The h Bar

imbAF

00:04

is the outer product of the vacuum state with itself equal to identity ?

ACuriousMind

@imbAF Why do you think it might be? Do you know any case in linear algebra where the outer product of a vector with itself is the identity? What would that imply?

imbAF

chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/ethz.ch/content/dam/ethz/special-interest/phys/…

At page 32

In the last equation, he goes from the state expression to the ladder operators

and I know that $a^\dagger_n|0\rangle=|n\rangle$

So the expression in the end must be $a^\dagger_i|0\rangle\langle 0|a_j^\dagger$

And because I dont see the vacuum state

I asked whether the outer product gives identity matrix

ACuriousMind

And I'm asking you to think a bit about that idea: What would it mean if the outer product of a vector with itself was the identity? What do we usually use the outer product of a vector with itself for and what do we call it?

imbAF

Projector?

ACuriousMind

that's not even a sentence :P

imbAF

00:16

Usually when we consider a basis of kets, the linear combination of the outer product of some of them, is called a projector, as it projects in the sub-space spanned by them

And also, of course, completeness, if the vectors are basis kets

ACuriousMind

Yes. And what would it mean if such a projector onto a single vector was the identity? Can you exhibit any vector in $\mathbb{R}^n$ for which this is the case?

imbAF

I don't know any such vector

You are asking what are the implications of the case when the identity can be gained by a single vector, right?

And whether this type of vector exists

The projector is an identity only within the sub space.

@ACuriousMind That would imply that any other vector could be expressed via this one vector. So it would be equal to it up to some constant. That is not possible

ACuriousMind

00:38

Almost correct - it's only possible in a one-dimensional vector space.

imbAF

In $\mathbb{R}^3$

in 1D you could

by taking 1

@ACuriousMind yes

And how does this translates to physics

regarding what I asked

ACuriousMind

It means clearly the outer product of the vacuum is not the identity anywhere except in the space that consists only of multiples of the vacuum

imbAF

I mean yeah, if you were to translate it 1 to 1 from linear algebra

But if you had to incorporate things such as Hamiltonian, system, single particle operator (which I don't know what it is), eigenstates of H and ladder operators, how would that translate

space that consists only of multiples of the vaccum? What does this mean?

ACuriousMind

@imbAF I don't understand the question. You asked "is the outer product of the vacuum state with itself equal to identity?" We've found the unambiguous and clear answer to that question.

imbAF

I understood that, that it is identity for a reason that makes sense in a linear algebra way

I wanted to understand " space that consists only of multiples of the vacuum"

But maybe another time.

hmmm

ACuriousMind

00:46

It's just words for the 1-dimensional vector space spanned by the vacuum vector

imbAF

ah

ok

ACuriousMind

Since we just discussed that outer products of single vectors can only be the identity in the one-dimensional space spanned by themselves

imbAF

Yes

mathematically I understand

I don't know how to picture or comprehend the notion of 1D in the physical world

where you consider particles of some type occupying different possible eigenstates

ACuriousMind

I don't know why you're trying to. The question you asked was a pure mathematical question, and we've got a pure mathematical answer.

imbAF

Because, that question was related with the scenario where we consider a multi particle system

But ok

If it doesn't make sense to have a physical understanding and the mathematical one suffices

ok

One more thing, it's more about notation and ladder operators in 2nd quantization

Only two questions

ACuriousMind

00:50

I don't know what you're talking about. If you wanted to ask a question about physics, you shouldn't have asked a purely mathematical question about the outer product of the vacuum vector :P

imbAF

did you see the link I gave?

ACuriousMind

I'm not reading it because you need to stop expecting other people to do the work of figuring out what your actual question is

imbAF

lol

It's not about reading anything

The link, discusses 2nd quantization and in the section about single operators

when trying to express an arbitrary operator via the fock states

it does this "trick" or however you want to call it, where the outer product of the vacuum state in the next step of the equation "vanishes", which implies identity matrix

hence my question

I actually thought, that I am doing a better job by providing the source material

How is that bad?

Anyway I have two more questions about 2nd quantization, notes of which we took yesterday and briefly spoke about it

1.Does it make sense to speak about ladder operators of the eigenstates and ladder operators of states, that are not necessarily eigenstats, but perhaps linear combinations of them.

What it is meant with first here? Because even after changing the operator he still ain't first, He is between its counterpart annihilation operator and some other ladder operators

qwerty