Einstein summation convention in QM

[dyn]

Hi
For an operator A we have Aψ_n = a_nψ_n ; the matrix elements of the operator A are given by A_mn= a_nδ_mn
My question is : is this an abuse of Einstein summation convention or is that convention not used in QM ?
Thanks

 

[PeroK]

The Einstein convention is not a universal standard.

In general, linear algebra textbooks have not been rewritten to conform to the Einstein summation convention.

 

[dyn]

Thanks
The equation i wrote was from a modern QM textbook so it seems that i can assume the convention is not always followed in QM

 

[andresB]

Authors tend to clearly state when they are (and when they are not) using Einstein convention, generally speaking.

 

[Haborix]

If you wanted to keep the summation convention in this example, then $a_n$ would be better written as $a(n)$, a function depending on a discrete variable ranging over the dimension of the underlying Hilbert space.

 

[dyn]

Hi
For an operator A we have Aψ_n = a_nψ_n ; the matrix elements of the operator A are given by A_mn= a_nδ_mn
My question is : is this an abuse of Einstein summation convention or is that convention not used in QM ?
Thanks

Is there a way to write this equation using Einstein convention or are some equations impossible to write using that convention ?

 

[George Jones]

Hi
For an operator A we have Aψ_n = a_nψ_n ; the matrix elements of the operator A are given by A_mn= a_nδ_mn
My question is : is this an abuse of Einstein summation convention or is that convention not used in QM ?
Thanks

Is there a way to write this equation using Einstein convention or are some equations impossible to write using that convention ?

This equation does not involve a sum, so why should a summation convention be used?

In more detail: it looks like $A$ is a selfadjoint operator, and that $\left\{ \left| \psi_i \right> \right\}$ is a complete set of eigenstates of $A$. Then,
$$A_{mn} = \left< \psi_m |A| \psi_n \right> = a_n\left< \psi_m | \psi_n \right> = a_n \delta_{mn} .$$
No sums or summation convention anywhere.

 

[hilbert2]

The part of QM where the summation convention is actually needed, is the relativistic QM with Dirac equation. There you have 4-vectors with matrix-valued coefficients, and only the 4-vector components are handled with this convention, not the spinor components.

 

[dyn]

This equation does not involve a sum, so why should a summation convention be used?

In more detail: it looks like $A$ is a selfadjoint operator, and that $\left\{ \left| \psi_i \right> \right\}$ is a complete set of eigenstates of $A$. Then,
$$A_{mn} = \left< \psi_m |A| \psi_n \right> = a_n\left< \psi_m | \psi_n \right> = a_n \delta_{mn} .$$
No sums or summation convention anywhere.

If the summation convention was followed on the RHS then the index n would be summed over because it appears twice leaving the index m which should be the only index appearing on the LHS. As far as i am aware that it how it works in SR with the summation convention

 

[George Jones]

If the summation convention was followed on the RHS then the index n would be summed over because it appears twice leaving the index m which should be the only index appearing on the LHS. As far as i am aware that it how it works in SR with the summation convention

But the equation in your original post does not involve a sum!

 

[dyn]

But the equation in your original post does not involve a sum!

If an index appears twice in a term does that not imply summation ?

 

[DrClaude]

If an index appears twice in a term does that not imply summation ?

No. In non-relativistic QM implicit summations are never employed.

 

[topsquark]

If an index appears twice in a term does that not imply summation ?

If you look back to post #7, the sum has already been taken. The Kronecker delta picks out the n term of the expansion and we get $a_n \mid \psi _n >$. The two indicies just happen to have the same value after the summation.

-Dan

 

[PeroK]

If an index appears twice in a term does that not imply summation ?

No. You're going to run into problems if you assume the Einstein summation is the universal default in mathematics and physics.

I don't know where you've got that idea from.

 

[vanhees71]

I'd also not recommend to introduce the Einstein summation convention in QT in this way, because you quite often need formulae as discussed here. I think, it's useful in the manipulation of tensor components in the Ricci calculus and should be used only for this purpose, and that's indeed where it is used in the literature.

 

[malawi_glenn]

If an index appears twice in a term does that not imply summation ?

Not always

 

[jbergman]

Hi
For an operator A we have Aψ_n = a_nψ_n ; the matrix elements of the operator A are given by A_mn= a_nδ_mn
My question is : is this an abuse of Einstein summation convention or is that convention not used in QM ?
Thanks

The QM Fields book I have by Schwartz says they don't bother with upstairs and downstairs indexes and anything with the same letter is summed over.

 

[topsquark]

The QM Fields book I have by Schwartz says they don't bother with upstairs and downstairs indexes and anything with the same letter is summed over.

But this equation would create a contradiction: $A \psi _n = a_n \psi _n$. If the n's were summed on the RHS then the n's go away. Why is there still an n on the LHS? It would make no sense. Clearly there must be exceptions. Any operation with a trace involved would cause this problem.

-Dan

 

[vanhees71]

As I said, it would be very inconvenient to introduce the summation convention in cases, where it makes things more complicated. In QM I'd not introduce it at all. Of course, if you deal with tensor operators, including field operators in QFT, I'd use them for the indices for these objects but not for bras and kets.

 

[jbergman]

But this equation would create a contradiction: $A \psi _n = a_n \psi _n$. If the n's were summed on the RHS then the n's go away. Why is there still an n on the LHS? It would make no sense. Clearly there must be exceptions. Any operation with a trace involved would cause this problem.

-Dan

Yeah, maybe I misspoke. Let me look up the exact wording.