Understanding covariant and contravarient components of vector in Ket notation

Question

I learned about covariant and contravariant vectors in the context of Vector and Tensor analysis and Now I'm learning about it in the context of Linear vector spaces in Dirac ket notation. I'm having difficulty in relating the two to each other.

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We know that any vector can be expanded in term of basis $\{e_i\}$ set, then $$\vec{A}=\sum_iA^ie_i$$ To find the component of the vector, One defines the reciprocal basis $\{e^i\}$ as $$e_i\cdot e^k=\delta_{ik}$$ Now we can find the component by simply taking the dot product with reciprocal basis: $$\vec{A}\cdot e^{i}=\sum_jA^je_j\cdot e^i=A^i$$ We can also expand the same vector in term of reciprocal basis as $$\vec{A}=\sum_iA_ie^{i}$$ We call components $A^i$ contravariant while $A_i$ covarient component of vectors.

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Now in the context of Linear Vector Spaces, It says that any vector can be expanded as $$|a\rangle =\sum_i a^i|i\rangle $$ And they directly define the components $$a_i=\sum_j \bar{a}^j\langle j|i\rangle $$ the last line object should be $\langle a|i\rangle $. I don't find any relation between the two. How does the reciprocal basis look in ket notation? Can any help me relating the two?

Answer 1

Take any two vectors $|a\rangle$ and $|b\rangle$ and expand them in the basis $\{ |i \rangle \}$.

Write $$|a\rangle=\sum_i a^i |i\rangle$$ and $$|b\rangle=\sum_j b^j |j\rangle$$ where $a^i$ and $b^i$ are the contravariant components of the vectors in the stated basis.

By the definition of a bra, $$\langle b |=\sum_j \bar {b^j} \langle j|$$

Taking the inner product of $|b\rangle$ and $|a\rangle$ we get-

$$\langle b|a\rangle=\sum_j\sum_i \bar{b^j}a^i \langle j|i\rangle$$

Rearranging the summation we get:

$$\langle b|a\rangle=\sum_i\Big\{\sum_j \bar{b^j}\ \langle j|i\rangle \Big \} a^i$$

The term inside the curly brackets is known as the covariant component of vector $|b\rangle$ with respect to the basis vector $|i\rangle$ and is written as $b_i$

This thus makes the inner product much less cumbersome:

$$\langle b | a \rangle= \sum_i b_i a^i$$

without even caring about whether the basis is orthonormal. If the basis is orthonormal then $b_i=\bar{b^i}$ and you get the ubiquitious relation:

$$\langle b | a \rangle =\sum_i \bar{b^i}a^i$$

Does that solve your problem?