# Unknown well-known equation in Sakurai

## Main Question or Discussion Point

Unknown "well-known" equation in Sakurai

On p391 in Modern Quantum Mechanics is stated: Now we use the well-known equation:

$$\frac{1}{E - H_0 -i \epsilon} = Pr. \left( \frac{1}{E-H_0} \right) + i \pi \delta \left(E-H_0 \right)$$

Frankly I've never heard of the equation or the notation Pr. My first thought is that the Pr. stands for Cachy principal value integral, but that's denoted $$P \int$$ or P.v. So I'm lost here...

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Your guess is correct Pr does stand for the principal value.

A. Neumaier
2019 Award

Your guess is correct Pr does stand for the principal value.
and the equation must be understood as one involving distributions rather than functions. I.e., it becomes a normal identity only after multiplying it with a nice function and then integrating it. This provides the integral found missing by the OP.

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Avodyne

Here's one way to think about it. Start with $1/(x-i\epsilon)$, and multiply the numerator and denominator by $x+i\epsilon$ to get

$${1\over x-i\epsilon}={x\over x^2+\epsilon^2} +i{\epsilon\over x^2+\epsilon^2}$$

For small $\epsilon$, the imaginary part is a narrow spike at $x=0$, and the area under it is $\pi$ (do the integral over $x$ to get this). So, as $\epsilon\to 0$, the imaginary part is becoming $i\pi\delta(x)$.

It takes a little more work to see that the real part is becoming the principal value, which I leave as an exercise.

t!m

This bothered me for quite a while too, as far too many authors claim this to be a well-known equation. Mattuck was the first to clear it up for me (in A Guide to Feynman Diagrams...) where he even sarcastically calls it "the so-called 'well-known theorem from complex function theory' ". Importantly, the written form

$$\frac{1}{x+i\epsilon} = P\frac{1}{x} - i\pi\delta(x)$$

is only shorthand for

$$\int dx \frac{f(x)}{x+i\epsilon} = P\int dx \frac{f(x)}{x} - i\pi \int dx f(x) \delta(x)$$

as pointed out by A. Neumaier.

tom.stoer

It should be noted that all these derivations should be questioned when applying the equation for operators instead of complex numbers :-) That doesn't mean that the equations necessarily become wrong; it only says that these simple derivations do not prove that they are right.

A. Neumaier
2019 Award

This bothered me for quite a while too, as far too many authors claim this to be a well-known equation. Mattuck was the first to clear it up for me (in A Guide to Feynman Diagrams...) where he even sarcastically calls it "the so-called 'well-known theorem from complex function theory' ". Importantly, the written form

$$\frac{1}{x+i\epsilon} = P\frac{1}{x} - i\pi\delta(x)$$

is only shorthand for

$$\int dx \frac{f(x)}{x+i\epsilon} = P\int dx \frac{f(x)}{x} - i\pi \int dx f(x) \delta(x)$$

as pointed out by A. Neumaier.
It is well-known, though perhaps not to everyone. The formula can be found in the Appendices of the book by Messiah, but also in the wikipedia article
http://en.wikipedia.org/wiki/Dirac_delta_function under the name ''Sokhatsky–Weierstrass theorem'' (a name I didn't know before looking it up).

A. Neumaier
2019 Award

It should be noted that all these derivations should be questioned when applying the equation for operators instead of complex numbers :-) That doesn't mean that the equations necessarily become wrong; it only says that these simple derivations do not prove that they are right.
They remain correct when H is self-adjoint. The proof is by diagonalization and reduction to the scalar case.

tom.stoer

They remain correct when H is self-adjoint. The proof is by diagonalization and reduction to the scalar case.
I know. But usually the the one-to-one translation of matrix stuff to operators requires the operators to be compact - which is rarely the case in QM

A. Neumaier
2019 Award

I know. But usually the the one-to-one translation of matrix stuff to operators requires the operators to be compact - which is rarely the case in QM
No compactness is needed.

Cauchy's integral formula is valid for analytic functions of an arbitrary self-adjoint operator, and provable by using a representation where it acts diagonally (which exists according to the spectral theorem, in one of its formulations). One gets ''our'' formula from this.

I know. But usually the the one-to-one translation of matrix stuff to operators requires the operators to be compact - which is rarely the case in QM
Tongue-in-cheek: all physical operators are compact... otherwise they're not physical!

I've heard at least one very slick lecturer state: "and this integral obviously converges because it's physical" --- former head of theoretical physics at Bell...

DarMM
Gold Member

Tongue-in-cheek: all physical operators are compact... otherwise they're not physical!
Actually an operator need only be bounded and self-adjoint to be physical. Or in truth it need only have bounded projections and be self-adjoint.

A. Neumaier
2019 Award

Actually an operator need only be bounded and self-adjoint to be physical.
Do you want to rob momentum its character of being physical?
(This is why I don't like C^* algebras as foundation....)

Or in truth it need only have bounded projections and be self-adjoint.
So the former was not in truth! But is the latter?

Self-adjoint alone is enough for being physical.

(Bounded projections is ambiguous. ''some projections are bounded'' is a nearly empty statement; ''all projections are bounded'' is far too strong, excluding momentum.)

tom.stoer

Tongue-in-cheek: all physical operators are compact... otherwise they're not physical!
No counter examples?

"and this integral obviously converges because it's physical" --- former head of theoretical physics at Bell...
I know a similar theorem that states that "numbers like 11, 13, 17 etc. are unphysical and do not exist in nature". There is one case where this theorem fails, namely the beta-function of QCD which reads

β(g) ~ g³ (11NC - 2NF) with NC=3

The factor "11" immediately proves that QCD is unphysical :-)

DarMM
Gold Member

Do you want to rob momentum its character of being physical?
(This is why I don't like C^* algebras as foundation....)

So the former was not in truth! But is the latter?

Self-adjoint alone is enough for being physical.

(Bounded projections is ambiguous. ''some projections are bounded'' is a nearly empty statement; ''all projections are bounded'' is far too strong, excluding momentum.)
Ah yes, good point! To explain more accurately, in Algebraic QFT using C*-algebras, one normally says that observables are self-adjoint bounded operators contained in a region of spacetime. So very strictly speaking AQFT would say that momentum is not an observable. However when I say very strictly, I mean very strictly. AQFT does not say that momentum is not observable, just that no complete measurement of it (enough to specify the point in the spectrum completely) is possible in a finite region. All you can observe in a finite region is operators whose eigenstates are ones whose support in momentum space is finite. The width of these regions is the resolution of the equipment and their number is determined by the highest and lowest momentum states the device can measure.

So momentum is not an observable in a very technical, strict manner. However more truthfully this is just a mathematical way of encoding that you cannot measure momentum "to a point", not that you cannot actually measure momentum.

t!m

It is well-known, though perhaps not to everyone. The formula can be found in the Appendices of the book by Messiah, but also in the wikipedia article
http://en.wikipedia.org/wiki/Dirac_delta_function under the name ''Sokhatsky–Weierstrass theorem'' (a name I didn't know before looking it up).
Ah, thanks! Now I can at least refer to it by name instead of carrying on the tradition of calling it 'a well-known theorem.'

I know a similar theorem that states that "numbers like 11, 13, 17 etc. are unphysical and do not exist in nature". There is one case where this theorem fails, namely the beta-function of QCD which reads

β(g) ~ g³ (11NC - 2NF) with NC=3

The factor "11" immediately proves that QCD is unphysical :-)
And what of the fractional quantum Hall effect? Those filling fractions are fantastically unphysical. I've never even heard of 7/15 ...

A. Neumaier
2019 Award

Ah yes, good point! To explain more accurately, in Algebraic QFT using C*-algebras, one normally says that observables are self-adjoint bounded operators contained in a region of spacetime. So very strictly speaking AQFT would say that momentum is not an observable. However when I say very strictly, I mean very strictly. AQFT does not say that momentum is not observable, just that no complete measurement of it (enough to specify the point in the spectrum completely) is possible in a finite region. All you can observe in a finite region is operators whose eigenstates are ones whose support in momentum space is finite. The width of these regions is the resolution of the equipment and their number is determined by the highest and lowest momentum states the device can measure.

So momentum is not an observable in a very technical, strict manner. However more truthfully this is just a mathematical way of encoding that you cannot measure momentum "to a point", not that you cannot actually measure momentum.
This happens to be true for momentum but has nothing to do with the problem of boundedness. One component of the electromagnetic field strength at a point x is local (and can in principle be measured arbitrarily well) but is not a bounded variable.

Whereas the projection of a momentum component to a bounded interval is bounded but cannot be measured exactly to the point (only arbitrarily well). But we'd discuss this in a new thread....

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tom.stoer

And what of the fractional quantum Hall effect? Those filling fractions are fantastically unphysical. I've never even heard of 7/15 ...
Yeah, must be wrong, too :-)

DarMM
Gold Member

This happens to be true for momentum but has nothing to do with the problem of boundedness. The electromagnetic field strength at a point x is local (and can in principle be measured arbitrarily well) but is not a bounded variable.

Whereas the projection of a momentum component to a bounded interval is bounded but cannot be measured exactly to the point (only arbitrarily well).
Perhaps I'm missing something, please correct me if I am, but the electromagnetic field strength at a point x is not an observable, since it is not an operator, it is only an operator valued distribution.

A. Neumaier