x,t), which is a function of time and space . A wavefunction contains all assessable physical information about a system in a particular state. For example, the energy of the stationary state of a system is obtained by solving the time-independent Schrodinger equation \hat{H}\psi=E\psi.

2) Every measurable physical property of a system is described by a Hermitian operator \hat{O} that acts on a wavefunction representing the state of the system.

In other words, to every observable quantity in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics. The most well-known Hermitian operator in quantum mechanics is the Hamiltonian \hat{H}, which is the energy operator. The wavefunctions that \hat{H} acts on are called eigenfunctions. Eigenfunctions of a Hermitian operator in quantum mechanics are further postulated to form a complete set. Other Hermitian operators frequently encountered in quantum mechanics are the momentum operator \hat{p} and the position operator \hat{x}.

3) The result of the measurement of a physical property of a system must be one of the eigenvalues of an operator \hat{O}.

The state of a system is expressed as a wavefunction, which can be a single basis wavefunction or a linear combination of a complete set of basis wavefunctions. Since basis wavefunctions of a Hermitian operator form a complete set, all wavefunctions can be written as a linear combination of basis wavefunctions.

 

Question

What about a system described by a stationary state?

Answer

The wavefunction can be expressed, though trivially, as \psi=\sum_{n=0}^{\infty}c_n\phi_n, where c_i=1 and c_{n\not\equiv i}=0 (we have assumed that the spectrum is discrete, i.e. the eigenvalues are separated from one another).

 

It is generally accepted by scientists that the values of the coefficients c_n are unknown prior to a measurement. Upon measurement, the result obtained is an eigenvalue associated with one of the eigenfunctions, and hence the phrase ‘the initial wavefunction collapses to one of the eigenfunctions’. This implies that the measurement alters the initial wavefunction such that a 2nd measurement, if made quickly, yields that same result (this obviously refers to wavefunctions describing non-stationary states, as wavefunctions of stationary states are independent of time). If we prepare a large number of identical systems and simultaneously measure them, the values of c_n are found; with \sum_{n=0}^{\infty}\vert c_n\vert^{2}=1, and the expectation value of the measurements being \langle\psi\vert\hat{O}\vert\psi\rangle.

4) The probability of obtaining an eigenvalue E_i upon measuring a system is given by the square of the inner product of the normalised \psi with the orthonormal eigenfunction \phi_i.

In other words,

\vert\langle\phi_i\vert\psi\rangle \vert^{2}=\vert\langle\phi_i\vert\sum_{n=0}^{\infty}c_n\phi_n\rangle \vert^{2}=\vert\langle c_i\phi_i\vert\phi_i\rangle \vert^{2}=\vert c_i\vert^{2}

If the spectrum is continuous, \psi=\int_{-\infty}^{\infty}c_n\phi_n\, dn and the probability of obtaining an eigenvalue in the range dn is \vert\langle\phi_n\vert\psi\rangle \vert^{2}\, dn.

5) The state of a system immediately after a measurement yielding the eigenvalue E_i is described by the normalised eigenfunction \phi_i.

We have explained in the postulate 3 that this is commonly known as the collapse of the wavefunction \psi to one of the eigenfunctions \phi_i. It is also known as the projection of \psi onto \phi_i, i.e., \hat{P}_i\vert\psi\rangle; or if \psi is not normalised:

\frac{\hat{P}_i\vert\psi\rangle}{\sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle}}

 

Question

Show that \sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle} is the normalisation constant.

Answer

To normalise a wavefunction,

NN\int c_{i}^{*}\phi_{i}^{*}c_i\phi_i\, d\tau=1\; \; \; \Rightarrow \; \; \; N^{2}=\frac{1}{\vert c_i\vert^{2}\langle\phi_i\vert\phi_i\rangle}\; \; \; \Rightarrow \; \; \; N=\frac{1}{ c_i\sqrt{\langle\phi_i\vert\phi_i\rangle}}

So,

\frac{1}{ \sqrt{\langle\psi\vert\hat{P}_i\vert\psi\rangle}}=\frac{1}{ \sqrt{\langle\psi\vert c_i\phi_i\rangle}}=\frac{1}{ \sqrt{\langle c_i\phi_i\vert\psi\rangle^{*}}}=\frac{1}{ \sqrt{(c_{i}^{*}c_i\langle \phi_i\vert\phi_i\rangle)^{*}}}

=\frac{1}{ \sqrt{c_{i}^{*}c_i\langle \phi_i\vert\phi_i\rangle}}=\frac{1}{ c_i\sqrt{\langle \phi_i\vert\phi_i\rangle}}

 

6) The time evolution of the wavefunction \psi(t) is governed the time-dependent Schrodinger equation i\hbar\frac{d}{dt}\vert\psi(t)\rangle=\hat{H}(t)\vert\psi(t)\rangle.

7) The total wavefunction of a system of identical particles must be either symmetric (for bosons) or antisymmetric (for fermions) under exchange of any two particles. This postulate is often referred to as the symmetrisation postulate.

 

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