I am working through Shankar's Introduction to Quantum Mechanics. I have come across exercise 1.10.1, which asks the reader to show that:


I can understand it intuitively, in the sense that $x\to ax$ will 'compress' the delta function in the x-axis direction, and it has to be $|a|$ because the delta function is an even function.

However, how can this relationship be shown mathematically? I have been playing with it for a few days and I seem to be struggling.


To expand on the method I have been trying, I am starting with the following completeness relation that Shankar gives:

$$\int_{-\infty}^{\infty}\vert x'\rangle\langle x'\vert dx'=I$$

I then create a new variable, $x''$, such that:

$$x''=ax'$$ $$dx''=adx'$$

then I substitute that in:

$$\int_{-\infty}^{\infty}\vert x''\rangle\langle x''\vert \frac{dx''}{a}=I$$

then I perform his operation (that he uses to show $\int\delta(x)=1$), where he takes the inner product on one side with $\vert f \rangle$ and on the other with $\langle x \vert$:

$$\int_{-\infty}^{\infty}\langle x \vert x''\rangle\langle x''\vert f \rangle\frac{dx''}{a}=\langle x \vert I \vert f \rangle = f(x)$$

$$f(x)\int_{-\infty}^{\infty}\delta(x,x'') dx'' = a f(x)$$

However, this seems to lead to the opposite conclusion, that $\delta(ax)=a\delta(x)$.

Where have I gone wrong?

  • $\begingroup$ Put it in an integral, make a variable substitution, consider cases for when $a>0$ and $a<0$ $\endgroup$ – Aaron Stevens Sep 19 at 18:46
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Sep 19 at 18:48
  • $\begingroup$ @AaronStevens I have been trying to do that, but I don't quite see how. $\endgroup$ – Time4Tea Sep 19 at 19:01
  • $\begingroup$ I thought $$\int_{-\infty}^{\infty}\langle x \vert x''\rangle\langle x''\vert f \rangle\frac{dx''}{a}=f(x)\int_{-\infty}^{\infty}\delta(x,x'') \frac{dx''}{a}$$ $\endgroup$ – Aaron Stevens Sep 19 at 19:21
  • $\begingroup$ @AaronStevens that seems equivalent to what I got. However, doesn't that mean that $\delta(ax)$ is a factor of $a$ bigger than $\delta(x)$, because you have to divide it by $a$ to get the same result? Perhaps I'm interpreting it wrong ... $\endgroup$ – Time4Tea Sep 19 at 19:24

You're working much too hard. The delta function is defined by

$$\int_a^b \delta(x) f(x) dx = \begin{cases}f(0) & 0 \in (a,b) \\ 0 & \text{otherwise}\end{cases}$$

Start with $\int_a^b \delta(ax)f(x) dx$, perform a variable substitution with $u= ax$, and then apply the definition of the delta function. As Aaron said, you should separately consider the cases when $a<0$ and $a>0$, and you should arrive at the result right away.

  • $\begingroup$ Thanks a lot, this is really helpful. This problem has been bugging me for days ... $\endgroup$ – Time4Tea Sep 19 at 21:27

J. Murray's answer was very helpful. So, just to make sure I've got this right, this is the definition of the delta function:

$$\int_a^b \delta(x) f(x) dx = \begin{cases}f(0) & 0 \in (a,b) \\ 0 & \text{otherwise}\end{cases}$$

and we want:

$$\int_b^c \delta(ax) f(x) dx$$

So, substitute $u=ax$, $du=adx$:

$$\int_{ab}^{ac} \delta(u) f(u/a) \frac{du}{a}=?$$

$$\frac{1}{a}\int_{ab}^{ac} \delta(u) f(u/a) du=\begin{cases}\frac{f(0)}{a} & 0 \in (ab,ac) \\ 0 & \text{otherwise}\end{cases}$$

by the definition of the delta function. Then, we have to switch the variables back:

$$\frac{1}{a}\int_{b}^{c} \delta(ax) f(x) a dx=\begin{cases}\frac{f(0)}{a} & 0 \in (b,c) \\ 0 & \text{otherwise}\end{cases}$$

$$\int_{b}^{c} \delta(ax) f(x) dx=\begin{cases}\frac{f(0)}{a} & 0 \in (b,c) \\ 0 & \text{otherwise}\end{cases}$$

which is what we want; however, that assumes that $a>0$. If $a<0$:

$$\frac{1}{a}\int_{ab}^{ac} \delta(u) f(u/a) du=\begin{cases}-\frac{f(0)}{a} & 0 \in (ac,ab) \\ 0 & \text{otherwise}\end{cases}$$

because $ab$ is now larger than $ac$, so the integral is 'reversed'. So, that is what gives the general case of:

$$\int_{b}^{c} \delta(ax) f(x) dx=\begin{cases}\frac{f(0)}{|a|} & 0 \in (b,c) \\ 0 & \text{otherwise}\end{cases}$$

Is that correct?

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    $\begingroup$ Looks right to me! $\endgroup$ – J. Murray Sep 19 at 21:32

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