I have just started QM and one thing that keeps bugging me is that whenever we have a continuous summation we take it as an integral (like in the formula below)... So why can we do this why summation of a function f(x) which is [f(x₁) +f(x₂)+f(x₃)...] equals to integral which is the area under f(x)...

$$\langle f | g\rangle=\int_{a}^{b} f(x) \overline{g(x)} w(x)$$

This formula is the redefinition of inner product in an infinite basis while in a finite basis inner product is defined as ∑f(x)g(x)* so why in an infinite basis we take it as an integral.... I was told that in infinite basis we have continuous functions which I understand however why for continuous functions we have to take summation as integrals...the book I refer says this is so because for infinite summation the sum goes to infinity so the next logical approach is integration...and that's my problem why is it the next logical approach? Sure it doesn't give infinity and gives us a good nice answer...but why can we substitute integration for summation when the two are very different operations mathematically.

  • 3
    $\begingroup$ Are you familiar with how integration is derived from Riemann sums? $\endgroup$ Jul 16, 2019 at 11:08
  • $\begingroup$ yes i am but in reinmann summs or integral there is an extra element ie the dx which makes it summation times dx hence giving area $\endgroup$
    – kay
    Jul 16, 2019 at 11:09
  • $\begingroup$ Related / duplicate question by the same author: physics.stackexchange.com/q/491876 $\endgroup$
    – Noiralef
    Jul 16, 2019 at 11:11
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    $\begingroup$ Actually, the argument is usually that sums are hard to compute. Tehre are only a few series whose solution is known with a simple formula. On the other hand, integrals are easily computable, because we have the table of derivatives next to us (or in our heads). $\endgroup$
    – FGSUZ
    Jul 16, 2019 at 11:20
  • $\begingroup$ yeah i have heard that...but just because integratiin is easier we cannot substitute it for summation unless its a proper approximation..so how is it a proper approximation? $\endgroup$
    – kay
    Jul 16, 2019 at 12:09

2 Answers 2


Suppose you have two states in a finite orthonormal (i.e. orthogonal and normalized) basis, $|a\rangle = \sum_nc_n|x_n\rangle$ and $|b\rangle = \sum_n k_n|x_n\rangle$, where $n$ ranges over some finite set. Computing the inner product of these states is straightforward: it's just $\langle a|b\rangle=\sum_n\sum_m c_n^*k_m\langle x_n|x_m\rangle$. Since the basis is orthonormal, we know that $\langle x_n|x_m\rangle=\delta_{mn}$, where $\delta_{mn}$ is the Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta), so we can simplify:

$$\langle a|b\rangle = \sum_m\sum_n c_n^*k_m\delta_{mn}=\sum_n c_n^*k_n$$

Now, suppose we want to construct a similar kind of basis for position space (in one dimension). We'll start by dividing up the $x$ axis into a countable number of intervals of length $\Delta x$, choosing as a set of endpoints $x_n=n\Delta x$, where $n\in\mathbb{Z}$. Once we have a set of intervals, we can construct a set of states that are uniformly distributed over each interval (very much like the rectangles usually employed in Riemann summation). Since each such state occupies the interval $[x_m,x_m+\Delta x)$ for some $m$, we can label them by that interval, constructing the set of states $|[x_m,x_m+\Delta x)\rangle$.

To make this set of states an orthonormal basis, we require two things:

  • Each state must be normalized: in other words, $\langle [x_m,x_m+\Delta x)|[x_m,x_m+\Delta x)\rangle=1$ for any $m$. This means that the total probability of finding each state in its respective interval must always be 1.
  • The states must be orthogonal: in other words, $\langle [x_m,x_m+\Delta x)|[x_n,x_n+\Delta x)\rangle=0$ for any $m\neq n$. This means that the probability of finding each state anywhere else besides its interval must be 0.

We can now specify a position state in this basis as

$$|f\rangle = \sum_n f_n |[x_n,x_n+\Delta x)\rangle$$

For convenience, instead of labeling the coefficients $f_n$, let's instead make them the outputs of a discrete-domain function $f:\{x_n\}_{n\in\mathbb{Z}}\to\mathbb{C}$, so that the state is now written

$$|f\rangle = \sum_n f(x_n)|[x_n,x_n+\Delta x)\rangle$$

Of course, the space of position states that can be constructed using this basis is quite limited, since the probability of finding the state within any of the intervals of size $\Delta x$ that we defined earlier must be constant for any function constructed using this basis. The smaller we make $\Delta x$, the more position states are accessible. In the end, though, what we really want is to construct any arbitrary position state, no matter how much its amplitude varies as a function of position. What this means is that we need to take the limit $\Delta x\to 0$.

With a finite $\Delta x$, over any finite length $L$, there are at most $\frac{L}{\Delta x}$ basis states; in contrast, with a $\Delta x$ approaching zero, there are now an infinite number of basis states over any finite interval. In taking $\Delta x$ to zero, we have turned our countable basis into an uncountable one, and our discrete basis into a continuous one.

Notice that the last expression we constructed looks nearly exactly like the Riemann sum usually used in integration: we're adding together a bunch of "rectangles" with "height" $f(x_n)$ and "width" $|[x_n,x_n+\Delta x)\rangle$. This means that, when we transfer from a discrete to a continuous basis, we can use the notation and machinery that is usually used when working with Riemann sums: namely, the interval $[x_n,x_n+\Delta x)$ turns into $[x_n,x_n+dx)$ for infinitesimal $dx$, and the sum turns into an integral:

$$|f\rangle = \int f(x)|[x,x+dx)\rangle $$

The state $|[x,x+dx)\rangle$ is uniformly distributed over an interval of infinitesimal length (i.e. length approaching zero). This means it's equivalent to a state which is localized entirely at the point $x$, and so we identify it as simply $|x\rangle$. From this, we get the usual expression:

$$|f\rangle = \int f(x)|x\rangle dx$$

If you want to take the inner product of two such states, remember that $\langle x|y\rangle$ is the zero-width limit of the interval states that we defined earlier, and you'll find, by taking a similar limit, that $\langle x|y\rangle = \delta(x-y)$, where $\delta(x-y)$ is the Dirac delta function. This should make the procedure from above directly applicable.


As @probably_someone has mentioned, you need to review the definition of Riemann sum. If you take $\sum_{all~ x_i} f(x_i)\Delta x$ to be $\sum_{all~ x_j, \Delta x\rightarrow 0 } f(x)\Delta x$, then $\Delta x$ become $dx$, and $\sum_{all~ x_j, \Delta x\rightarrow 0 } f(x)\Delta x=\int_{domain~of ~ x} f(x) dx$. There's a chapter explaining this concept in james stewart's calculus. You may consider review the concept of Calculus II, or read some posts in Math.stack.exchange or Wolfram.

  • $\begingroup$ but in the summation above ie ∑f(x)g(x)* there is no Δx term so how can ee change it to riemanns sums? $\endgroup$
    – kay
    Jul 16, 2019 at 12:06
  • $\begingroup$ @kay when you convert $\langle f|g\rangle $ to $f(x)g(x)$ you probably used summation of $|x\rangle\langle x|$ over all $x$, which is equivalent to $\int_x dx |x\rangle\langle x| $ because $x$ was over $R$, I guess you may consider that's where it comes from. $\endgroup$ Jul 16, 2019 at 12:11
  • $\begingroup$ iam sorry i didnt get what you mean by |x〈〉x| $\endgroup$
    – kay
    Jul 16, 2019 at 12:13
  • $\begingroup$ ok now i get it thanks...but then again why ∑|x〉〈x |is changed with an integral? as ∫dx |x〉〈x |? $\endgroup$
    – kay
    Jul 16, 2019 at 12:14

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