Continuous resolution of identity in a discrete Hilbert-space

[FredMadison]

In a Hilbert-space whose dimensionality is either finite or countably infinite, we have the discrete resolution of identity

$$\sum_n |n\rangle \langle n| = 1$$

In many cases, for example to obtain the wavefunctions of the discrete states, one employs the continuous form of the resolution of identity, namely

$$\int dx |x\rangle \langle x| = 1$$

It doesn't seem quite valid to apply a continuous operator in a discrete Hilbert space. How can one justify it?

 

[Ben Niehoff]

It's not really a Hilbert space, and $\lvert x \rangle$ is not really a vector in the Hilbert space (let alone a basis!). The norm of $\lvert x\rangle$ is undefined:

$$\int dx \, \langle x \mid x \rangle \equiv \int dx \, \Big( \delta(x) \Big)^2 = \infty$$
So $\lvert x \rangle$ is not in $L^2(\mathbb{R})$.

In order to talk about $\lvert x \rangle$ and $\lvert p \rangle$ as "basis vectors", you need a "rigged Hilbert space".

In any case, a proper basis, as you say, should be countably infinite.

 

[bhobba]

$$\int dx \, \langle x \mid x \rangle \equiv \int dx \, \Big( \delta(x) \Big)^2 = \infty$$ So $\lvert x \rangle$ is not in $L^2(\mathbb{R})$.

That's applying it naively.

In fact the Dirac delta function squared isn't even defined in the area of math its part of called distribution theory (but evidently that can be extended so it makes sense, but whether you can integrate it is another matter), but is sometimes used illegitimately for example in QFT.

Welcome to applied math

Seriously if you want to start to understand this stuff you first need to come to grips with distribution theory:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

The Rigged Hilbert spaces of QM are basically Hilbert spaces with distribution theory stitched on, adding rigging so to speak - hence the name Rigged.

But even aside from that knowledge of this area of math is of great value in many areas eg Fourier transforms are very elegant and is well worth the effort.

Thanks
Bill

 

[Jazzdude]

$$\int dx \, \langle x \mid x \rangle \equiv \int dx \, \Big( \delta(x) \Big)^2 = \infty$$
So $\lvert x \rangle$ is not in $L^2(\mathbb{R})$.

I believe there's an error in your argument here. $\langle x \mid x \rangle$ does not evaluate to $\delta^2(x)$ but to $\delta(x-x)=\delta(0)$, which is fundamentally undefined and certainly non-finite.

Cheers,

Jazz

 

[Ben Niehoff]

I believe there's an error in your argument here. $\langle x \mid x \rangle$ does not evaluate to $\delta^2(x)$ but to $\delta(x-x)=\delta(0)$, which is fundamentally undefined and certainly non-finite.

Cheers,

Jazz

You're right.

 

[bhobba]

You're right.

Yes he is, but don't feel bad - at a naive level so were you.

<x|x> = ∫<x|x'><x'|x>dx' = ∫δ(x-x')δ(x-x')dx then do a change of variable and, like I said naively, you get ∫δ(x)δ(x) dx.

It just goes to show why Von Neumann was so dead against Diracs approach, being the mathematician he was.

Thanks
Bill