2
$\begingroup$

I am unable to comprehend the following lines given in page 657 of Shankar's Principles of Quantum mechanics:

One tricky point: The cross product is defined to be orthogonal to the vectors in the product with respect to an inner product $${\bf A} \cdot {\bf B}=\sum A_i B_i$$ and not $${\bf A} \cdot {\bf B} =\sum {A_i}^* B_i$$ even when the components of ${\bf A}$ are complex. There is no contradiction here, for the vectors ${\bf A}_1, {\bf A}_2,...,{\bf A}_n$ are fictitious objects that enter a mnemonic and not the elements of the space $\mathbb V^n(C)$ on which the operator acts.

Firstly, I am finding it hard to understand what meaning "with respect to an inner product" is adding to the first lines. Secondly, the axioms lead us to the mathematical form of the inner product ${\bf A} \cdot {\bf B} =\sum {A_i}^* B_i$(in orthonormal basis). When the vectors are defined over a real scalar field, the complex conjugates yield nothing new; real numbers stay real numbers. Here, the author has considered the components to be complex. Despite this consideration, no complex conjugates are taken in the inner product. The reason behind this is confusing. What does he mean when he tells us that the vectors are fictitious and that they do not enter the space on which the operator acts?

Could someone guide me through them by giving me a few hints or explaining what the lines mean?

$\endgroup$
  • $\begingroup$ This is why I prefer a coordinate free inner product: $ A\cdot B\equiv \frac 1 4 (||A+B||^2-||A-B||^2)$. $\endgroup$ – JEB May 12 '18 at 4:59
  • $\begingroup$ @JEB could you guide me through your comment? $\endgroup$ – R004 May 12 '18 at 10:21
  • $\begingroup$ The point is that that defines an inner product without referencing sums over components, or possible complex conjugation of them. $\endgroup$ – JEB May 12 '18 at 14:39
0
$\begingroup$
  1. At the level of formulas, Shankar is stating the fact that $$ {\bf A} \cdot ({\bf A} \times {\bf B}) ~=~0 ~=~{\bf B} \cdot ({\bf A} \times {\bf B}) $$ only holds for vectors ${\bf A}$ and ${\bf B}$ with complex-valued components if the dot product is bilinear rather than sesquilinear wrt. complex numbers $\mathbb{C}$.

  2. The notation $\mathbb{V}^n(C)$ denotes an $n$-dimensional complex vector space, cf. Def. 4 on p. 5. Its elements $|V\rangle$ are called vectors.

  3. We speculate that the sentence

    ... ${\bf A}_1, {\bf A}_2,...,{\bf A}_n$ are fictitious objects that enter a mnemonic and not the elements of the space $\mathbb{V}^n(C)$.

    try to make a distinction between a vector $|V\rangle$ and an $n$-tuple $(A_1, \ldots, A_n)$ of components.

$\endgroup$
  • $\begingroup$ He also say why--something about fictitious mnemonics. Is he just referring to the fact a cross product of 2 vectors is not really a vector, so it's OK to use a different inner product? $\endgroup$ – JEB May 12 '18 at 14:50
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic May 12 '18 at 16:03
  • $\begingroup$ The concepts bilinear and sesquilinear are new to me. I shall give them a read and try to comprehend the first point. The second is a definition. Point two tells us that the vectors are elements of a complex vector space $\mathbb V^n(C)$. From my reading, I see that Shankar is excluding the vectors from $\mathbb V^n(C)$. This now leaves me confused. I'm not sure if I am thinking it right. Lastly, is the speculation "make a distinction between $| V\rangle$ and an $n$-tuple" of any importance? The components of $| V\rangle$ can be better put in the form of a column vector. $\endgroup$ – R004 May 12 '18 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.