Understanding the Hermitian Conjugate in Inner Products

[SebastianRM]

Hey, I am currently reading over the linear algebra section of the "introduction to quantum mechanics" by Griffiths, in the Inner product he notes: "The inner product of two vector can be written very neatly in terms of their components: <a|B>=a1* B1 + a2* B ... " He also took upon the generalisation that any scalar is a complex number.
I do not understand how is it that the components of a are conjugates of the complex number , because of the * symbol means. When he writes usual dot products he does not use * as a multiplication symbol, in the book context it has only been treated as the conjugate symbol, so I don't consider the possibility of * standing for 'multiplication'.

 

[fresh_42]

The explanations here are pretty good: https://en.wikipedia.org/wiki/Dot_product

It looks as if you have complex vector spaces in which case we don't have a symmetric bilinear form, but a sesquilinear. That is one of the components is the complex conjugate, in your case probably the first component. This depends on the author and you should be sure whether $\langle\lambda a\,|\, \mu B \rangle = \bar{\lambda}\mu \langle a\,|\,B\rangle$ or $\langle\lambda a\,|\,\mu B \rangle = {\lambda}\bar{\mu}\langle a\,|\,B\rangle$. Mathematicians and physicists tend to make it differently and I cannot memorize which one does it which way.

From what you wrote, it looks like $\vec{a}\cdot \vec{b} = \langle\vec{a}\,|\,\vec{b} \rangle = \sum_i \bar{a_i}b_i \in \mathbb{C}$, i.e. sesquilinear in the first and linear in the second factor.

If vector spaces are real, then there is no need for a conjugate.

 

[BvU]

Hi,

components of a are conjugates of the complex number

they are not. If $\ \vec a =(a_1, a_2, ...)$ and $\ \vec b = (b_1, b_2, ...)$ then the inner product is defined as $a_1^* b_1 + a_2^* b_2 + ...$

 

[SebastianRM]

Thank you guys, I think I just got confused with the notation since we have been using * for conjugates. And he normally does in the book too, thank you!

 

[fresh_42]

Thank you guys, I think I just got confused with the notation since we have been using * for conjugates. And he normally does in the book too, thank you!

I've seen $\bar{z}\; , \;z^*\; , \;z^t\; , \;z^\tau\; , \;z^\dagger$ so you should get it right which one Griffiths uses. Only the one correct notation $\iota(z)$ is never used.

 

[SebastianRM]

Hey guys, as an update, I continued reading on the Linear Algebra Apendix, as in the Matrix section he states: "The inner prodect of two vectors cab be written very neatly as a matrix product: <a|B>=(a+)b ."
The plus is intended to be a dagger, that is the Hermitian Conjugate. Which meas that indeed <a|B> = a1*b1 +a2*b2 ...; where '*' does stand for the conjugate on each element on a, which brings me to the original question, why is it that the inner product of 2 vectors with complex scalars implies conjugates.

 

[fresh_42]

I already told you that you will find many answers here: https://en.wikipedia.org/wiki/Dot_product

"For vectors with complex entries, using the given definition of the dot product would lead to quite different properties. For instance the dot product of a vector with itself would be an arbitrary complex number, and could be zero without the vector being the zero vector (such vectors are called isotropic); this in turn would have consequences for notions like length and angle. Properties such as the positive-definite norm can be salvaged at the cost of giving up the symmetric and bilinear properties of the scalar product, through the alternative definition $a \cdot b = \sum_i \bar{a_i}b_i\,$"

 

[SebastianRM]

So from what I understand, the Inner product definition for complex numbers is just a different definition from the actual dot product? Since properties are lost if the 'usual definition' of dot product

 

[fresh_42]

Yes. The reason to define such an operation is to be able to do geometry, i.e. have lengths and angles. For a length we use the Euclidean norm, i.e. Pythagoras:
$\left\| \vec{a} \right\|^2=\sum_i a_i^2$ for real vectors. We want to have a real number as a result. Now if the $a_i$ were complex, we would get a complex number as length, which is hard to interpret. However, if we take the one-sided conjugate, then we get $\left\| \vec{a} \right\|^2=\sum_i a_i^*\,a_i \in \mathbb{R}$.

 

[SebastianRM]

Thank you so much for the help!

 

[vanhees71]

I really come more and more to the conclusion that Griffiths's book on QM does more harm than good to beginning students. The explanation, why one needs a sesquilinear form for complex vector spaces to define a scalar product that induces a norm, should be given clearly. It helps a lot to understand the subject, why a definition is used!

 

[PeroK]

I really come more and more to the conclusion that Griffiths's book on QM does more harm than good to beginning students. The explanation, why one needs a sesquilinear form for complex vector spaces to define a scalar product that induces a norm, should be given clearly. It helps a lot to understand the subject, why a definition is used!

That's a a bit harsh on Griffiths. There's no obligation on a QM text to teach elementary linear algebra. And, if the student's knowldege of linear algebra is deficient, then any text on QM is going to be problematic.

A student with the same knowldege of linear algebra as the OP would be all at sea with someone like Sakurai, for example. His mathematics is pretty wild and woolly from the outset!

 

[vanhees71]

Well, yes, but I've the impression that in this forum the most fundamental problems with QT by beginning students arise when this textbook is used. I don't know the book myself, it's just an impression. I'm a bit surprised by this since Griffiths's classical electrodynamics book is obviously excellent. I also heard this book praised by many of our students, but I never heard that anybody used Griffiths's QM book. So I can't say whether they like it or not.

 

[PeroK]

Well, yes, but I've the impression that in this forum the most fundamental problems with QT by beginning students arise when this textbook is used. I don't know the book myself, it's just an impression. I'm a bit surprised by this since Griffiths's classical electrodynamics book is obviously excellent. I also heard this book praised by many of our students, but I never heard that anybody used Griffiths's QM book. So I can't say whether they like it or not.

I liked Griffiths's book. That's where I learned QM, before I moved on to Sakurai.

It is very orthodox, in the sense that I've not had to unlearn anything from Griffiths in order to conform to the PF orthodoxy!

 

[SebastianRM]

Well, yes, but I've the impression that in this forum the most fundamental problems with QT by beginning students arise when this textbook is used. I don't know the book myself, it's just an impression. I'm a bit surprised by this since Griffiths's classical electrodynamics book is obviously excellent. I also heard this book praised by many of our students, but I never heard that anybody used Griffiths' QM book. So I can't say whether they like it or not.

Personally I liked the QM book better that the Electrodynamics one. I found that in the Edynamics one, he normally skips a lot of steps during his examples or proofs, and also makes a lot of references to problems that are provided; however, one does not normally have the time to work all of them out, so it becomes a little annoying.

 

[SebastianRM]

That's a a bit harsh on Griffiths. There's no obligation on a QM text to teach elementary linear algebra. And, if the student's knowldege of linear algebra is deficient, then any text on QM is going to be problematic.

A student with the same knowledge of linear algebra as the OP would be all at sea with someone like Sakurai, for example. His mathematics is pretty wild and woolly from the outset!

I assume Griffiths was aware of this and added a whole Appendix, explaining linear algebra applied to complex numbers. However it was in this section where my question came from. As a reminder, elementary linear algebra does not involve complex numbers, which is what triggers the use of the hermitian conjugate.