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Verifying a Vector Identity (BAC-CAB) using Levi-Civita

Author: CHRISTINE ADELLE L. RICO

Here is another method of verifying a vector formula using the Levi-Civita symbol. Levi-Civita symbol $\epsilon_{ijk}$ is a tensor of rank three and is defined by +1 if the indices $i,j,k$ are in even permutation of $1,2,3$ , -1 if the indices are in odd permutation, and 0 if any two indices are the same.

Prove that $\vec{a} \times (\vec{b}\times\vec{c}) = \vec{b} (\vec{a}\cdot\vec{c}) - \vec{c} (\vec{a}\cdot\vec{b})$ .

Proof:

Let $\vec{a} = a_i, \vec{b} = b_j, \vec{c} = c_k$ .

Using the Levi-Civita symbol, we rewrite the cross products,

$(\vec{b}\times\vec{c})_m = b_j c_k \epsilon_{jkm}$

$(\vec{a}\times(\vec{b}\times\vec{c})_m)_n = a_i \epsilon_{imn} b_j c_k \epsilon_{jkm}$ .

Since each term is only scalar, they can be rearranged such that,

$(\vec{a}\times(\vec{b}\times\vec{c})_m)_n = a_i b_j c_k \epsilon_{imn} \epsilon_{jkm}$ where $m$ is summed over.

Note that $\epsilon_{imn} \epsilon_{jkm}$ is only nonzero if $i,n,j, k$ are all different from $m$ . There are two possibilities of its implications, either,

$i=j$ and $n=k$ or

$i=k$ and $n=j$ .

Consider the case of $i=j$ and $n=k$ , which gives $\epsilon_{imn} \epsilon_{inm} = -1$ for any value of $m$ . If $i=k$ and $n=j$ , $\epsilon_{imn} \epsilon_{nim} = +1$ . Therefore,

$\epsilon_{imn} \epsilon_{jkm} = \delta_{ik} \delta_{nj} - \delta_{ij} \delta_{nk}$ .

We can now write the proof so that the $n$ th component is,

$\begin{array} {rcl} (\vec{a}\times(\vec{b}\times\vec{c}))_n &=& a_i b_j c_k (\delta_{ik} \delta_{nj} - \delta_{ij} \delta_{nk})\\&=& a_i b_j c_k \delta_{ik} \delta_{nj} - a_i b_j c_k \delta_{ij} \delta_{nk}\\&=& a_i b_n c_i - a_i b_i c_n\\&=& b_n a_i c_i - c_n a_i b_i\\ &=& \vec{b} (\vec{a}\cdot \vec{c}) - \vec{c} (\vec{a} \cdot \vec{b})\end{array}$ .

Therefore,

$\vec{a} \times (\vec{b}\times\vec{c}) = \vec{b} (\vec{a}\cdot\vec{c}) - \vec{c} (\vec{a}\cdot\vec{b})$ .

—————————–

Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.

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Vector Identity Formula #12 verified by using Levi-Civita

Hi, guys!
The following is my solution on verifying these two vector identity formulas.

Show that
(a) $\overrightarrow{\triangledown}\cdot \overrightarrow{x}=3$
and
(b)
$\overrightarrow{\triangledown}\times \overrightarrow{x}=0$

Solution:
Let
$\overrightarrow{\bigtriangledown }=\widehat{e_{i}}\frac{\partial }{\partial x_{i}}=\widehat{e_{i}}\partial _{i}$
and
$\overrightarrow{x}=\widehat{e_{j}}x_{j}$

(a)
$\overrightarrow{\bigtriangledown }\cdot \overrightarrow{x}=\widehat{e_{i}}\partial _{i}\cdot \widehat{e_{j}}x_{j} \\ = \delta _{ij}\partial _{i}x_{j}$

since
$\delta _{ij}=1 \ \textup{if} i=j \\ = 0 \textup{if} i\neq j$
and
$\partial _{i}x_{j}=1 \ \textup{if} \ i=j \\ \textup{where} \ i,j = 1, 2, 3$

Thus,
$\overrightarrow{\triangledown}\cdot \overrightarrow{x}=3$

and

(b)
$\overrightarrow{\bigtriangledown }\times \overrightarrow{x}=\widehat{e_{i}}\partial _{i}\times \widehat{e_{j}}x_{j} \\ = \epsilon _{ijk}\partial _{i}x_{j}\widehat{e_k}$

since
$\epsilon _{ijk}= +1 \if \odd\ permutation \\ -1 \; \if\ even \ permutation,\\ or \ 0 \ if \ there \ is \ repeated \ index$
and
$\partial _{i}x_{j}=0 \ if \ i=j \\ \; =1 \ if \ i\neq j$

Therefore,
$\overrightarrow{\triangledown}\times \overrightarrow{x}=0$

About Me:
I am Yusof-Den Jamasali, an MS Physics student of Mindanao State University – Iligan Institute of Technology. I like painting, and drawing. And loves to eat avocado and carrot.

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Verifying a vector identity using Levi-Civita

VERIFYING A VECTOR IDENTITY USING LEVI-CIVITA

Bianca Rae B. Sambo

Hello physics enthusiast! I am BR, a graduate student in Physics in Mindanao State University – Iligan Institute of Technology. Hope my solution will be of use to you. Keep visiting this site!

The vector identity to be verified using Levi-Civita is,

where we define the following vectors as,

$\vec{a} = a_{i} \hat{i}$
$\vec{b} = b_{j} \hat{j}$
$\vec{\nabla} = \partial{k } \hat{k}$

We start by expanding the left hand side of the identity.

$\vec{\nabla}(\vec{a}\bullet\vec{b})=\vec{\nabla} (\delta_{ij} a_{i} b_{j})=\partial{k} \hat{k} (\delta_{ij} a_{i} b_{j})$

Then we set i = j and apply the distributive property of the partial derivative.

Equation 1:

$\vec{\nabla}(\vec{a}\bullet\vec{b})=\partial{k}(a_{j} b_{j}) \hat{k} =a_{j} \partial{k} (b_{j})\hat{k} + b_{j}\partial{k}(a_{j}) \hat{k}$

Notice in the right hand side of the vector identity, we have the equations $(\vec{a}\bullet\vec{\nabla})\vec{b}$ and $(\vec{b}\bullet\vec{\nabla})\vec{a}$ . These two equations can be written as,

$(\vec{a}\bullet\vec{\nabla})\vec{b}= a_{k} \partial{k}(b_{j})\hat{j}$

and

$(\vec{b}\bullet\vec{\nabla})\vec{a}= b_{k}\partial{k}(a_{i}) \hat{i}$

So we incorporate this to Equation 1 above.

$\vec{\nabla}(\vec{a}\bullet\vec{b})= a_{k} \partial{k}(b_{j}) \hat{j}+ b_{k}\partial{k}(a_{i}) \hat{i}+ a_{j}\partial{k} (b_{j}) \hat{k}+b_{j}\partial{k}(a_{j})\hat{k} - a_{k}\partial{k} (b_{j}) \hat{j}- b_{k} \partial{k} (a_{i}) \hat{i}$

Equation 2:

$\vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]+[(b_{j} \partial{k}(a_{j}) \hat{k}- b_{k}\partial{k}(a_{i})\hat{i})]$

Notice also that:

$\vec{a}\times(\vec{\nabla}\times\vec{b})=a_{i}\hat{i}\times(\epsilon_{kjm}\partial{k}(b_{j})\hat{m})=\epsilon_{iml}\epsilon_{kjm}a_{i} \partial{k}(b_{j}) \hat{l}=\epsilon_{lim}\epsilon_{kjm}a_{i}\partial{k}(b_{j})\hat{l}=(\delta_{lk}\delta_{ij}-\delta_{lj}\delta_{ik})a_{i}\partial{k}(b_{j})\hat{l}$ $\vec{a}\times (\vec{\nabla}\times\vec{b})=[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]$

Similarly,

$\vec{b}\times(\vec{\nabla}\times\vec{a})=b_{j}\hat{j}\times(\epsilon_{kin}\partial{k}(a_{i})\hat{n})=\epsilon_{jnp}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=\epsilon_{pjn}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=(\delta_{pk}\delta_{ji}-\delta_{pi}\delta_{jk}) b_{j}\partial{k}(a_{i})\hat{p}$ $\vec{b}\times(\vec{\nabla}\times\vec{a})=[b_{j}\partial{k}(a_{j})\hat{k}- b_{k}\partial{k}(a_{i})\hat{i}]$
So plugging these equations to equation 2, we finally verified the vector identity because we get:

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Vector Identities formula #6

Prove:


let:



Solution:








note:



thus;

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LOWEST-ORDER RELATIVISTIC ENERGY CORRECTION OF 1-D HARMONIC OSCILLATOR

Lotis R. Racines and Edwin B. Fabillar

In quantum mechanics, relativistic correction to the energy levels of a system is used when it is introduced by a little disturbance we often recognized as $\lambda$ . Fine structure is an example of this where the splitting of spectral lines of atoms is due to its first-order relativistic corrections. Here is an example of finding the first-order relativistic corrections of a given system.

Our task is to find the lowest-order relativistic corrections to the energy levels of the one-dimensional harmonic oscillator.

Note: Our reference through all these is Jackson’s book of Quantum Mechanics

Start:

We begin by eq’n 6.52 ,

$E_{r}^' = - \frac {1}{2mc^2} [E^2 - 2 E\langle V \rangle + \langle V^2 \rangle]$

where $E = (n + \frac {1}{2}) \hbar \omega$

and $V = \frac {1}{2} m \omega ^2 x ^2$

So,

$E_{r}^' = - \frac {1}{2mc^2} [(n + \frac{1}{2})^2 \hbar^2 \omega ^2 - 2(n + \frac{1}{2}) \hbar \omega (\frac{1}{2}m \omega ^2) \langle x^2 \rangle + \frac{1}{4} m^2 \omega ^4 \langle x^4 \rangle]$

with $\langle x^2 \rangle = (n + \frac{1}{2}) \frac{\hbar}{m \omega}$

Substituting this, we get

$E_{r}^' = - \frac{1}{2mc^2} [(n + \frac{1}{2})^2 \hbar ^2 \omega ^2 - (n + \frac{1}{2})(n + \frac{1}{2}) \frac{\hbar}{m \omega} (\hbar \omega)(m \omega ^2)$

$+ \frac{1}{2} m^2 \omega^4 \langle x^4 \rangle]$

$E_{r}^' = - \frac{1}{2mc^2} [(n + \frac{1}{2})^2 \hbar^2 \omega ^2 - (n + \frac{1}{2})^2 \hbar^2 \omega ^2 + \frac{1}{4} m^2 \omega ^4 \langle x^4 \rangle]$

$E_{r}^' = - \frac{m \omega ^4}{8c^2} \langle x^4 \rangle$ (1)

We now introduce the ladder operators. That is,

$a_{+} = \sqrt {n+1}|n \rangle$

$a_{-} = \sqrt {n} |n-1 \rangle$

Using these, we could then derive $\langle x^4 \rangle$ basing on Eq’n 2.69,

$x^4 = \frac{\hbar^2}{4m^2 \omega ^2} (a_{+}^2 + a_{+}a_{-} + a_{-}a_{+} + a_{-}^2)(a_{+}^2 + a_{+}a_{-} + a_{-}a_{+} + a_{-}^2)$

$\langle x^4 \rangle = \frac{\hbar^2}{4m^2 \omega ^2} \langle m | (a_{+}^2 a_{-}^2 + a_{+}a_{-}a_{+}a_{-} + a_{+}a_{-}a_{-}a_{+} + a_{-}a_{+}a_{+}a_{-}$

$+ a_{-}a_{+}a_{-}a_{+} + a_{-}^2 a_{+}^2) | n \rangle$

Note that only equal numbers of raising and lowering operators will survive.

By eq’n 2.66, $h(\x) = h_{even} (\x) + h_{odd} (\x)$

$\langle x^4 \rangle = \frac{\hbar ^2}{4m^2 \omega ^2}{\langle n| a_{+}^2[\sqrt {n(n-1)}|n-2\rangle] + a_{+} a_{-} \langle n|n \rangle + a_{+} a_{-} \langle (n+1)|n \rangle$

$+ a_{-} a_{+} \langle n|n \rangle + a_{-} a_{+} \langle (n+1)|n \rangle + a_{-}^2 \langle \sqrt {(n+1)(n+2)}|n+2 \rangle}$

$\langle x^4 \rangle = \frac{\hbar^2}{4m^2 \omega^2} \{\langle n|[\sqrt{n(n-1)}\sqrt{n(n-1)}|n \rangle] + n \langle n|n \rangle + (n+1) \langle n|n \rangle$