# Going from Kronecker deltas to Dirac deltas

• A
• MaestroBach
MaestroBach
TL;DR Summary
Reading the intro to a book on non equilibrium green's functions, where it defines the orthonormality relation of two kets in the continuum formulation. Not sure how to understand the dirac deltas.
I'm reading Stefanucci's Nonequilibrium Many Body Theory of Quantum Systems.

In the first chapter, where it goes over basic quantum mechanics, it first defines the usual orthonormality condition I'm familiar with,

$$\langle n' | n \rangle = \delta_{n, n'}$$

where $$| n \rangle$$ is the ket describing a particle in the interval ##x_n \pm \Delta/2##

However, it then goes on to say that in the continuum formulation, the orthonormality relation becomes:

$$\langle n' | n \rangle = \lim_{\Delta\to0}\frac{\delta_{n, n'}}{\Delta} = \delta(x_{n'}-x_n)$$

I am not sure how to think about this, specifically in the case where ## x_{n'} = x_n ##

Obviously in all other cases, then the kronecker delta i'm used to is equivalent to the dirac delta, but in that one case, the book seems to be implying that instead of = 1, ##\langle n' | n \rangle = \infty##?

Kronecker's delta is for discrete physical quantity. Dirac's delta is for continuous physical quantity, e.g. coordinate x and momentum p.
$$<x'|x^">=\delta(x'-x^")$$
$$<x'|x'>=\delta(0)$$
infinity, so eigen vector of coordinate x whose eigenvalue is x’ cannot be normalized.

Last edited:
anuttarasammyak said:
Kronecker's delta is for discrete physical quantity and Dirac's delta is for continuous physical quantity, e.g. coordinate x and momentum p.
$$<x'|x^">=\delta(x'-x^")$$
$$<x'|x'>=\delta(0)$$
infinity, so eigen vector of coordinate x whose eigenvalue is x’ cannot be normalized.
I see...

Later in the same chapter, for many particles, this is given:

$$\psi(x) | y_1 ... y_N \rangle = \sum{k=1}^{N}(\pm)^{N + k} \delta(x - y_k) | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle$$,

where ##\psi(x)## is the annihilation operator and x is another coordinate that CAN equal y (To be more specific, ##y = r \sigma##). Does this imply that if ## x = y_k##, then the entire state essentially blows up when acted on by ##\psi(x)##? I guess I'm just confused because this seems different from the behavior that an annihilation operator would have in a discrete context.

My apologies if I'm not giving enough context here.

$$\psi(x) | y_1 ... y_N \rangle = \sum_{k=1}^{N}(\pm)^{N + k} \delta(x - y_k) | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle$$
Applying ##\int dx##
$$\int \ dx\ \psi(x) | y_1 ... y_N \rangle = \sum_{k=1}^{N}(\pm)^{N + k} | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle$$
It seems to make sense.

Mordred
I guess the issue was just that I was not aware that I should be applying ##\int dx## at all, the book wasn't clear about this (as far as I know). Are you doing it soley because of the dirac delta? Or for continuous quantities, is it implied that expressions are to be integrated over?

Also, thank you

You can learn about delta function in standard QM textbooks. How about reading other textbooks than yours to find its idea thoroughly.

MaestroBach said:
Are you doing it soley because of the dirac delta?
It’s more the other way around. The inner product of two continuous functions ##f## and ##g## is defined to be ##\int f(x)g(x)dx## so we’ll always be integrating. The delta functions are how we shoehorn the zero-width peak into that formalism.

There’s a good explanation of continuous-spectrum observables here although not all in one place - Cresser takes one step forward with the discrete problem, then covers the generally non-trivial modifications required for the continuous problem, then repeats for the next step. So you may want to skim through, reading in depth only when you hit the sections covering continuous observables. I expect that you’ll find it to be very helpful for this problem.

MaestroBach
Nugatory said:
It’s more the other way around. The inner product of two continuous functions ##f## and ##g## is defined to be ##\int f(x)g(x)dx## so we’ll always be integrating. The delta functions are how we shoehorn the zero-width peak into that formalism.

There’s a good explanation of continuous-spectrum observables here although not all in one place - Cresser takes one step forward with the discrete problem, then covers the generally non-trivial modifications required for the continuous problem, then repeats for the next step. So you may want to skim through, reading in depth only when you hit the sections covering continuous observables. I expect that you’ll find it to be very helpful for this problem.
I do understand that with an inner product we'll always be integrating, but I guess seeing I was only looking at a ket, I didn't think of the integration at all. Wouldn't the integration happen after doing the inner product with a bra? Is there no way to think about the ket multiplied by the dirac delta pre-inner product?

Or am I just not understanding something correctly?

MaestroBach said:
Wouldn't the integration happen after doing the inner product with a bra?
No. With a continuous spectrum, the inner product is the integral.

MaestroBach
MaestroBach said:
I do understand that with an inner product we'll always be integrating
With a discrete spectrum you do a sum, not an integral. (Yes, an integral is the continuous counterpart to a sum. But you seem to be drawing a distinction between the two.)

MaestroBach said:
Are you doing it soley because of the dirac delta? Or for continuous quantities, is it implied that expressions are to be integrated over?
Both.

The Dirac delta function is only well-defined when it appears in an integral.

And, when you're doing an inner product with a continuous spectrum, you have to do an integral anyway.

PeterDonis said:
No. With a continuous spectrum, the inner product is the integral.
Ah yes, of course- I don't know what I was thinking for a second.

I guess where I'm still stuck is that up to now, if I were given just a ket state, I'd be able to look at it and draw some kind of information from just it, without having to do the inner product of it and a bra or integrating.

With the ket above i have with the dirac delta, this seems impossible because the dirac delta isn't well defined when not appearing in an integral. The user anuttarasammyak has a response where they integrate just the ket and then note that the ket seems to make sense. Is that the general way to consider kets with continuous observables?

Nugatory said:
It’s more the other way around. The inner product of two continuous functions ##f## and ##g## is defined to be ##\int f(x)g(x)dx## so we’ll always be integrating. The delta functions are how we shoehorn the zero-width peak into that formalism.

There’s a good explanation of continuous-spectrum observables here although not all in one place - Cresser takes one step forward with the discrete problem, then covers the generally non-trivial modifications required for the continuous problem, then repeats for the next step. So you may want to skim through, reading in depth only when you hit the sections covering continuous observables. I expect that you’ll find it to be very helpful for this problem.
Thank you very much!

MaestroBach said:
The user @anuttarasammyak has a response where they integrate just the ket and then note that the ket seems to make sense. Is that the general way to consider kets with continuous observables?
Integrating the ket doesn't make a lot of sense formally, but in practice it is often a convenient way of seeing what will happen when we close it with a bra - that's the case here. If you encounter a mental speedbump doing that you could think about evaluating something like (##J## arbitrary) $$\langle y_J|\psi(x) | y_1 ... y_N \rangle$$ where it's clear that you are calculating an inner product so will be integrating and the role of the delta function in the ket will be apparent.

Last edited:
MaestroBach
MaestroBach said:
up to now, if I were given just a ket state, I'd be able to look at it and draw some kind of information from just it
You can.

MaestroBach said:
With the ket above i have with the dirac delta
The delta function is not part of the ket. It is part of one particular representation of the ket. The ket exists independently of any representation, and you can infer representation-independent facts about the ket without having to use any particular representation. For example, you can say that the ket ##\ket{x}## represents a particle whose position is exactly ##x## with probability ##1##, without having to look at any delta functions or do any integrals.

MaestroBach
PeterDonis said:
You can.

The delta function is not part of the ket. It is part of one particular representation of the ket. The ket exists independently of any representation, and you can infer representation-independent facts about the ket without having to use any particular representation. For example, you can say that the ket ##\ket{x}## represents a particle whose position is exactly ##x## with probability ##1##, without having to look at any delta functions or do any integrals.
Nugatory said:
Integrating the ket doesn't make a lot of sense formally, but in practice it is often a convenient way of seeing what will happen when we close it with a bra - that's the case here. If you encounter a mental speedbump doing that you could think about evaluating something like (##J## arbitrary) $$\langle y_J|\psi(x) | y_1 ... y_N \rangle$$ where it's clear that you are calculating an inner product so will be integrating and the role of the delta function in the ket will be apparent.
Thank you both!

• Quantum Physics
Replies
8
Views
2K
• Quantum Physics
Replies
6
Views
1K
• Quantum Physics
Replies
2
Views
299
• Quantum Physics
Replies
7
Views
978
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
8
Views
1K
• Quantum Physics
Replies
14
Views
4K
• Quantum Physics
Replies
7
Views
1K