Category: Advanced Chemistry

Posted on

Russell Saunders coupling

Russell Saunders coupling, or L-S coupling, is a method to derive the total angular momentum of a light multi-electron atom from its component angular momenta. It assumes that spin-spin coupling > orbit-orbit coupling > spin-orbit coupling. This coupling scheme leads to the grouping of degenerate eigenstates into terms and levels, both of which correspond well with spectroscopic data of elements in the first three periods of the periodic table.

The total angular momentum \boldsymbol{\mathit{j}} of a 1-electron atom is defined as the vector sum \boldsymbol{\mathit{j}}=\boldsymbol{\mathit{l}}+\boldsymbol{\mathit{s}}, where \boldsymbol{\mathit{l}} and \boldsymbol{\mathit{s}} are the orbital angular momentum and spin angular momentum, respectively, of the electron in the atom. For an n-electron atom, Henry Russell and Frederick Saunders proposed that the total angular momentum \boldsymbol{\mathit{J}} of the atom is \boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}, where \boldsymbol{\mathit{L}}=\sum_{i=1}^n\boldsymbol{\mathit{l}}_i and \boldsymbol{\mathit{S}}=\sum_{i=1}^n\boldsymbol{\mathit{s}}_i. If we use a 2-electron atom as an example, the corresponding operator is:

\hat{J}=\hat{L}\otimes I+I\otimes\hat{S}\; \; \; \; \; \; \; \; 290

Since spin-spin coupling > orbit-orbit coupling > spin-orbit coupling, the spin angular momenta couple amongst themselves and the orbital angular momenta couple amongst themselves, before the total spin angular momentum couples with the total angular momentum. Hence, we have \hat{L}=\hat{l}^{(1)}\otimes I+I\otimes\hat{l}^{(2)} and \hat{S}=\hat{s}^{(1)}\otimes I+I\otimes\hat{s}^{(2)}. The eigenstates of \hat{L} are coupled representations of the basis vectors of \hat{l}^{(1)} and \hat{l}^{(2)}, while the eigenstates of \hat{S} are coupled representations of the basis vectors of \hat{s}^{(1)} and \hat{s}^{(2)}. The eigenstates of \hat{J} can either have the form of an uncoupled representation \vert L,M_L,S,M_S\rangle or a coupled representation \vert J,M_J,L,S\rangle. The uncoupled representation is used when spin-orbit coupling is weak.

 

Next article: Term symbols
Previous article: relativistic Many-electron Hamiltonian
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Relativistic many-electron Hamiltonian

The multi-electron Hamiltonian including the relativistic spin-orbit coupling term (but excluding other relativistic corrections) is

\hat{H}_R=\hat{H}_T+\hat{H}_{so}\; \; \; \; \; \; \; \; 283

where \hat{H}_T and \hat{H}_{so} are given by eq240 and eq261a, respectively.We shall now show that \hat{J}_z and \hat{J}^2 commute with \sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i and therefore with \hat{H}_R. Assuming n=2,

\left[\hat{J}_z,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=\left [\hat{L}_z^T+\hat{S}_z^T ,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right ]=\left [\hat{L}_{1z}+\hat{L}_{2z}+\hat{S}_{1z}+\hat{S}_{2z} ,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right ]

Expanding the RHS of the second equality of the above equation and using eq99, eq100, eq101, eq165, eq166 and eq167, \left[\hat{J}_z,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0. In general,

\left[\hat{J}_z,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 284

Similarly,

\left[\hat{J}_x,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; and\; \; \; \; \;\left[\hat{J}_y,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 285

Furthermore, [\hat{J}_z,\hat{H}_T]=[\hat{L}_z^T,\hat{H}_T]+[\hat{S}_z^T,\hat{H}_T]. Using eq243, [\hat{J}_z,\hat{H}_T]=0. Combining this with eq284,

\left[\hat{J}_z,\hat{H}_R\right]=0\; \; \; \; \; \; \; \; 286

Next,

\left[\hat{J}^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=\left[\hat{J}_x^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]+\left[\hat{J}_y^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]+\left[\hat{J}_z^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]

Using the identity [\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}] and substituting eq284 and eq285 in the above equation, \left[\hat{J}^2,\sum_{i=1}^2\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0. In general,

\left[\hat{J}^2,\sum_{i=1}^n\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\right]=0\; \; \; \; \; \; \; \; 287

Using eq244, eq245, \left[\hat{J}^2,\hat{H}_T\right]=\left[\hat{{L}^2}^T+\hat{{S}^2}^T+2\hat{\boldsymbol{\mathit{S}}}^T\cdot\hat{\boldsymbol{\mathit{L}}}^T,\hat{H}_T\right]=0. So,

\left[\hat{J}^2,\hat{H}_R\right]=0\; \; \; \; \; \; \; \; 288

Therefore, \hat{J}^2 and \hat{J}_z commute with \hat{H}_R. We have also shown in an earlier article that \hat{P}_{ij} commutes with \hat{J}^2, \hat{J}_z and \hat{H}_R. This implies that we can select a common complete set of eigenfunctions for \hat{J}^2, \hat{J}_z, \hat{P}_{ij} and \hat{H}_R. . Unlike the case of the hydrogen atom, \hat{{L}^2}^T and \hat{{S}^2}^T no longer commute with \hat{H}_{so} for a multi-electron atom (see eq182 and eq182a). Despite that, the quantum numbers J, L and S are used to characterise the states of elements in the first three periods of the periodic table, because spin-orbit coupling is weak for these atoms.

Similar to hydrogen, the eigenvalue equation of \hat{H}\psi=E\psi for a multi-electron atom is solved by treating \hat{H}_{so} as a perturbation if spin-orbit coupling is weak, with the unperturbed part of the eigenvalue equation solved using the Hartree-Fock method.

 

Question

Evaluate E_{so} using the first order perturbation theory.

Answer

For a multi-electron system, \hat{J}^2=\hat{L}^2+\hat{S}^2+2\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}. So, \hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}=\frac{1}{2}\left ( \hat{J}^2-\hat{L}^2-\hat{S}^2\right ), which when substituted in E_{so}=\langle\psi\vert\xi(r)\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\vert\psi\rangle and using eq133, eq160 and eq205a, gives

E_{so}=A\frac{\hbar^2}{2}[J(J+1)-L(L+1)-S(S+1)]\; \; \; \; \; \; \; \; 289

where A=\langle\psi\vert\xi(r)\vert\psi\rangle=\biggl\langle\psi\left | \sum_{i=1}^n\frac{1}{2m_e^{\;2}c^2r_i}\frac{dV_i}{dr_i}\right |\psi\biggr\rangle.

A is evaluated empirically to be a constant for a given term.

 

 

Next article: Russell Saunders coupling
Previous article: relativistic one-electron Hamiltonian
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Relativistic one-electron Hamiltonian

The one-electron Hamiltonian including the relativistic spin-orbit coupling term (but excluding other relativistic corrections) adds the term in eq261 to eq45 to give:

\hat{H}=-\frac{\hbar^2}{2m_e}\nabla^2-\frac{Ze^2}{4\pi\varepsilon_0r}+\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\; \; \; \; \; \; \; \; 282

or

\hat{H}=\frac{\hat{p}^2}{2m_e}-\frac{Ze^2}{4\pi\varepsilon_0r}+\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}

Since V=-\frac{Ze^2}{4\pi\varepsilon_0r}

\hat{H}=\hat{H}_0+\frac{1}{2m_e^{\;2}c^2}\frac{Ze^2}{4\pi\varepsilon_0r^3}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}

where \hat{H}_0=\frac{\hat{p}^2}{2m_e}-\frac{Ze^2}{4\pi\varepsilon_0r}.

From eq106 and eq107, we know that [\hat{L^2},\hat{H}_0]=0. We have also shown in an earlier article (Pauli matrices) that [\hat{L^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0 and [\hat{S^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0. Consequently, \left\[\hat{L^2},\frac{\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}}{r^3}\right\]=0 and \left\[\hat{S^2},\frac{\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}}{r^3}\right\]=0.

 

Question

Show that [\hat{J}^2,\hat{p}^2]=0, [\hat{J}^2,\frac{1}{r}]=0, [\hat{J^2},\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0 and [\hat{J}_z,\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}]=0, where \hat{\boldsymbol{\mathit{J}}}=\hat{\boldsymbol{\mathit{L}}}+\hat{\boldsymbol{\mathit{S}}}.

Answer

Substituting \hat{J}^2=\boldsymbol{\mathit{J}}\cdot\boldsymbol{\mathit{J}}=\hat{L}^2+\hat{S}^2+2\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}  in [\hat{J}^2,\hat{p}^2], expanding the expression and using eq104, we have [\hat{J}^2,\hat{p}^2]=0. Similarly, [\hat{J}^2,\frac{1}{r}]=0. Clearly, [\hat{J}^2,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =0. Finally, expanding the RHS of [\hat{J}_z,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =[\hat{J}_z,\hat{L}_x\hat{S}_x]+[\hat{J}_z,\hat{L}_y\hat{S}_y]+[\hat{J}_z,\hat{L}_z\hat{S}_z], noting that \hat{L}_i and \hat{S}_i act on different vector spaces and using eq100, eq101, eq166 and eq167, [\hat{J}_z,\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}] =0.

 

Since \hat{L}^2, \hat{S}^2, \hat{J}^2 and \hat{J}_z all commute with \hat{H}, we can select a complete set of eigenfunctions in the form of \vert\hat{J}^2,\hat{J}_z,\hat{L}^2,\hat{S}^2\rangle for all operators. The eigenvalue equation of \hat{H}\psi=E\psi is solved by treating \hat{H}_{so}=\frac{1}{2m_e^{\;2}c^2}\frac{Ze^2}{4\pi\varepsilon_0r^3}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}} as a perturbation.

 

Question

Evaluate E_{so} using the first order perturbation theory.

Answer

For a one-electron system, \hat{J}^2=\hat{L}^2+\hat{S}^2+2\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}. So, \boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}=\frac{1}{2}(\hat{J}^2-\hat{L}^2-\hat{S}^2). Using this equation and eq133, eq160 and eq205a,

E_{so}=\langle\psi\vert\xi(r)\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}\vert\psi\rangle=\langle\xi(r)\rangle\frac{\hbar^2}{2}[J(J+1)-L(L+1)-S(S+1)]

where \langle\xi(r)\rangle=\biggl\langle\psi\left | \frac{1}{2m_e^{\;2}c^2}\frac{ze^2}{4\pi\varepsilon_0r^3} \right |\psi\biggr\rangle.

 

 

Next article: Relativistic many-electron Hamiltonian
Previous article: Non-crossing rule
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Degenerate perturbation theory (an example)

Consider the eigenvalue equation of eq262 for a three-dimensional infinite cubical well, where

\hat{H}^{(0)}(x,y,z)=\left\{\begin{matrix} 0 & if\: 0<x<a,0<y<a,0<z<a\\ \infty & otherwise \end{matrix}\right.\; \; \; \; \; \; \; \; 277

\psi_{n_x,n_y,n_z}^{(0)}=\left ( \frac{2}{a} \right )^{\frac{3}{2}}sin\left ( \frac{n_x\pi}{a}x \right )sin\left ( \frac{n_y\pi}{a}y \right )sin\left ( \frac{n_z\pi}{a}z \right )\; \; \; \; \; \; \; \; 278

E_{n_x,n_y,n_z}^{(0)}=\frac{\pi^{2}\hbar^2}{2ma^2}(n_x^{\;2}+n_y^{\;2}+n_z^{\;2})\; \; \; \; \; \; \; \; 279

where n_x,n_y,n_z\in \mathbb{Z}^+.

From eq278, the ground state is \psi_{111}^{(0)}, while the first excited state is characterised by \psi_1=\psi_{112}^{(0)}, \psi_2=\psi_{121}^{(0)} and \psi_3=\psi_{211}^{(0)}. It is obvious, from eq279, that the first excited state is degenerate. Let’s assume that the potential in the well is perturbed such that first-order correction to \hat{H}^{(0)} is

\hat{H}^{(1)}=\left\{\begin{matrix} V_0,&if\: 0<x<\frac{a}{2},0<y<\frac{a}{2},0<z<a\\0& otherwise \end{matrix}\right.\; \; \; \; \; \; \; \; 280

We shall analyse the degenerate subspace of the first excited state. The first step is to compute the matrix elements W_{jl}=\langle\psi_{j,D1}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{l,D1}^{(0)}\rangle and substitute them in eq274, with the assumption that \Psi_n^{(0)}=\alpha_1\psi_1+\alpha_2\psi_2+\alpha_3\psi_3:

\begin{pmatrix} \frac{V_0}{4} &0 &0 \\ 0 & \frac{V_0}{4} &\frac{16V_0}{9\pi^2} \\ 0 &\frac{16V_0}{9\pi^2} &\frac{V_0}{4} \end{pmatrix}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}=U_n^{(1)}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix} \; \; \; \; \; \; \; \; 281

Finding the non-trivial solutions of the above linear homogeneous equation,

\begin{vmatrix} \frac{V_0}{4}-U_n^{(1)}&0 &0 \\ 0 & \frac{V_0}{4}-U_n^{(1)}&\frac{16V_0}{9\pi^2} \\ 0 &\frac{16V_0}{9\pi^2} &\frac{V_0}{4}-U_n^{(1)} \end{vmatrix}=0

\left ( \frac{V_0}{4}-U_n^{(1)}\right ) \left [\left ( \frac{V_0}{4}-U_n^{(1)}\right )^2-\frac{256V_0^2}{81\pi^4} \right ] =0

The roots are U_1^{(1)}=\frac{V_0}{4}, U_2^{(1)}= \frac{V_0}{4}+\frac{16V_0}{9\pi^2} and U_3^{(1)}= \frac{V_0}{4}-\frac{16V_0}{9\pi^2}, which means that the perturbation \hat{H}^{(1)} lifts the degeneracy of the unperturbed states. To find the exact eigenstates \Psi_n^{(0)}, we substitute the roots back in eq281. For the first root, we have

\begin{pmatrix} \frac{V_0}{4}\alpha_1\\\frac{V_0}{4}\alpha_2+ \frac{16V_0}{9\pi^2}\alpha_3\\ \frac{16V_0}{9\pi^2}\alpha_2+ \frac{V_0}{4}\alpha_3 \end{pmatrix}=\frac{V_0}{4}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}

Comparing both sides of the above equation, \alpha_1 can be any number but \alpha_2=0 and \alpha_3=0. So, the normalised wavefunction associated with U_1^{(1)} is \Psi_1^{(0)}=\psi_1. For the second root, we have

\begin{pmatrix} \frac{V_0}{4}\alpha_1\\\frac{V_0}{4}\alpha_2+ \frac{16V_0}{9\pi^2}\alpha_3\\ \frac{16V_0}{9\pi^2}\alpha_2+ \frac{V_0}{4}\alpha_3 \end{pmatrix}=\left ( \frac{V_0}{4}+\frac{16V_0}{9\pi^2}\right )\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}

Comparing both sides, \alpha_1=0 since V_0\neq0, and \alpha_2=\alpha_3=1 (\alpha_2 and \alpha_3 are not equal to zero because that will result in \Psi_2^{(0)} being a zero eigenfunction). The normalised wavefunction associated with U_2^{(1)} is \Psi_2^{(0)}=\frac{1}{\sqrt{2}}(\psi_2+\psi_3). Similarly, for U_3^{(1)}, we have \Psi_3^{(0)}=\frac{1}{\sqrt{2}}(\psi_2-\psi_3).

Finally, by substituting \Psi_1^{(0)}, \Psi_2^{(0)} and \Psi_3^{(0)} in eq273, we find that they diagonalise \hat{H}^{(1)}. We call these linearly combined states that diagonalise \hat{H}^{(1)} “good” states.

 

Next article: Time-dependent perturbation theory
Previous article: Degenerate Perturbation theory
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Degenerate perturbation theory

The degenerate perturbation theory, an extension of the perturbation theory, is used to find an approximate solution to a quantum-mechanical problem involving non-perturbed states that are degenerate.

In the non-degenerate case, it is unambiguous that \Psi_n^{\;(0)}=\psi_n^{\;(0)}, i.e. \Psi_n^{\;(0)} is described by individual basis states \psi_n^{\;(0)}. However, for a set of degenerate orthonormal states \left \{\psi_{j,D1}^{\;(0)}\right \} (\small D1 denotes the same eigenvalue for all states in the set) of the unperturbed Hamiltonian \hat{H}^{(0)}, where \hat{H}^{(0)}\psi_{j}^{\;(0)}=E_{D1}^{\;(0)}\psi_{j}^{\;(0)}, \Psi_{n}^{\;(0)} can represent any linear combination of \left \{\psi_{j,D1}^{\;(0)}\right \} or the non-degenerate states \psi_{n}^{\;(0)}, where \psi_{n}^{\;(0)}\notin \left \{\psi_{j,D1}^{\;(0)}\right \}. If there are more than one set of degenerate states, each set with a distinct energy level, \Psi_{n}^{\;(0)} represents linear combinations of \left \{\psi_{j,D1}^{\;(0)}\right \}, linear combinations of \left \{\psi_{j,D2}^{\;(0)}\right \}, and so on, and individual states of \psi_{n}^{\;(0)}, where \psi_{n}^{\;(0)}\notin \left \{\psi_{j,DX}^{\;(0)}\right \}.

For each set of degenerate orthonormal states, e.g. \left \{\psi_{j,D1}^{\;(0)}\right \}, of the unperturbed Hamiltonian \hat{H}^{(0)}, the linear combination \small \Psi_{n}^{\;(0)}=\sum_{j=1}^{g}\alpha_j\psi_{j,D1}^{\;(0)} (where \small g is the degeneracy) is an eigenstate of \hat{H}^{(0)} with the same eigenvalue \small U_{n,D1}^{\;(0)}=E_{n,D1}^{\;(0)}:

\small \hat{H}^{(0)}\Psi_{n}^{\;(0)}=\sum_{j=1}^{g}\alpha_j\hat{H}^{(0)}\psi_{j,D1}^{\;(0)} =E_{n,D1}^{\;(0)}\sum_{j=1}^{g}\alpha_j\psi_{j,D1}^{\;(0)}=E_{n,D1}^{\;(0)}\Psi_{n}^{\;(0)} \; \; \; \; \; \; \; \; 271

Therefore, the derivation in the previous article branches off from eq267. While \small \Psi_{n}^{\;(0)} is no longer unambiguously described by individual basis states \small \psi_{n}^{\;(0)}, \small U_{n}^{\;(0)} is still \small E_{n}^{\;(0)}. So, eq267 becomes

\small \hat{H}^{(0)}\Psi_{n}^{\;(1)}+\hat{H}^{(1)}\Psi_{n}^{\;(0)}-E_{n}^{\;(0)}\Psi_{n}^{\;(1)}-U_{n}^{\;(1)}\Psi_{n}^{\;(0)}=0\; \; \; \; \; \; \; \; 272

Multiplying the above equation from the left by \small \Psi_{j}^{\;(0)*} and integrating,

( E_{j}^{\;(0)}-E_{n}^{\;(0)} )\langle\Psi_{j}^{\;(0)}\vert\Psi_{n}^{\;(1)}\rangle +\langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle =0

When E_{j}^{\;(0)}=E_{n}^{\;(0)}, we have.

\langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle =0\; \; \; \; \; \; \; \; 273

Since \langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle represents the matrix elements of \hat{H}^{(1)}, eq273 implies that in the presence of degenerate states, the first-order perturbation theory only works if we can select appropriate wavefunctions that diagonalise \hat{H}^{(1)}.

Consider the case of just one set of degenerate orthonormal states \left \{ \psi_{l,D1}^{(0)} \right \} of the unperturbed Hamiltonian \hat{H}^{(0)}, with the other states of \hat{H}^{(0)} being non-degenerate. If we further analyse the subspace of \left \{ \psi_{l,D1}^{(0)} \right \}, we substitute \Psi_{n}^{\;(0)}=\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)} in eq272 to give:

\hat{H}^{(0)}\Psi_{n}^{\;(1)}+\hat{H}^{(1)}\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)}-E_{n,D1}^{(0)}\Psi_{n}^{\;(1)}-U_n^{(1)}\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)}=0

Taking the inner product with \psi_{j,D1}^{\;(0)} and using eq36 on the first term,

\sum_{l=1}^gW_{jl}\alpha_l=U_n^{(1)}\alpha_l\; \; \; \; \; \; \; \; 274

where W_{jl}=\langle\psi_{j,D1}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{l,D1}^{(0)}\rangle  are the matrix elements of \hat{H}^{(1)}.

Rewriting the eigenvalue equation of eq274 in matrix form,

\left [ \begin{pmatrix} W_{11} &W_{12} &\cdots &W_{1g} \\ W_{21} &W_{22} &\cdots &W_{2g} \\ \vdots &\vdots & \ddots &\vdots \\ W_{g1} & W_{g2}&\cdots &W_{gg} \end{pmatrix}-U_n^{(1)}I \right ]\begin{pmatrix} \alpha_1\\\alpha_2 \\ \vdots \\ \alpha_g \end{pmatrix}=0\; \; \; \; \; \; \; \; 275

where I is the g\times g identity matrix.

Eq275 is a linear homogeneous equation, which has non-trivial solutions if

\left | \begin{pmatrix} W_{11} &W_{12} &\cdots &W_{1g} \\ W_{21} &W_{22} &\cdots &W_{2g} \\ \vdots &\vdots & \ddots &\vdots \\ W_{g1} & W_{g2}&\cdots &W_{gg} \end{pmatrix}-U_n^{(1)}I \right |=0\; \; \; \; \; \; \; \; 276

Expanding the determinant (also known as a secular equation), we get the characteristic equation, which can be solved for the roots of U_n^{(1)}. Each of these roots is then substituted into eq274 to find the set of coefficients \alpha_l and hence, the exact eigenstates \Psi_n^{(0)}. If these exact eigenstates diagonalises \hat{H}^{(1)}, we refer to them as “good” states, which allow U_n^{(1)} to be calculated (see next article for an example).

 

Question

What is a linear homogeneous equation? Why does a linear homogeneous equation have non-trivial solutions if the determinant of the operator is zero?

Answer

A linear homogeneous equation is a matrix equation of the form A\boldsymbol{\mathit{X}}=0, where A is an n\times n matrix and \boldsymbol{\mathit{X}} is column vector. Let A be a 3\times3 matrix representing three position vectors  and \boldsymbol{\mathit{X}}=(x,y,z). We have

\begin{pmatrix} a_1 &a_2 &a_3 \\ b_1 & b_2 &b_3 \\ c_1 &c_2 &c_3 \end{pmatrix}\begin{pmatrix} x\\y \\ z \end{pmatrix}=\begin{pmatrix} a_1x+a_2y+a_3z\\b_1x+b_2y+b_3z \\ c_1x+c_2y+c_3z \end{pmatrix}=0

So, \boldsymbol{\mathit{a}}\cdot\boldsymbol{\mathit{X}}=0, \boldsymbol{\mathit{b}}\cdot\boldsymbol{\mathit{X}}=0 and \boldsymbol{\mathit{c}}\cdot\boldsymbol{\mathit{X}}=0. Now, a non-trivial solution of A\boldsymbol{\mathit{X}}=0 occurs when \boldsymbol{\mathit{X}}\neq0. Therefore, we require any non-zero position vector \boldsymbol{\mathit{X}} that is orthogonal to \boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}. This is fulfilled if the position vectors \boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}} lie in a plane, which happens when the scalar triplet product \boldsymbol{\mathit{a}}\cdot(\boldsymbol{\mathit{b}}\times\boldsymbol{\mathit{c}}), which is equal to the volume of the parallelepiped spanned by \boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}, is zero. Since the scalar triple product is also equal to the determinant of A, i.e. \vert A\vert=\boldsymbol{\mathit{a}}\cdot(\boldsymbol{\mathit{b}}\times\boldsymbol{\mathit{c}}) (which can be easily proven by expanding both sides and showing that LHS equals to RHS), A\boldsymbol{\mathit{X}}=0 has non-trivial solutions if \vert A\vert=0.

 

Since there are multiple roots of U_n^{(1)}, the perturbation \hat{H}^{(1)} usually lifts the degeneracy of the unperturbed states. An example is the relativistic Hamiltonian’s spin-orbit coupling perturbation \hat{H}^{(1)}=\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}, which splits terms ^{2S+1}\textrm{\textit{L}} into levels ^{2S+1}\textrm{\textit{L}}_J. If there are equal roots, we would need to further analyse the problem with higher order perturbations to completely lift the degeneracy (as is the case when considering the Zeeman effect).

 

Question

Find the characteristic equation for the degenerate set \left \{ \psi_1^{(0)},\psi_2^{(0)} \right \}.

Answer

Eq276 becomes,

\left | \begin{pmatrix} W_{11} &W_{12} \\ W_{21} &W_{22} \end{pmatrix}-U_n^{(1)}I \right |=0

{U_n^{(1)}}^{2}-U_n^{(1)}(W_{11}+W_{22})+W_{11}W_{22}-W_{12}W_{21} =0

 

 

Next article: Degenerate perturbation theory (an example)
Previous article: Perturbation theory
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Perturbation theory

Perturbation theory is a method for finding an approximate solution to a problem by building on exact solutions of a simpler and related problem.

Consider the eigenvalue equation of

\hat{H}^{(0)}\psi_n^{\;(0)}=E_n^{\;(0)}\psi_n^{\;(0)}\; \; \; \; \; \; \; \; 262

where \hat{H}^{(0)} is the non-relativistic Hamiltonian, \psi_n^{\;(0)} is a complete set of orthonormal eigenfunctions that spans a Hilbert space, and E_n^{\;(0)} represents the exact eigenvalue solutions. We call \hat{H}^{(0)} and \psi_n^{\;(0)} the unperturbed Hamiltonian and unperturbed set of wavefunctions respectively.

If we extend the non-relativistic Hamiltonian to include certain interactions like spin-orbit coupling or a potential due to a ligand field, our eigenvalue equation in general is:

\hat{H}\Psi_n=U_n\Psi_n\; \; \; \; \; \; \; \; 263

where \hat{H} and \Psi_n are the perturbed Hamiltonian and perturbed set of wavefunctions respectively.

One common method of approximation is to write the new perturbed Hamiltonian as a power series in an arbitrary parameter \lambda:

\hat{H}=\hat{H}^{(0)}+\lambda\hat{H}^{(1)}+\lambda^{2}\hat{H}^{(2)}+\cdots\; \; \; \; \; \; \; \; 264

where \hat{H}^{(1)} and \hat{H}^{(2)} are the first-order and second-order corrections respectively to \hat{H}^{(0)}.

Similarly, the associated perturbed wavefunctions and energies of perturbed states are:

\Psi_n=\Psi_n^{\;(0)}+\lambda\Psi_n^{\;(1)}+\lambda^{2}\Psi_n^{\;(2)}+\cdots\; \; \; \; \; \; \; \; 265

U_n=U_n^{\;(0)}+\lambda U_n^{\;(1)}+\lambda^{2}U_n^{\;(2)}+\cdots\; \; \; \; \; \; \; \; 266

If we are satisfied with using just the first two terms in each of the power series of eq264, eq265 and eq266 for the approximation, we are dealing with what is known as the first-order perturbation theory. Substituting the first two terms in each of the three equations in eq263 and ignoring second-order quantities,

Since the parameters are arbitrary, they can be non-zero. If so, the only way to satisfy eq266 when U_n(\lambda)=0 is for U_n^{\;(0)}, U_n^{\;(1)}, U_n^{\;(2)} and so on to be zero. This implies that the powers of \lambda are independent variables. Consequently, the coefficients in eq266a must be zero:

\hat{H}^{(0)}\Psi_n^{\;(0)}-U_n^{\;(0)}\Psi_n^{\;(0)} =0

\hat{H}^{(0)}\Psi_n^{\;(1)}+\hat{H}^{(1)}\Psi_n^{\;(0)} -U_n^{\;(0)}\Psi_n^{\;(1)}-U_n^{\;(1)}\Psi_n^{\;(0)} =0\; \; \; \; \; \; \; \; 267

Substituting \Psi_n^{\;(0)}=\psi_n^{\;(0)} and U_n^{\;(0)}=E_n^{\;(0)} in eq267,

\hat{H}^{(0)}\Psi_n^{\;(1)}+\hat{H}^{(1)}\psi_n^{\;(0)} =E_n^{\;(0)}\Psi_n^{\;(1)}+U_n^{\;(1)}\psi_n^{\;(0)}\; \; \; \; \; \; \; \; 268

To find U_n^{\;(1)}, we multiply eq268 by \psi_n^{\;(0)*} and integrate; and use eq36 and the orthonormality of \psi_n^{\;(0)},

\langle\hat{H}^{(0)}\psi_n^{\;(0)}\vert\Psi_n^{\;(1)}\rangle+\langle\psi_n^{\;(0)}\vert\hat{H}^{(1)}\vert\psi_n^{\;(0)}\rangle =E_n^{\;(0)}\langle\psi_n^{\;(0)}\vert\Psi_n^{\;(1)}\rangle+U_n^{\;(1)}

U_n^{\;(1)}=\langle\psi_n^{\;(0)}\vert\hat{H}^{(1)}\vert\psi_n^{\;(0)}\rangle \; \; \; \; \; \; \; \; 269

Eq269 reveals that we can compute U_n^{\;(1)} using the first-order Hamiltonian and the complete set of wavefunctions of \hat{H}^{(0)}. However, the equation only works if all the wavefunctions are non-degenerate. To explain this limitation, we have to find the expression for \Psi_n^{\;(1)}. Since , the unperturbed wavefunction  can be fully characterised by the complete set \left \{ \psi_n^{\;(0)} \right \}. We therefore expand \Psi_n^{\;(1)}, which belongs to the same Hilbert space, as a linear combination of \left \{ \psi_n^{\;(0)} \right \}.

 

Question

Show that writing the linear combination as \Psi_n^{\;(1)}=\sum_{m\neq n}c_m\psi_m^{\;(0)} (i.e. \Psi_n^{\;(1)} has no \psi_n^{\;(0)} component) ensures that \Psi_n in eq263 is normalised with respect to the first order of \lambda.

Answer

When \Psi_n is normalised, \langle\Psi_n\vert\Psi_n\rangle=1. Substituting eq265 in this equation and expanding up to \lambda,

\langle\Psi_n\vert\Psi_n\rangle=1+\lambda\langle\Psi_n^{\;(1)}\vert\psi_n^{\;(0)}\rangle+\lambda\langle\psi_n^{\;(0)}\vert\Psi_n^{\;(1)}\rangle

If \Psi_n^{\;(1)} has no \psi_n^{\;(0)} component, i.e. \Psi_n^{\;(1)}\neq a\psi_n^{\;(0)}+b\psi_m^{\;(0)}+\cdots,

\langle\Psi_n\vert\Psi_n\rangle=1+0+0=1

 

Rearranging eq268 to \left ( \hat{H}^{(0)}-E_n^{\;(0)}\right )\Psi_n^{\;(1)}=\left ( U_n^{\;(1)}-\hat{H}^{(1)}\right )\psi_n^{\;(0)} and substituting \Psi_n^{\;(1)}=\sum_{m\neq n}c_m\psi_m^{\;(0)} in the equation,

\sum_{m\neq n}\left (E_m^{\;(0)}-E_n^{\;(0)}\right )c_m\psi_m^{\;(0)}=\left (U_n^{\;(1)}-\hat{H}^{(1)}\right )\psi_n^{\;(0)}

Multiplying the above equation by \psi_i^{\;(0)} and integrating,

\sum_{m\neq n}\left (E_m^{\;(0)}-E_n^{\;(0)}\right )c_m\langle\psi_i^{\;(0)}\vert\psi_m^{\;(0)}\rangle=U_n^{\;(1)}\langle\psi_i^{\;(0)}\vert\psi_n^{\;(0)}\rangle-\langle\psi_i^{\;(0)}\vert\hat{H}^{(1)}\vert\psi_n^{\;(0)}\rangle

If l=n, the above equation reduces to eq269. If l\neq n, only the term with \langle\psi_m^{\;(0)}\vert\psi_m^{\;(0)}\rangle survives, giving:

Substituting the above equation in \Psi_n^{\;(1)}=\sum_{m\neq n}c_m\psi_m^{\;(0)},

\Psi_n^{\;(1)}=\sum_{m\neq n}\frac{\langle\psi_m^{\;(0)}\vert\hat{H}^{(1)}\vert\psi_n^{\;(0)}\rangle}{E_n^{\;(0)}-E_m^{\;(0)}}\psi_m^{\;(0)}\; \; \; \; \; \; \; \; 270

The total wavefunction is \Psi_n=\Psi_n^{\;(0)}+\sum_{m\neq n}\frac{\langle\psi_m^{\;(0)}\vert\hat{H}^{(1)}\vert\psi_n^{\;(0)}\rangle}{E_n^{\;(0)}-E_m^{\;(0)}}\psi_m^{\;(0)}. If E_n^{\;(0)}=E_m^{\;(0)}, we have a problem. So, eq269 only works when all the wavefunctions are non-degenerate. If some of the wavefunctions are degenerate, we require another method called degenerate perturbation theory to calculate the eigenvalues.

To derive the second-order energy correction, we consider the following Schrodinger equation:

where we have used just a first-order correction to the Hamiltonian to simplify the calculations.

Collecting the powers of  and letting the coefficient be zero, we have

Multiplying the above equation by  and integrating over all space,

Using eq35, . So,

For ,

Substituting \Psi_n^{\;(1)}=\sum_{m\neq n}c_m\psi_m^{\;(0)} and eq269a in the above equation and using eq35,

 

 

Next article: Degenerate perturbation theory
Previous article: Spin-orbit coupling
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Spin-orbit coupling

Spin-orbit coupling is the interaction between a particle’s spin angular momentum and orbital angular momentum. An electron orbiting around the nucleus ‘sees’ the nucleus circling it, just as a person on earth perceives the sun circling the earth while the latter orbits around the sun.

This apparent nuclear orbit creates a magnetic field B that exerts a torque on the electron’s spin magnetic dipole moment \mu_s, resulting in an additional term of \hat{H}_{so}=\sum_{i=1}^{n}A_i\hat{\boldsymbol{\mathit{S}}}_i\cdot\hat{\boldsymbol{\mathit{L}}}_i (where A_i=\frac{1}{2m_e^{\;2}c^{2}}\frac{1}{r_i}\frac{dV_i}{dr_i}) in the multi-electron Hamiltonian. To derive this term, we consider a 1-electron atom.

Let L=m_erv_{\perp} be the orbital angular momentum of the electron and I be the proton’s current loop, which generates a magnetic field of magnitude B=\frac{\mu_0I}{2r} given by the Biot-Savart law. Since \frac{1}{t}=\frac{v_{\perp}}{2\pi r}, we have

I=\frac{e}{t}=\frac{eL}{2\pi m_er^{2}}\; \; \; \; \; \; \; \; 259

Substitute eq259 in B, we have, B=\frac{\mu_0e}{4\pi m_er^{3}}L or \frac{\boldsymbol{\mathit{B}}}{\hat{\boldsymbol{\mathit{B}}}}=\frac{\mu_0e}{4\pi m_er^{3}}\frac{\boldsymbol{\mathit{L}}}{\hat{\boldsymbol{\mathit{L}}}}, where \hat{\boldsymbol{\mathit{B}}} and \hat{\boldsymbol{\mathit{L}}} are unit vectors. Since \boldsymbol{\mathit{B}} and \boldsymbol{\mathit{L}} point in the same direction, \hat{\boldsymbol{\mathit{B}}}=\hat{\boldsymbol{\mathit{L}}}=\hat{\boldsymbol{\mathit{M}}}. Multiplying both sides of \frac{\boldsymbol{\mathit{B}}}{\hat{\boldsymbol{\mathit{M}}}}=\frac{\mu_0e}{4\pi m_er^{3}}\frac{\boldsymbol{\mathit{L}}}{\hat{\boldsymbol{\mathit{M}}}} by \hat{\boldsymbol{\mathit{M}}}, we have \boldsymbol{\mathit{B}}=\frac{\mu_0e}{4\pi m_er^{3}}\boldsymbol{\mathit{L}}, which we substitute in eq65 (where \boldsymbol{\mathit{\mu}} is the spin analogue of eq61) to give U=-\frac{\gamma_e\mu_0e}{4\pi m_er^{3}}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}.

Substituting eq164 and \frac{1}{c^{2}}=\mu_0\varepsilon_0 in U yields

U=\frac{e^{2}}{4\pi\varepsilon_0 m_e^{\;2}c^{2}r^{3}}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}

For a 1-electron atom, V=-\frac{e^2}{4\pi\varepsilon_0r} and so

U=\frac{1}{ m_e^{\;2}c^{2}r}\frac{dV}{dr}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 260

Eq260 can be written in terms of the Larmor frequency of the electron. From eq149, \omega_L=\frac{eB}{m_e}=\frac{e}{m_e}\frac{\mu_0e}{4\pi m_er^{3}}L=\frac{1}{m_e^{\;2}c^{2}r}\frac{dV}{dr}L. So, U=\frac{\omega_L}{L}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}. Swapping \omega_L with the Thomas precession rate \omega_T, we obtain the correction term of U_{tp}=-\frac{1}{2 m_e^{\;2}c^{2}r}\frac{dV}{dr}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}.

The total spin-orbit Hamiltonian is

\hat{H}_{so}=U+U_{tp}=U_{tp}=\frac{1}{2 m_e^{\;2}c^{2}r}\frac{dV}{dr}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 261

For a multi-electron atom,

\hat{H}_{so}=\sum_{i=1}^n\frac{1}{2m_e^{\;2}c^2r_i}\frac{dV_i}{dr_i}\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{L}}_i\; \; \; \; \; \; \; \; 261a

 

Next article: Perturbation theory
Previous article: Thomas precession
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Thomas precession

Thomas precession is a relativistic kinematic correction to the spin-orbit coupling Hamiltonian.

Consider an electron orbiting counter-clockwise around a nucleus at O along the path of a regular n-sided polygon in the xy-plane (see diagram above). At t=0, the nucleus sees the electron making a turn at (x_0,y_0) at an angle \theta=tan^{-1}\frac{y_1}{x_1}. At subsequent vertices, the nucleus sees the electron turning at the same angle before arriving back to (x_0,y_0).

In the rest frame of the electron (denoted by blue axes), (x_1,y_1) becomes \left ( \frac{x_1}{\gamma},y_1 \right ) due to length contraction along the direction of travel, and the electron sees itself turning at angle of \theta'=tan^{-1}\left ( \gamma\frac{y_1}{x_1} \right ) at (x_0,y_0), where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}. At subsequent vertices, the electron sees the same turning angle before arriving back to (x_0,y_0). Substituting tan\theta=\frac{y_1}{x_1} in tan\theta'=\gamma\frac{y_1}{x_1}, we have tan\theta'=\gamma tan\theta. In the limit of large n, the polygon approaches a circle with each turning angle being very small. Using the small-angle approximation, \theta'=\gamma\theta. According to the nucleus and the electron, each turn corresponds to \theta=\frac{2\pi}{n} and \theta'=\frac{2\pi\gamma}{n} respectively. So, after completing a full orbit, the electron sees itself turning an angle of \theta'_T=2\pi\gamma.

In the rest frame of the nucleus (denoted by red axes), the orientation of the spin axis of the electron S remains unchanged (just like the spin axis of the earth being constant from the sun’s perspective). However, in the rest frame of the electron, the spin axis of the electron appears to rotate clockwise. After a completing a full orbit, the change in the orientation of S is \Delta\theta'=2\pi\gamma-2\pi. Since the rate of precession is defined as the rate of change in orientation of the rotational axis of a rotating body, the rate of \Delta\theta', called the Thomas precession, is

\omega_T=\omega(\gamma-1)\; \; \; \; \; \; \; \; 255

where \omega_T=\frac{\Delta\theta'}{t}, \omega=\frac{2\pi}{t} and t is the period of the orbit.

Substitute a=\frac{v_{\perp}^{\;2}}{r} and eq58, where \omega=\frac{v_{\perp}}{r}, in eq255 (note that the radius is perpendicular to the path of travel and is not contracted, and therefore common to both nucleus and electron), \omega_T=\frac{v_{\perp}}{r}(\gamma-1)=\frac{v_{\perp}a}{v_{\perp}^{\;2}}(\gamma-1). Since v\ll c, we have \left | -\frac{v_{\perp}^{\;2}}{c^{2}} \right |<1. Let x=-\frac{v_{\perp}^{\;2}}{c^{2}} in the binomial series (1+x)^{-\frac{1}{2}}=1-\frac{x}{2}+\frac{3x^{2}}{8}-\cdots. We have \omega_T\approx \frac{v_{\perp}a}{v_{\perp}^{\;2}}\left ( 1+\frac{v_{\perp}^{\;2}}{2c^{2}}-1 \right ) or

\omega_T= \frac{v_{\perp}a}{2c^{2}}\; \; \; \; \; \; \; \; 256

Substitute F=m_ea=-\frac{dV}{dr} and eq59, where L=rm_ev_{\perp}, in eq256

\omega_T= -\frac{1}{2m_e^{\;2}c^{2}r}\frac{dV}{dr}L\; \; \; \; \; \; \; \; 257

 

Next article: Spin-orbit coupling
Previous article: Why does a current flowing through a wire generate a magnetic field?
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Length contraction

Length contraction is the shortening of the length of an object moving relative to an observer, compared to its proper length, which is the length measured by an observer at rest relative to the object. Consider a moving train with a clock, which consists of a photon bouncing horizontally between two mirrors A and B (see diagram below).

For observer X, who is at rest on a platform and seeing the passing train, the total distance travelled by the photon from A to B in t_1 (indicated by the top purple arrow) is

ct_1=L+ut_1\; \; \; \; \; \; \; \; 250

where c is the speed of light, L is length of the clock according to X, and u is the speed of the train.

When the photon returns from B to A in t_2, A would have travelled a distance of ut_2. Thus, the distance travelled by the returning photon from B to A (indicated by the bottom purple arrow) is

ct_2=L-ut_2\; \; \; \; \; \; \; \; 251

The total time taken is t=t_1+t_2. Substituting eq250, where t_1=\frac{L}{c-u}, and eq251, where t_2=\frac{L}{c+u}, in t, yields

t=\frac{2L}{c}\frac{1}{1-\frac{u^{2}}{c^2}}\; \; \; \; \; \; \; \; 252

For observer Y, who is on the train, the total time t_0 (or proper time) measured by him is t_0=\frac{2L_0}{c}, where L_0 is called the proper length, which is the length measured by an observer who is at rest with respect to the object being measured. From eq249,

t=t_0\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{2L_0}{c}\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\; \; \; \; \; \; \; \; 253

Equating eq252 with eq253 gives

L=\frac{L_0}{\gamma}\; \; \; \; \; \; \; \; 254

where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}.

To illustrate eq254, consider a person on Earth observing a spaceship moving from Earth to the Moon. The proper length between the Earth and the Moon is the length measured by an observer at rest with respect to the two (i.e., the Earth frame), while the proper length of the spaceship is the length measured by an astronaut on the spaceship. The astronaut measures the proper length of the spaceship, which remains the same regardless of the spaceship’s motion. However, due to length contraction, the observer on Earth sees the spaceship as contracted in the direction of its motion. Meanwhile, the astronaut, traveling on the spaceship, observes the distance between the Earth and the moon as contracted, since, from his perspective, the Earth and the moon are moving.

 

Next article: Why does a current flowing through a wire generate a magnetic field?
Previous article: Time dilation
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Posted on

Time dilation

Time dilation is the difference in time measured by an observer who is moving relative to another observer.

Consider two frames of reference, each with an observer and a clock, which consists of a photon bouncing between two mirrors A and B. If the frame of observer X is at rest, while the frame of observer Y is moving away from it, observer Y sees the photon of his clock travelling vertically between the mirrors (diagram I) and measures t_0=\frac{2L}{c}, where t_0 is known as the proper time, L is the distance between the mirrors and c is the speed of light.

Observer X, however, sees the same photon moving in a triangular path (as observer Y’s clock moves away from him) and measures t=\frac{2D}{c} (diagram II), where t is the time recorded using his own clock.

Substituting D=\sqrt{\left ( \frac{vt}{2}\right )^{2}+L^{2}}, where v is the speed of the moving frame relative to the rest frame, and L=\frac{ct_0}{2} in t=\frac{2D}{c}, we have

t=\gamma t_0\; \; \; \; \; \; \; \; 249

where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} is called the Lorentz factor, which is useful in deriving the Thomas half.

Eq249 states that the time interval measured by an observer who is stationary relative to the moving clock for events occurring on the moving clock will always be longer than the time interval measured by someone at rest with respect to the clock. This longer time interval means that, from the perspective of the stationary observer, the moving clock is slowing down relative to the observer’s own clock. In other words, the stationary observer will see the moving clock as ticking more slowly, and the time measured by the moving clock will seem dilated or stretched from the viewpoint of the stationary observer.

 

Next article: Length contraction
Previous article: Many-electron systems
Content page of quantum mechanics
Content page of advanced chemistry
Main content page
Mono Quiz