Resolution of the identity

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Homework Statement


For the space Lw2(a,b), we can write the basis in a discrete fashion as {en|n∈ℤ} or in a continuous fashion as |x> (as we would in quantum mechanics for the position representation), such that we may write the identity operator as either
I=∑n|en><en|
or
I=∫abdxw(x)|x><x|

Show that these are identity operators, and hence show that δ(x-x0)/√[w(x)w(x0)]=∑nen(x)en*(x0).

The Attempt at a Solution


I have done the first bit. For the discrete form,
<f|I|g>=<f|∑n|en><en|g>
=<f|∑ngn|en>
where gn is the coordinate of |g> with respect to the orthonormal basis vector |en>
=<f|g>
For the continuous form,
<f|I|g>=<f|∫abdxw(x)|x><x|g>
=<f|∫abdxg(x)w(x)|x>
as <x|g>=g(x) (note the |g>=∫abg(x)w(x)|x>dx in this representation)
=<f|g>

I'm struggling with the second bit. The RHS makes me think I need to consider <x|x0> and write
<x|x0>=<x|I|x0>
and then use the discrete form of I so that
=∑n<x|en><en|x0>
=∑nen(x)en*(x0)
which sorts out the RHS. This should equal <x|I|x0> with the continuous identity inserted, but I get
<x|x0>=<x|∫abdx'w(x')|x'><x'|x0>
=<x|∫abdx'w(x')|x'>δ(x'-x0)
using the continuous orthonormality condition so <x|x'>=δ(x-x')
=<x|w(x)|x0>
because of the delta function property, so then I obtain
=w(x)δ(x-x0)
again using orthonormality. So I have
w(x)δ(x-x0)=∑nen(x)en*(x0)
so my LHS has failed somewhere. Can anyone help, thanks :)
 

Answers and Replies

  • #2
Fredrik
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I think that when you define ##|x\rangle## so that ##I=\int \mathrm d x\, w(x)|x\rangle\langle x|## (with ##w\neq 1##) rather than ##I=\int \mathrm dx\, |x\rangle\langle x|##, it's simply not true that ##\langle x|x'\rangle=\delta(x-x')##.
 
  • #3
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I think that when you define ##|x\rangle## so that ##I=\int \mathrm d x\, w(x)|x\rangle\langle x|## (with ##w\neq 1##) rather than ##I=\int \mathrm dx\, |x\rangle\langle x|##, it's simply not true that ##\langle x|x'\rangle=\delta(x-x')##.
Oh I see. However how would I go about modifying this orthonormality condition - I can't think how it would work :/

Thinking about this question confuses me a lot - how can we say that the space has a discrete basis (which presumably implies finite dimensionality) and a continuous basis (implying infinite dimensionality) when obviously the dimensionality is fixed regardless of basis.
 
  • #4
Fredrik
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Oh I see. However how would I go about modifying this orthonormality condition - I can't think how it would work :/

Thinking about this question confuses me a lot - how can we say that the space has a discrete basis (which presumably implies finite dimensionality) and a continuous basis (implying infinite dimensionality) when obviously the dimensionality is fixed regardless of basis.
The set of square-integrable functions is an infinite-dimensional vector space, with a semi-inner product. It has a countable basis. However, the relationship between a square-integrable function ##f## and its Fourier transform ##\tilde f##,
$$\tilde f(p)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} f(x)e^{-ipx}\mathrm dx,\qquad f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \tilde f(p)e^{ipx}\mathrm dp,$$ is a reason to think of the set ##\{u_p|p\in\mathbb R\}##, where each ##u_p## is defined by ##u_p(x)=\frac{1}{\sqrt{2\pi}}e^{ipx}## for all ##x\in\mathbb R##, as a kind of "basis" for the space, even though none of its elements is actually in the space. So apparently we are now considering a larger vector space, that includes not only the square-integrable functions, but also things like momentum eigenfunctions.

We can also consider things like ##\delta(x-x')## (position "eigenfunctions"...which aren't even functions). Now things are getting pretty complicated, and it's hard to see how it can even make sense. To see how to make sense of it all, you would have to study rigged Hilbert spaces, but that shouldn't be necessary to understand QM at this level.
 
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  • #5
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The set of square-integrable functions is an infinite-dimensional vector space, with a semi-inner product. It has a countable basis. However, the relationship between a square-integrable function ##f## and its Fourier transform ##\tilde f##,
$$\tilde f(p)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} f(x)e^{-ipx}\mathrm dx,\qquad f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \tilde f(p)e^{ipx}\mathrm dp,$$ is a reason to think of the set ##\{u_p|p\in\mathbb R\}##, where each ##u_p## is defined by ##u_p(x)=\frac{1}{\sqrt{2\pi}}e^{ipx}## for all ##x\in\mathbb R##, as a kind of "basis" for the space, even though none of its elements is actually in the space. So apparently we are now considering a larger vector space, that includes not only the square-integrable functions, but also things like momentum eigenfunctions.

We can also consider things like ##\delta(x-x')## (position "eigenfunctions"...which aren't even functions). Now things are getting pretty complicated, and it's hard to see how it can even make sense. To see how to make sense of it all, you would have to study rigged Hilbert spaces, but that shouldn't be necessary to understand QM at this level.
Thanks for explaining that :)

With regards to the original problem then, I have managed to work out by going backwards from the answer that
<x|x'>=δ(x-x')/√[w(x)w(x')]
would be an orthonormality condition that would make things work. However, if this is correct, I don't know how I would be able to know this for myself at all, as it doesn't seem very obvious. Would you be able to explain this to me? Thankyou.

Edit: According to my understanding, for unit weight, we did (using the fact that the eigenfunctions are delta functions)
<x|x0>=∫-∞dx'δ(x'-x)δ(x'-x0)
=δ(x-x0).
So now, we should have
<x|x0>=∫abdx'δ(x'-x)δ(x'-x0)w(x')
=w(x)δ(x-x0)
which doesn't make this problem work.
 
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  • #6
Fredrik
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In the problem statement, you're saying "show that these are identity operators". So it seems that we're supposed to show in particular that ##\int \mathrm dx\, w(x)|x\rangle\langle x|=I##. But how can we, unless the function ##w## is given in the problem? If we don't know what ##w## is, then we don't know what operator we're supposed to show is equal to the identity. So are we supposed to show that there exists a ##w## such that ##\int \mathrm dx\, w(x)|x\rangle\langle x|=I##? Have you quoted the problem exactly as it was given to you, or are you paraphrasing?

Edit: And how is ##|x\rangle## defined? An orthonormality condition is usually part of the definition.

I'm going to move this thread to the advanced physics homework forum, because we're doing non-rigorous "physicist's math" here.
 
  • #7
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In the problem statement, you're saying "show that these are identity operators". So it seems that we're supposed to show in particular that ##\int \mathrm dx\, w(x)|x\rangle\langle x|=I##. But how can we, unless the function ##w## is given in the problem? If we don't know what ##w## is, then we don't know what operator we're supposed to show is equal to the identity. So are we supposed to show that there exists a ##w## such that ##\int \mathrm dx\, w(x)|x\rangle\langle x|=I##? Have you quoted the problem exactly as it was given to you, or are you paraphrasing?

Edit: And how is ##|x\rangle## defined? An orthonormality condition is usually part of the definition.

I'm going to move this thread to the advanced physics homework forum, because we're doing non-rigorous "physicist's math" here.
The question follows on from a section of notes, so I have paraphrased to an extent otherwise it would sound a bit weird. So the actual wording is

Let |en> with n∈ℤ be a (discrete) orthonormal basis for the space Lw2(a,b). Explain why the identity operator can be expressed as
I=∑n|en><en|
and also, applying the formal notation introduced in this subsection, as
I=∫abdxw(x)|x><x|.
Hence show that δ(x-x0)/√[w(x)w(x0)]=∑n<x|en><en|x0>
=∑nen(x)en*(x0).

When it says 'formal notation introduced in this subsection', well in the section, first
f(x0)=<x0|f>
=∫-∞dxδ(x-x0)f(x)
was written, and then it was proven that the identity can be written as
I=∫-∞dx|x><x|.
 
  • #8
Fredrik
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I still can't make sense of it. For a statement that involves a variable to make sense, the variable must be assigned a value, or be the target of a "for all" or "there exists". The ##w## in the problem appears to be neither of those things. Sure, we can say that there's a ket ##|w\rangle## such that ##w(x)=\langle x|w\rangle## for all ##x##, but that just moves the problem from ##w## to ##|w\rangle##. Maybe the idea is to prove that the equality in the problem holds for all ##w## such that ##I=\int_{-\infty}^{+\infty}\mathrm dx\, w(x)|x\rangle\langle x|##. (But is there really more than one such ##w##?)

I'm very curious about how your book "proves" that ## I=\int_{-\infty}^{+\infty}\mathrm dx\, |x\rangle\langle x|##. Maybe that proof could be a clue about what you're supposed to do (or maybe not...this result seems to contradict the problem). What book is it?
 
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  • #9
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I still can't make sense of it. For a statement that involves a variable to make sense, the variable must be assigned a value, or be the target of a "for all" or "there exists". The ##w## in the problem appears to be neither of those things. Sure, we can say that there's a ket ##|w\rangle## such that ##w(x)=\langle x|w\rangle## for all ##x##, but that just moves the problem from ##w## to ##|w\rangle##. Maybe the idea is to prove that the equality in the problem holds for all ##w## such that ##I=\int_{-\infty}^{+\infty}\mathrm dx\, w(x)|x\rangle\langle x|##. (But is there really more than one such ##w##?)

I'm very curious about how your book "proves" that ## I=\int_{-\infty}^{+\infty}\mathrm dx\, |x\rangle\langle x|##. Maybe that proof could be a clue about what you're supposed to do (or maybe not...this result seems to contradict the problem). What book is it?
Hmm I've pretty much given up on it now.

The proof is as follows:
<f|I|g>=<f|(∫-∞dx|x><x|)|g>
=∫-∞dx<f|x><x|g>
=∫-∞dxf*(x)g(x)
=<f|g>
if that helps.

It's not actually a book, just some notes written designed for a maths course that contains a few exercises...

If the above is of no help, then thanks for your time anyway :)
 
  • #10
Fredrik
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Hmm I've pretty much given up on it now.

The proof is as follows:
<f|I|g>=<f|(∫-∞dx|x><x|)|g>
=∫-∞dx<f|x><x|g>
=∫-∞dxf*(x)g(x)
=<f|g>
if that helps.

It's not actually a book, just some notes written designed for a maths course that contains a few exercises...

If the above is of no help, then thanks for your time anyway :)
A math course doing super-non-rigorous math. :confused: I didn't expect that. I assumed that this was homework for a QM course based on Sakurai or a similar book.

The calculation above looks very strange as it's written. The first equality is essentially the result that the calculation is supposed to prove, and the third line is just showing what the second line would look like when we use the ##f(x)=\langle x|f\rangle## notation. Things make more sense if we write these results like this:
$$\langle f|I|g\rangle =\langle f|g\rangle =\int\mathrm dx\, \langle f|x\rangle\langle x|g\rangle =\langle f|\left(\int\mathrm dx\, |x\rangle\langle x|\right)|g\rangle.$$ Now the first equality follows from the definition of the identity operator. The second should be viewed as a definition of ##\langle f|g\rangle## (which is a notation for something similar to an inner product of ##|f\rangle## and ##|g\rangle##. The third is something that we'd expect to hold, but can't prove rigorously without an adequate definition of kets, bras and the integral. If these equalities hold for all ##|f\rangle,|g\rangle##, then we should have ##I=\int\mathrm dx\, |x\rangle\langle x|##. We would certainly be able to conclude this if we could be sure that ##|f\rangle##, ##|g\rangle## and ##\left(\int\mathrm dx\, |x\rangle\langle x| \right)|g\rangle## have well-defined norms.

Even when we write things this way, and explain it as I did, I find this argument kind of bizarre. I think it's more standard to take the existence of an operation that ensures that ##\langle f|g\rangle## is a complex number, and the result ##I=\int\mathrm dx\,|x\rangle\langle x|##, as axioms, and then use them, and the definition of the ##f(x)## notation, to show that ##\langle f|g\rangle## can be written as ##\int\mathrm dx\, f^*(x)g(x)##. Maybe this is what your text is trying to do? The calculation would be written almost as you wrote yours. I would just remove the =<f|g> at the end and put <f|g>= at the start instead.

Unfortunately I still can't make sense of the problem statement, and in particular that w.
 
  • #11
vela
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Oh I see. However how would I go about modifying this orthonormality condition - I can't think how it would work :/
If you consider
$$\lvert x_0 \rangle = I \lvert x_0 \rangle = \int dx\, w(x) \lvert x \rangle\langle x \vert x_0 \rangle,$$ it appears you need ##\delta(x-x_0) = w(x) \langle x \vert x_0 \rangle##.
 

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