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Quantum Science Philippines

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 nth 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)
and
(b)

Solution:
Let

and

(a)

since

and

Thus,

and

(b)

since

and

Therefore,

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,

 \vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a}) 

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:

 \vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a}) 

 

 

 

 

 

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Vector Identities formula #6

Prove:
\vec{\nabla}\times(\vec{\nabla}\times\vec{a})=\vec{\nabla}(\vec{\nabla}\cdot\vec{a})-\nabla^2\vec{a}

let:
\vec{\nabla}=\partial_l\widehat{e_l}
\vec{a}=a_i\widehat{e_i}

Solution:
=\vec{\nabla}\times(\vec{\nabla}\times\vec{a})
=\partial_l\widehat{e_l}\times[\partial_l\widehat{e_l}\times a_i\widehat{e_i}]
=\partial_l\widehat{e_l}\times[\partial_la_i\in_{lij}(\widehat{e_l}\times\widehat{e_i})]
=\partial_l\widehat{e_l}\times[\partial_l a_i\in_{lij}\widehat{e_j}]
=\partial_l\partial_la_i\in_{lji}\in_{ljn}(\widehat{e_l}\times\widehat{e_j})
=\partial_l\partial_la_i\in_{jil}\in_{jln}\widehat{e_n}
=\partial_l\partial_la_i\delta_{il}\delta_{ln}\widehat{e_n}-\partial_l\partial_la_i\delta_{in}\delta_{ll}\widehat{e_n}

note:
\delta_{ll}=1
\delta_{il}=1,i=l

thus;
=\partial_l\partial_la_i\widehat{e_l}-\partial_l\partial_la_i\widehat{e_i}
=\partial_ia_i\partial_l\widehat{e_l}-\partial_l\partial_la_i\widehat{e_i}
=(\vec{\nabla}\cdot\vec{a})\vec{\nabla}-(\vec{\nabla}\cdot\vec{\nabla})\vec{a}
=\vec{\nabla}(\vec{\nabla}\cdot\vec{a})-\nabla^2\vec{a}



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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 [eq] \lambda [/eq]. 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 ,

[eq] E_{r}^’ = – \frac {1}{2mc^2} [E^2 – 2 E\langle V \rangle + \langle V^2 \rangle] [/eq]

where [eq] E = (n + \frac {1}{2}) \hbar \omega [/eq]

and [eq] V = \frac {1}{2} m \omega ^2 x ^2 [/eq]

So,

[eq] 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] [/eq]

with [eq] \langle x^2 \rangle = (n + \frac{1}{2}) \frac{\hbar}{m \omega} [/eq]

Substituting this, we get

[eq] 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) [/eq]

[eq]+ \frac{1}{2} m^2 \omega^4 \langle x^4 \rangle] [/eq]

[eq] 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] [/eq]

[eq] E_{r}^’ = – \frac{m \omega ^4}{8c^2} \langle x^4 \rangle [/eq]                    (1)

We now introduce the ladder operators. That is,

[eq] a_{+} = \sqrt {n+1}|n \rangle [/eq]

[eq] a_{-} = \sqrt {n} |n-1 \rangle [/eq]

Using these, we could then derive [eq] \langle x^4 \rangle[/eq] basing on Eq’n 2.69,

[eq] 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)[/eq]

[eq] \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_{-} [/eq]

[eq] + a_{-}a_{+}a_{-}a_{+} + a_{-}^2 a_{+}^2) | n \rangle [/eq]

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

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

[eq] \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 [/eq]

[eq] + a_{-} a_{+} \langle n|n \rangle + a_{-} a_{+} \langle (n+1)|n \rangle + a_{-}^2 \langle \sqrt {(n+1)(n+2)}|n+2 \rangle} [/eq]

[eq] \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 [/eq]

[eq] + n \langle (n+1)|n \rangle + (n+1) \langle (n+1)|n \rangle + \sqrt{(n+1)(n+2)} \langle \sqrt {(n+1)(n+2)}|n \rangle \} [/eq]

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

[eq] \langle x^4 \rangle = (\frac{\hbar}{2m \omega})^2 [n^{2} – n + n^2 + n^2 + n + n^2 + n + n^2 + 2n + 1 + n^2 + 3n +2] [/eq]

[eq] \langle x^4 \rangle = (\frac{\hbar}{2m \omega})^2 [6n^2 + 6n + 3] [/eq]

Going back to (1) to get [eq] E_{r}^’ [/eq] ,

[eq] E_{r}^’= – \frac{m \omega ^4}{8c^2} (\frac{\hbar ^2}{4 m^2 \omega ^2}) (3)(3n^2 + 2n + 1) [/eq]

Thus, the lowest-order relativistic correction of one-dimensional harmonic oscillator is

[eq] E_{r}^’ = – \frac{3}{32} (\frac {\omega ^2 \hbar ^2}{mc^2}) (3n^2 + 2n +1) [/eq]

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