Fundamentals
Postulates of QM
[States and wavefunctions] The state of a system is fully described by a function \(\Psi(r_1,r_2,...,t)\). Describe meaning it contains all the informations about properties that can be measured.
[Prescription] Observables are represented by Hermitian operators chosen to satisfy the commutation relations
\[[q,p_{q'}]=i\hbar \delta_{qq'} \\ [q,q']=0 \\ [p_q,p_{q'}]=0\]
where q are coordinates and p are linear momenta.
[Outcomes of Measurements] A system described by \(\psi\), mean value of an observable \(\Omega\) in a series of measurements is equal to the expectation value of the corresponding operator \(\hat{\Omega}\).
\(<\Omega>=\frac{<\psi|\Omega|\psi>}{<\psi|\psi>}\)
ALT If \(\psi\) is an eigenfunction of operator \(\Omega\), the determination of \(\Omega\) always yields its eigenvalue \(\omega\). Else, a single measurement will yield one of the eigenvalues of \(\Omega\), and the probability of getting \(\omega_n\) is equal to \(|c_n|^2\), where \(c_n\) is the coefficient of \(\psi_n\) in the eigenfunction expansion. [Collapse of the wavefuntion] Immediately after the measurement, the state of the system will be \(\psi_n\).
[Born interpretation] The probability that a particle will be found in the volume element \(d\tau\) at the point \(\vec{r}\) is \(|\psi(\vec{r})|^2d\tau\).
[TDSE] The wavefuntion \(\Psi\) evolves with time according to TDSE:
\(i\hbar \frac{\partial \Psi}{\partial t}=H\Psi\)
Operators
Common Operators
| Operator | Operation | Note |
|---|---|---|
| \(\hat{x}\) | \(\times x\) | |
| \(\hat{V}\) | \(\times V\) | |
| \(\hat{p}\) | \(\frac{\hbar}{i}\frac{\partial}{\partial x}\) | Component-wise, applies to all coordiate systems |
Derivation of Operators
- In QM, construct operators with position and momentum - write down the expression of the quantity and then substitute in the position and momentum operators.
- The expressions of the quantities are identical to classical ones.
For example, the kinetic energy operator: \[ \begin{align*} T &= \frac{p^2}{2m}\\ \implies \hat{T}&=\frac{1}{2m}\hat{p}\hat{p}\\ &=\frac{1}{2m} \frac{\hbar}{i}\frac{\partial}{\partial x} \frac{\hbar}{i}\frac{\partial}{\partial x}\\ &=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \end{align*} \]
Eigenvalues and Eigenfunctions
\[ \hat{\Omega}f = \omega f \]
- Eigenfunction expansion: \(g=\sum_n c_n f_n\)
- Degeneracy: linear combinations of degenerate e-functions is an e-function with the same e-value. Note this is not true for a combination of non-degenerate e-functions.
Commutators
\[ [A,B]=AB-BA \]
- Note: \([\hat{x},\hat{p_x}]=i\hbar \mathbf{I}\)
- Property: If two observables are to have simultaneously precisely defined values, their corresponding operators must commute.
- Pairs that can not commute are said to be complementary.
Notations
Dirac Bra-Ket: \(<m|\hat{\Omega}|n>=\int f^*_m \hat{\Omega} f_n d\tau\)
Matrix: \(\mathbf{\Omega}_{mn}=<m|\hat{\Omega}|n>\)
Completeness relation: \(\sum_s |s><s|=1\)
Overlap intergral: \(S=<m|n>\)
Orthonormality: \(<m|n>=\delta_{mn}\)
Hermitian Operators
\[ <m|\Omega|n>=<n|\Omega|m>^* \]
- E-values of Hermitian operators are real.
- E-functions for different E-values of a Hermitian operator are orthogonal.
Uncertainty
This is the same as the standard deviation. \[ \Delta x=\sqrt{<\!\!x^2\!\!>-<\!\!x\!\!>^2} \]
- The expectation value for an eigenfunction has zero uncertainty - the measurement is exact.
Heisenberg uncertainty principle
\[ \Delta x \Delta p \geq \frac{1}{2}\hbar \]
The uncertainties in two measurements \(P\) and \(Q\) satisfy: \[ \Delta P \Delta Q\geq\frac{1}{2}|<[\hat{P},\hat{Q}]>| \]
Wavefunctions
Schroedinger Equations
Hamiltonian operator is the total energy operator: \[ \hat{H}=\hat{T}+\hat{V} \] TISE is: \[ \hat{H} \psi = E \psi \] TDSE is: \[ \hat{H}\Psi=i\hbar\frac{\partial \Psi}{\partial t} \] A special class of solutions, \(\Psi(x,t) = \psi(x) e^{-iEt/\hbar}\), are called stationary states. For these, the time-dependence cancels out.
Characteristics of Wavefunctions
- \(\psi^*\psi\) is single-valued
- Must be square integratable, i.e. \(\psi^*\psi\) can not be infinite for a finite range
- Continuous everywhere
- Have a continuous derivative, except where potential is ill-behaved.
1-D Free Particles
Zero potential everywhere - expect to have definite momentum. \[ -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} =E\psi \] The solution is \(\psi=A\,\mathrm{exp}(ipx/\hbar)+B\,\mathrm{exp}(-ipx/\hbar)\), with \(p=\sqrt{2mE}\). Each term can be verified with momentum eigenvalue equation, \[ p\psi(x)=\hat{p}\psi(x)=\frac{\hbar}{i}\frac{\partial \psi}{\partial x} \]
that \(p\) and \(-p\) are e-values. Hence the general solution is a superposition of the positive and negative directing terms.
Note:
- the resultant linear combination is not an eigenfunction of \(\hat{p}\).
- can be written in terms of sin and cos.
Further: Flux density is defined to be
\(J_x=\frac{1}{2m}(\psi^*p_x\psi+\psi
p_x^*\psi^*)\)
Further: A wavepacket is a superposition
of wavefunctions with different energies - the energy of the particle is
not precisely specified.
An infinitely thick potential wall
The potential energy has the form: \[ \begin{align*} x<0: \,\,&V(x)=0\\ x\geq 0: \,\, & V(x)=V \end{align*} \] Hence the hamiltonians are: \[ \begin{align*} x<0: \,\,&H=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\\ x\geq 0: \,\, & H=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V \end{align*} \] The solution to the corresponding S-eqns are: \[ \begin{align*} x<0: \,\,& \psi = A \mathrm{e}^{ikx}+B \mathrm{e}^{-ikx}, \,\,\,\,\,\,\,k\hbar=\sqrt{2mE}\\ x\geq 0: \,\, & \psi = A' \mathrm{e}^{ik'x}+B' \mathrm{e}^{-ik'x}, \,k'\hbar=\sqrt{2m(E-V)} \end{align*} \] For cases of \(E<V\), \(k'\) is imaginary, so write as \(k'=i\kappa\). \[ x\geq 0: \,\, \psi = A' \mathrm{e}^{-\kappa x}+B' \mathrm{e}^{\kappa x}, \,\kappa\hbar=\sqrt{2m(V-E)} \] Since the wall is infinitely thick, \(B'\) must be 0 to avoid infinite amplitude. Therefore, \(\psi\) is an exponential decay inside the wall.
Penetration: the effect that a particle may be found inside a classically forbidden region. The penetration depth is the distance \(1/\kappa\).
A barrier of finite width
Particle in a box
\[ H=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V(x)\\ V(x)=\left\{ \begin{array}{ll} 0, &0\leq x\leq L\\ \infty, & \mathrm{otherwise} \end{array} \right. \]
General solution for within the box is the same as a free particle.
Due to infinite potential, amplitude is 0 outside the box. Wavefunctions are continuous, so the B.C.s are \(\psi(0)=0\) and \(\psi(L)=0\). Therefore, \[ \begin{array}{lll} \psi(x)=A\, \mathrm{sin}\,kx & k=\frac{n\pi}{L} & n \in Z_+ \end{array} \] or after normalisation, \[ \psi(x)=\sqrt{\frac{2}{L}}\, \mathrm{sin}\,(\frac{n\pi x}{L}) \] Energy is also known by \(E_n=\frac{k^2 \hbar^2}{2m}=\frac{n^2h^2}{8mL^2}\).
- The nth wavefunction has \(n-1\) nodes.
- At high n, the distribution corresponds to the classical behaviour that spends equal time on average at all points. Correspondence principle: classical mechanics emerges from quantum mechanics at high quantum numbers.
- Lowest energy state has \(n=1\), with non-zero energy called zero point energy.
- The energy difference between states, \(\Delta E =(2n+1)\frac{h^2}{8mL^2}\) decreases as L increases. In the limit \(L\to \infty\), it becomes a free particle, no quantization.
Harmonic Oscillator
The potential is \(V=\frac{1}{2}kx^2\), and the hamiltonian is \[ H=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+\frac{1}{2}kx^2 \]
Ground state and ZPE
Using trail function \(\psi_0=e^{-\frac{1}{2}\alpha x^2}\), \(E_0=\frac{1}{2}\hbar\omega\), where \(\omega=\sqrt{k/m}\).
Evaluated: \(\Delta x =[\hbar^2/4km]^{1/4}\), \(\Delta p_x=(\hbar^2km/4)^{1/4}\), so \(\Delta x\Delta p=\hbar/2\). ZPE is the minimum energy.
Scaled Coordinate
Use \(s=1/\sqrt{\alpha}=[\hbar^2/km]^{1/4}\), and let \(q=x/s\) to be the new scaled coordinate. Thus Hamiltonian simplifies to \(H=-\frac{1}{2}\frac{\mathrm{d}^2}{\mathrm{d}q^2}+\frac{1}{2}q^2\).
Higher order solutions
Form: \(\psi=H(q)e^{-q^2/2}\)
where \(H(q)\) are Hermite polynomials, the first few are:
\[ \begin{array}{ll} 0 & 1\\ 1 & 2q\\ 2 & 4q^2-2\\ 3 & 8q^3-12q\\ v & (-1)^v e^{q^2}\frac{\mathrm{d}^v}{\mathrm{d} q^v}e^{-q^2} \end{array} \]
Properties of solutions
- Equal spacing of energy
- separation decreases with mass
- increases with force const
- limit of 0 force const - does not confine the particle at all - no quantization
- Virial theorem: if a potential can be expressed as \(V=ax^s\), then \(2<E_k>=s<E_p>\).
- For low energy solutions, particle stays close to \(x=0\); for high energy states, dominant maxima are close to turning points, which resembles classical distribution.
- Wavefunctions continues past \(V=E_v\), hence enters classically forbidden region. Probability of quantum tunnelling decreases with energy.
- Orthogonal to each other, as for any eigenfunctions of \(\hat{H}\).
Morse Oscillator
A more realistic model for real bonds than H.O. Potential energy reaches an asymptotic value as bond-length increases. \[ V_M(r)=D_e[1-\mathrm{exp}(-\beta(r-r_e))]^2 \] The force constant can be obtained by expanding exp. \(k=2\beta^2 D_e\)
The Morse wavefunctions can be solved exactly. Energies take the form \[ E_v=(v+\frac{1}{2})\hbar\omega-(v+\frac{1}{2})^2\hbar\omega x_e \] with \(\omega=\sqrt{k/m}\) and \(x_e=\hbar\beta^2/2m\omega=\hbar\omega/4D_e\) called anharmonicity constant.
Thus the energy separation decreases by \(2\hbar\omega x_e\) as \(v\) increases, and reaches a maximum value. The states beyond that value is considered physically meaningless.
Angular Momentum
Operators
\[ \vec{J}=\vec{r}\times\vec{p} \]
Component-wise, \[ J_x=yp_z-zp_y , etc. \] In terms of QM operators, \[ \hat{J_x}=\frac{\hbar}{i}(y\hat{\partial_z}-z\hat{\partial_y}), etc. \] The remaining is the magnitude squared operator \[ \hat{J^2}=\hat{J_x^2}+\hat{J_y^2}+\hat{J_z^2} \]
Commutators
The components of angular momenta do not commute. For example, \[ [\hat{J_x},\hat{J_y}]=-\hbar^2(y\hat{\partial_x}-x\hat{\partial_y})=i\hbar\hat{J_z} \]
Proof required. Remember components are independent of each other, and cross derivatives are equal.
However, any one of the component commute with \(\hat{J^2}\). \[ [\hat{J_z},\hat{J^2_x}]=i\hbar(\hat{J_y}\hat{J_x}+\hat{J_x}\hat{J_y})\\ [\hat{J_z},\hat{J^2_y}]=-i\hbar(\hat{J_y}\hat{J_x}+\hat{J_x}\hat{J_y})\\ \therefore [\hat{J_z},\hat{J^2}]=0 \] Therefore, we can not know simultaneously exactly more than one component of angular momenta, but can know one component and the magnitude.
Spherical polar coordinates
2D Laplacian: \(\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}\)
3D Laplacian: \(\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}=\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{1}{r^2}\Lambda^2\), where the legendrian, \(\Lambda^2=\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}+\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\sin \theta\frac{\partial}{\partial \theta}\)
These gives:
z-Angular momentum: \(\hat{J_z}=\frac{\hbar}{i}\frac{\partial}{\partial \phi}\)
magnitude squared operator: \(\hat{J^2}=-\hbar^2 \Lambda^2\)
Eigenfunctions
J_z
First find the Eigenfunction of \(\hat{J_z}\), by trying \(\exp(ik\phi/\hbar)\). The Eigenvalue equation is \[ \hat{J_z}\psi=-i\hbar\frac{\partial}{\partial \phi}\psi=k\psi. \] The boundary condition is \(\psi(\phi+2\pi)=\psi(\phi)\), as the wavefunction is single-valued.
Thus, \(k=M\hbar\), where \(M\) is an integer. Normalised wavefunction is \(\psi_M=\sqrt{\frac{1}{2\pi}}\exp(iM\phi)\).
The angular momentum is an integer multiple of \(\hbar\).
J^2
Eigenfunctions of \(\hat{J^2}\) are separable functions of \(\phi\) and \(\theta\), \(Y_{JM}=\Theta_{JM}(\theta)\Phi_M(\phi)\), called spherical harmonics. The e-value turns out to be \(\hbar^2J(J+1)\).
Properties:
- The seperation increases with increasing \(J\).
- The energy levels are \(2J+1\) degenerate.
- No ZPE.
Shift Operators
Define: \[ \hat{J_+}=J_x+iJ_y \\ \hat{J_-}=J_x-iJ_y \] It follows that: \[ \begin{array}{lll} [J_z,J_+]=\hbar J_+ & [J_z,J_-]=-\hbar J_- & [J_+,J_-]=2\hbar J_z \\ [J^2,J_+]=[J^2,J_-]=0 \end{array} \]
Rigid rotor
The rotational behaviour of a diatomic molecule is equivalent to a single particle of reduced mass, \(\mu=\frac{m_1 m_2}{m_1+m_2}\), moving on a surface of sphere with radius \(R\) equal to the bond length.
Rotational KE: \(\frac{1}{2}I\omega^2=\hat{J^2}/2I\), (because \(J=I\omega\)).
In zero field, \[ \hat{H}\psi=\frac{\hat{J^2}}{2I}\psi=\frac{\hbar^2J(J+1)}{2I}\psi=BJ(J+1)\psi \] where \(B\) is the rotational constant.
Rotation-Vibration
The total Hamiltonian is the sum of vibration and rotation Hamiltonians. \[\begin{align} \hat{H}\Psi &=(\hat{H}_{vib}+\hat{H}_{rot})\psi^{vib}_v\psi^{rot}_{JM}\\ & =\psi^{rot}_{JM}\hat{H}_{vib}\psi^{vib}_v +\psi^{vib}_v\hat{H}_{rot}\psi^{rot}_{JM}\\ &=[E_v+BJ(J+1)]\Psi \end{align}\] Note that \(\Delta E_{vibes} \sim 1000\Delta E_{rot}\).
Getting Real
Born-Oppenheimer approximation
Idea: nuclei are very heavy, so they move slowly.
When studying electrons, the nuclei can be treated as stationary; when studying nuclei, the electronic potential energy is used in the nuclear Hamiltonian. \[\begin{align} &\mathrm{electronic\,SE} & \hat{H}_{elec}\psi_{elec}(\mathbf{q};\mathbf{Q})&=E_{elec}(\mathbf{Q})\psi_{elec}(\mathbf{q};\mathbf{Q})\\ &\mathrm{electronic\,Hamiltonian} & \hat{H}_{elec}&=\hat{T}_E+\hat{V}(\mathbf{q};\mathbf{Q})\\ &\mathrm{nuclear\,SE} & \hat{H}_{nuc}\psi_{nuc}(\mathbf{Q})&=E_{nuc}(\mathbf{Q})\psi_{nuc}(\mathbf{Q})\\ &\mathrm{nuclear\,Hamiltonian} & \hat{H}_{elec}&=\hat{T}_N+E_{elec}(\mathbf{Q}) \end{align}\]
Hydrogen atom
For the electron in the hydrogen atom, it is in a spherical Coloumbic field. \[ \hat{V}=-\frac{Ze^2}{4\pi\epsilon_0r}\\ \hat{T}=-\frac{\hbar^2}{2m_e}\nabla^2=-\frac{\hbar^2}{2m_e}(\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{1}{r^2}\Lambda^2) \]
Atomic units
The purpose is to get rid of the cluster of constants.
| Unit | Symbol | Name | Definition |
|---|---|---|---|
| Length | \(a_0\) | bohr | \(4\pi\epsilon_0\hbar^2/m_ee^2\) |
| Mass | \(m_e\) | Elec mass | |
| Charge | \(e\) | proton charge | |
| Energy | \(E_h\) | Hartree | \(e^2/4\pi\epsilon_0a_0\) |
| Angular Momentum | \(\hbar\) |
With these, the hamiltonian of the hydrogen atom can be simplified as follows. \[ \hat{H}=-\frac{1}{2r_2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{\hat{l}^2}{2r^2}-\frac{Z}{r} \]
Wavefunctions
The wavefunctions are separable functions of \(r\) and spherical harmonics, \(\psi(r,\theta,\phi)=R(r)Y_{lm}(\theta,\phi)\).
The eigenvalue equation becomes \[\begin{align} \hat{H}RY_{lm}&=[-\frac{1}{2r_2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{\hat{l}^2}{2r^2}-\frac{Z}{r}]RY_{lm}\\ &=[-\frac{1}{2r_2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}+\frac{l(l+1)}{2r^2}-\frac{Z}{r}]RY_{lm}\\ &=ERY_{lm} \end{align}\] It is apparent that the radial part depends parametrically on \(l\). In addition, there are infinitely many solutions for each \(l\), labelled by principal quantum number \(n\). The complete label therefore is \(\psi_{nlm}=R_{nl}Y_{lm}\).
By writing \(\rho=Zr\), the radial function is the same for any H-like atoms, with the energy scaled with \(Z^2\).
The radial wavefunctions \(R_{nm}\) are associated Laguerre functions.
- Energy is \(E_{n}=-\frac{Z^2}{2n^2}\)
- n controls the values of \(l\in(0,n-1)\)
- Wavefunctions have total \(n-1\) nodes: \(l\) angular nodes and \(n-l-1\) radial nodes.
- Orbitals specified by \(m\) (complex) is useful for spectroscopy, while taking linear combinations to form \(x/y\) (real) orbitals are useful for bonding explanation.
Radial probability density
When multiplied by \(dr\) it gives the probability of finding the particle between \(r\) and \(r+dr\). \[ P(r)dr=\int\int R^2|Y|^2r^2\sin \theta \,dr\,d\theta\,d\phi \\P(r)dr=R^2r^2dr \] because spherical harmonics are normalized by \(\int\int |Y|^2\sin\theta\,d\theta \,d\phi=1\).
Many-electron atoms
Helium
The Hamiltonian for He with clamped nucleus is: \[ \hat{H}=-\frac{\hbar^2}{2m_e}\nabla_1^2-\frac{\hbar^2}{2m_e}\nabla_2^2-\frac{Ze^2}{4\pi\epsilon_0r_1}-\frac{Ze^2}{4\pi\epsilon_0r_2}+\frac{e^2}{4\pi\epsilon_0r_{12}} \] Separable solutions no longer work because of the \(r_{12}\) term.
Central field approximation
Put the spherical average of the repulsion from the other electron into the potential energy term of one electron Hamiltonian. In order for this simplification to work, the more general potential \(V\) still only depends on \(r\).
Self-Consistent Field method
- Start with approximate orbitals
- Calculate average field of electron 1
- Put this into the potential of electron 2
- Calculate new function of electron 2
- Calculate the averate field of electron 2
- Put this into the potential of electron 1
- …
- Until a preset convergence standard is met.
Orbital energies
As \(\hat{V}\neq -\frac{1}{r}\), the orbital energies now depend on \(l\) in addition to \(n\).
Near the nucleus, \(V\to -\frac{Z}{r}\); further from the nucleus, \(V\to -\frac{1}{r}\). Hence, orbitals with more radial density close to the nucleus will have a potential weighed more by the \(-Z/r\) term, hence have lower energy. E.g., 2s penetrates more effectively than 2p; 2p is more effectively screened than 2s.
Slater’s rule of effective nuclear charge: \(Z_{eff}=Z-s_{nl}=Z-(0.35(N_n-1)+0.85N_{n-1}+N_{core})\).
Spin
An electron has an intrinsic angular momentum called spin, \(\hat{s}\), associated with quantum numbers \(s\) and \(m_s\).
Integer s: Bosons;
Half-integer s: Fermions
And \(m_s\) takes integers steps from \(-s\) to \(s\).
Electron spin
Electrons have \(s=1/2\), hence \(m_s = 1/2\) or \(-1/2\), corresponding to spin ‘up’ and ‘down’. Or noted \(\alpha\) and \(\beta\).
Spin angular momenta obeys the same equations as orbital angular momentum, \[ \begin{array}{ll} \hat{s_z}\alpha=\frac{1}{2}\hbar \alpha & \hat{s}^2\alpha=\frac{3}{4}\hbar^2\alpha\\ \hat{s_z}\beta=-\frac{1}{2}\hbar \beta & \hat{s}^2\beta=\frac{3}{4}\hbar^2\beta \end{array} \] Note, \(\alpha\) and \(\beta\) are orthonormal; the spins of different particles are independent.
Overall wavefunctions
Now a state can be described as \(|nlm_lm_s>\), obeying \[\begin{align} \hat{H}|nlm_lm_s>&=E_{nl}|nlm_lm_s>\\ \hat{l}^2|nlm_lm_s>&=l(l+1)\hbar^2|nlm_lm_s>\\ \hat{l_z}|nlm_lm_s>&=\hbar m_l|nlm_lm_s>\\ \hat{s}^2|nlm_lm_s>&=s(s+1)\hbar^2|nlm_lm_s>\,,\,s=1/2\\ \hat{s_z}|nlm_lm_s>&=\hbar m_s|nlm_lm_s>\\ \end{align}\]
Spins of two electrons
Four possible states \[ \begin{array}{ll} \alpha_1\alpha_2 & \alpha_1\beta_2 & \beta_1\beta_2 &\beta_1\alpha_2 \end{array} \] Use operator \(\hat{S}_z = \hat{s_{1z}}+ \hat{s_{2z}}\) to calculate the e-value \(M_s\) \[ \begin{array}{ll} \alpha_1\alpha_2 & M_s=1\\ \alpha_1\beta_2 ,\beta_1\alpha_2 & 0\\ \beta_1\beta_2 & -1 \end{array} \] But it is more useful to have \[ \begin{array}{lll} M_S&S=1&S=0\\ 1&\alpha_1\alpha_2 \\ 0&\sqrt{1/2}(\alpha_1\beta_2 +\beta_1\alpha_2) & \sqrt{1/2}(\alpha_1\beta_2 -\beta_1\alpha_2)\\ -1&\beta_1\beta_2 \end{array} \] Note: The \(S\) values are inferred from \(M_s\) values, as \(M_s\) takes integer steps from \(-S\) to \(S\).
Pauli Principle
In a 2-electron system, since electrons are indistinguishable, the probability density must be unaffected by exchanging the electrons. \[ |\Psi(1,2)|^2=|\Psi(2,1)|^2 \] Taking the square root, \[ \Psi(1,2)=\pm\Psi(2,1) \] But empirically, it is always \[ \Psi(1,2)=-\Psi(2,1) \] i.e. Electronic wavefunctions are always antisymmetric with respect to the exchange of the electrons (orbital and spin).
Antisymmetric functions
To satisfy Pauli Principle, only two ways are possible
- Use symmetric orbial functions and antisymmetric spin functions;
- Use antisymmetric orbital functions with symmetric spin functions.
If 2 is used, when the two orbital functions are the same, it vanishes. Hence in the same orbital, spin functions must be antisymmetric, i.e. opposite spins.
Fermi Holes
If 2 is used, when the two electrons are at the same place, the orbital function again vanishes. Hence, QMly the probability of putting two triplet electrons close to each other is small (0 if on top of each other), hence the time experiencing such high potential energy is short, hence triplet electrons have lower energy at and around Fermi holes.
Coulombic and Exchange integrals
Calculate the electron repulsion for a singlet state: \[ <\frac{1}{r_{12}}>_+=<\Psi_+|\frac{1}{r_{12}}|\Psi_+>\\=\int\!\!\!\int \sqrt{1/2}(\psi_{a}(1)\psi_b(2)+\psi_{a}(2)\psi_b(1))*\frac{1}{r_{12}}*\sqrt{1/2}(\psi_{a}(1)\psi_b(2)+\psi_{a}(2)\psi_b(1))\,d^3r_1\,d^3r_2\\ =\int\!\!\!\int\psi_{a}(1)\psi_b(2)\frac{1}{r_{12}}\psi_{a}(1)\psi_b(2)\,d^3r_1\,d^3r_2 \\+ \int\!\!\!\int\psi_{a}(1)\psi_b(2)\frac{1}{r_{12}}\psi_{a}(2)\psi_b(1)\,d^3r_1\,d^3r_2\\ =J+K \] J is the Coulombic integral and K is the exchange integral.
For the triplet state, \[ <\frac{1}{r_{12}}>_{-}=J-K \]
Multi-electron atom obeying Pauli
- Make a SCF approximation - electrons interact with each other, but only with spherically averaged fields.
- To accommodate Pauli, take the antisymmetric orbitals (Slater determinants). Now electrons have further interaction, e.g. Fermi holes.
- 1 and 2 gives Hartree-Fock approximation - generally accurate to 1% of the energy.
- The remaining 1% comes from direct Coulombic interactions between the electrons called electron correlation. This part of energy has a major effect on the chemistry.
Electron correlation
In SCF, all angular momentum quantum numbers of each atom is well-defined.
In real atom, electrons repel one another and exchange orbital angular momentum - that of each electron is no longer well-defined. Total orbital angular momentum quantum numbers are \(L\) and \(M_L\).
As wavefunctions obeys Pauli, orbital and spin wavefunctions are connect, hence individual spin quantum numbers are no longer well-defined either. Total spin are \(S\) and \(M_S\).
Total orbital and total spin can interact, to give total angular momentum \(J\) and \(M_J\), while making \(M_L\) and \(M_S\) no longer well-defined.
Total wavefunction can be written \(|LSJM_J>\), with the energy depending only on \(LSJ\), and is \(2J+1\) degenerate.
Hund’s Rule
Terminology
Configuration: which orbital shells contain electrons.
Term: L and S specified.
Level: L, S, J specified.
Multiplicity: \(2S+1\).
Term symbol: \(^{2S+1}L_J\)
Rules
Used to determine the lowest \(E_{LSJ}\) configuration.
- For the ground configuration, maximize \(S\). Maximize the number of Fermi holes.
- Then maximize \(L\).
- For shells more than half-full, maximize J; for shells less than half-full, minimize J.
Explaining ionisation energy trend
N has 3 electrons with parallel spin; N+ has 2. - 2 less Fermi holes
O has 3 parallel spins and 1 paired; O+ has 3 parallel spins. - same number of Fermi holes
Spin-orbit coupling
This coupling interaction is not described in Schroedinger Hamiltonian due to its relativistic origin.
For single electron: \(E_J=\zeta \vec{l}\cdot\vec{s}, \,\zeta>0\)
For multi electron, single term: \(E_J=\lambda \vec{L}\cdot\vec{S}, \,\lambda=\pm\zeta/2S\) \[ \because\hat{J}^2=(\hat{L}+\hat{S})^2=\hat{L}^2+\hat{S}^2+2\hat{L}\cdot\hat{S}\\ \therefore \hat{L}\cdot\hat{S}=(\hat{J}^2-\hat{L}^2-\hat{S}^2)/2\\ \therefore E_J=\frac{\lambda}{2}[J(J+1)-L(L+1)-S(S+1)] \] Also, \(\lambda\) is positive for less than half-full shell, negative for more than half full shell, Hund’s third rule results.
Lande intervel rule: the splitting between adjacent levels is proportional to the higher J value. \[ E_{J+1}-E_J=\lambda(J+1) \]
Approximation Methods
Variation Method
Variation principle
Any arbitrary function satisfying B.C. of the problem, the expectation value of its energy is not less than the lowest eigenvalue of the Hamiltonian.
Proof required. Use Eigenfunction expansion.
For wavefunctions with one parameter
Use a trail function with a parameter and minimise the expectation value of energy w.r.t. the parameter.
For multiple parameters
Linear combination of atomic orbitals (LCAO)
Simplified notation \[ \alpha=\int s_aHs_a\,d\tau\\ \beta = \int s_aHs_b\,d\tau\\ \beta = \int s_as_b\,d\tau \mathrm{\,\,-overlap\,integral} \] Find stationary points for both \(c_a\) and \(c_b\). The determinant equation is called the secular equation.
Always sketch the resultant orbitals to illustrate better.
Without ignoring the overlap integral, it can be worked out that the energy raised for the antibonding orbital is larger than the energy lowerd for the bonding orbital. \[ E_{ab}=\alpha +\frac{\beta}{1-S}\\ E_{b}=\alpha-\frac{\beta}{1+S} \]