# B Summation vs. integral

1. Feb 8, 2017

### Isaac0427

When taking the superposition of wavefunctions with definite values of any observable (I'll just use momentum, but I am assuming it would work for any variable), I have seen the integral be used:
$\psi = \int_{-\infty}^{\infty}\phi(k)e^{ikx}dk$
and the sum be used:
$\psi = \sum_{j=0}^{\infty}c_je^{ik_jx}$
where k is the wavenumber ($\frac{p}{\hbar}$). It seems to me like these two are one-and-the-same no mater if the possible values of momentum are discrete or continuous, though the integral is used more when the allowed momenta are continuous and the sum is used when momenta are discrete. Is this correct? I know that if we were dealing with some other observable, I would have to replace the plane wave with a different function.

Last edited: Feb 8, 2017
2. Feb 8, 2017

### mikeyork

In your integral $\psi$ is a function of $k$. In your sum, $\psi$ is a function of $x$.

I suspect that in your integral you should replace $dx\rightarrow dk$.

Then the answer to your question is that in general they are not the same. But they become the same in the special case that $\phi(k) = \sum_j c_j \delta(k-k_j)$ where $\delta(.)$ is the Dirac delta function.

3. Feb 8, 2017

### SunRay-dvsh

ur 1st ψ is in continuous time and 2nd ψ is in frequency domain

4. Feb 8, 2017

### Isaac0427

The dx was a mistake. I fixed that. But, why are they not the same?

5. Feb 8, 2017

### Isaac0427

I get that my last statement was rather vague. When looking at those equations, I see that both are summations, and that k can be any value with either, continuous or discrete. What am I seeing wrong?

6. Feb 8, 2017

### Isaac0427

Now that I fixed the integral, both are functions of x, correct?

7. Feb 8, 2017

### mikeyork

You need to revisit calculus and see that you need the limit $k_{j+1} - k_j \rightarrow dk$.

8. Feb 8, 2017

### Isaac0427

Why can't that be the case here? I don't see any reason why, say, $k_5$ can't be discribed by $k_5=k_4+dk$. There are an infinite number of possible values of $k_j$, and thus $k_j$ can represent every value of k from negative infinity to infinity, even two values that are infinitely close to each other. I'm using j as an arbitrary subscript, not the value of $k_j$.

9. Feb 8, 2017

### mikeyork

$dk$ has to be infinitesimally small for the sum to become an integral. Your sum doesn't specify that. As I said, revisit calculus.

10. Feb 8, 2017

### Isaac0427

I guess that is what I don't understand. Why can't the sum represent an integral? The $k_j$s don't *have* to be infinitesimally close to one another, but they can be, right? If you set up the summation so that the $k_j$s span every possible value (including multiple values infinitesimally close to each other) from negative infinity to positive infinity, don't you essentially have an integral? In other words, you define $k_j$ so that $k_{j+1}-k_j=dk$. That would turn the summation effectively into an integral, correct?

11. Feb 8, 2017

### blue_leaf77

Example? I suspect that you encountered it when learning periodic potential. In such case the wavefunction is also a periodic function and hence can be expanded into Fourier series, but the value of $k_j$'s must follow certain pattern.
Why do you think so?

12. Feb 9, 2017

### Isaac0427

An integral is essentially an infinite sum over a continuous variable, correct? The summation is clearly an infinite sum. If the allowed values of $k_j$ were continuous, we would essentially have an integral, right?

13. Feb 9, 2017

### blue_leaf77

Yes, but that doesn't mean you can say the two forms are mathematically identical, without further use of additional notation. You know that integral is a limit of a sum when the interval approaches zero, this condition should be explicitly included in the expression. That's the formal way to write an integral as a sum as far as I know.

14. Feb 9, 2017

### Isaac0427

Speaking in a general sense, would it be correct to say that any wavefunction $\psi(x)$ can be expressed as a sum of wavefunctions with definite momenta by
$\sum_{j=0}^{\infty}c_je^{ik_jx}$
where the probability density of $\psi(x)$ having the momentum $\hbar k_j$ is $|c_j|^2$? This is in general, meaning that if the possible momenta are continuous, then the interval between two $k_j$s would approach zero.

Edit:
Fixing a few mistakes in my equations.

15. Feb 9, 2017

### mikeyork

No. As I already explained in #2, the equivalence is possible only when $\phi(x)$ is a sum of delta functions. You are now trying to suggest that you have an infinite sum of delta-functions, one for each possible value of $k$.

If you have an infinite count of $c_j$ values, how will you normalize them? If your $k_j$ intervals approach $dk$, then you will have to make each $c_j$ proportional to $dk$ and so they will vanish as $dk\rightarrow 0$ and standard calculus tells you that is how you get an integral.

But this whole discussion is silly. If you have continuous $k$ use the integral form. If you have discrete $k_j$, use the sum form.

16. Feb 10, 2017

### Staff: Mentor

No.

You need to study Rigged Hilbert Spaces.

As a buildup you should study Distribution Theory:
https://www.amazon.com/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

Then, with some background in analysis (at upper undergradute level) you can get the full technical detail - see attached document.

If you lack such knowledge then the following will supply what you need to know if you know calculus to calculus BC level:
http://matrixeditions.com/UnifiedApproach4th.html
http://matrixeditions.com/FunctionalAnalysisVol1.html

But at just 14 you simply do not have the background to attempt it. Patience grasshopper, patience - sometimes questions cant be answered until you know a lot more than a 14 yo would.

Thanks
Bill

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17. Feb 10, 2017

### Staff: Mentor

I must respectfully disagree with what you were told - this is WRONG or at the most partially correct.

Exactly what an integral is is the subject of an area called analysis that you may, just may encounter in grade 12 - but most probably at university - provided you don't wimp out and not take it because most people find it HARD.

But if you really really want to know see:
http://press.princeton.edu/titles/8221.html

But you should know calculus at an intuitive level first. Fortunately that is easily done - see:
https://www.amazon.com/Quick-Calculus-Self-Teaching-Guide-2nd/dp/0471827223

If you undertake the journey at 14 you are to be congratulated. Most doing college math try to avoid it like the plague. I did a math degree and it was a required subject, but they removed it because too many people complained. A pity - it's absolutely essential to answer questions like you posed - and much more advanced ones like the links I gave in my previous answer.

Thanks
Bill

Last edited by a moderator: May 8, 2017
18. Feb 10, 2017

### Staff: Mentor

No - its much more complicated than that - but way beyond a 14yo. I have given the links for the correct answer - take a look - don't be discouraged if its gobbbly-goook at your level - that's only to be expected.

Thanks
Bill

19. Feb 11, 2017

### kith

In standard calculus, $dx$ is not a number. The integral is only a shorthand notation for a certain limiting process. Its definition is something like
$$\int_a^b f(x) dx := \lim_{n \rightarrow \infty} \sum_{i=1}^n f(x_i) \Delta x$$

People like Leibniz who were involved with the creation of the calculus thought of $dx$ as an "infinitely small" number. It is important to note that this kind of thinking has left no trace in rigorous standard calculus and $dx$ is always used as a shorthand notation and never as a number.

(More recently, the idea of the "infinitely small" has been made rigorous in the small field of non-standard analysis. I think @micromass has written an insight about this but I couldn't find it right away.)

That's it as far as mathematics is concerned. In physics, $dx$ is used more informally and for example in geometric derivations as if it was a number. But physicists still wouldn't say that the integral is "the same" as a sum.

Last edited: Feb 11, 2017
20. Feb 11, 2017

### Staff: Mentor

You are of course correct.

But IMHO the best way is what you find in an analysis textbook that sneakily avoids the idea of limits - but of course you can do it that way if you like. It also hones in on the key property of the real number system - the LUB axiom (also known as completeness - but that is just by the by). The axioms of the rationals is exactly the same as the axioms of real numbers - to the OP look them up. Except, and this is of crucial importance, the completeness axiom - that only applies to the reals. It is what resolves Zeno's paradox and is a very interesting thing to point out to philosophers that argue about this - I just love 'poking' philosophers a little bit - although I usually come off second best - but not on this one - it is clear cut.

One divides the interval to be integrated over into a partition ie a number of sub-intervals. You take the minimum of the function in any sub-interval and multiply it by the length of that particular sub-interval. Sum it up. Call it the lower sum. Do the same with the maximum value in the partition and call it the upper sum. Now lower sum <= upper sum and will always remain so regardless of partition size ie the number of intervals in the partition.

If the partition is made finer say by dividing each sub-interval in half all the time - but any procedure that makes the sub-intervals smaller will work - we have an increasing sequence of lower sums Ln bounded above by any upper sum. We also have a decreasing sequence of upper sums Un bounded below by any lower sum. Ln is an increasing sequence and one of the defining properties of the real numbers is the least upper bound property. This says any sequence of increasing real numbers bounded above has a least upper bound. Thus Ln is bounded above hence has a LUB I will call Rl. There is a simple corollary to the LUB axiom, the greatest lower bound property, that says any sequence of decreasing real numbers that is bounded below has a greatest lower bound. So Un has a greatest lower bound I will call Ru. Now Rl <= Ru. However if Ru = Rl then the function is said to be Riemann integrable with Riemann integral Ru=Rl.

Another advantage of this method is there are other definitions of the integral eg the Lebesgue integral that is defined a bit differently - but defining it the way above allows you to clearly see the difference between the two and to show, in the case of Lebesgue integrals that if a function is Riemann integrable it is Lebesgue integrable.

Why are Lebesgue integrals important - well that's another thread but its intimately tied up with the Hilbert space of QM and we have the rather important and very interesting Fubini theorem that does not apply to Riemann integrals:
https://en.wikipedia.org/wiki/Fubini's_theorem

Physicists, and applied mathematicians in general, often exchange integral signs with gay abandon. You cant do that for Riemann Integrals, but you can do that for Lebesgue integrals. Forgetting this can on occasion can lead to problems.

In fact that's why analysis is so important - if you aren't careful - and in analysis you are very careful - you will sometimes come unstuck.

Thanks
Bill

Last edited: Feb 11, 2017