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Yes I have seen the explanation of why this is so in quantum mechanical textbooks. However, let's use the identity operator and do the following:

$$\langle x|x'\rangle =\langle x|I|x'\rangle =\int\langle x|x''\rangle\langle x''|x\rangle dx''=\int\delta(x-x'')*\delta(x''-x)dx'\tag{1}$$

(the integral is from $-\infty$ to $+\infty$. Now, let's look at what we have inside the integral. Clearly, the term inside

$$\delta(x-x'')*\delta(x''-x)\tag{2}$$

is nonzero only if $x''=x'=x$. We can restate this as follows:

$$\delta(x-x'')*\delta(x''-x)=0\text{, if } x\neq x'\tag{3}$$

and

$$\delta(x'-x'')*\delta(x''-x)=\infty \text{ (for certain $x''$ in the region from $-\infty$ to $+\infty$)}\tag{4}$$

if $x=x'$; this means that

$$\delta(x-x'')*\delta(x''-x)=\delta(x'-x).\tag{5}$$

However, if we put $\delta(x'-x)$ instead of $\delta(x-x'')*\delta(x''-x)$ we get the following:

$$\int\delta(x-x'')*\delta(x''-x)dx''=\int\delta(x'-x)dx''=\delta(x'-x)\int dx''\tag{6}$$

(I remind that the integral is from $-\infty$ to $+\infty$). So I get that

$$\langle x|x'\rangle=\delta(x'-x)\int dx''\tag{7}$$

which is not just $\delta(x'-x)$. I would be grateful if someone could tell me where my assumptions are wrong and why.

Qmechanic
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Volodymyr
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2 Answers2

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Performing the manipulation you suggest, $$\langle x|x'\rangle=\langle x|\mathbb 1 |x'\rangle = \langle x|\left(\int dx'' |x'' \rangle\langle x''|\right)|x'\rangle $$ $$= \int dx '' \langle x|x''\rangle \langle x''|x'\rangle = \int dx'' \delta(x-x'')\delta(x''-x')$$

The defining property of the delta function is that $\int dx'' f(x'') \delta (x''-x') = f(x')$. Comparing that to what we have above, we get $$ \langle x|x'\rangle = \int dx'' \underbrace{\delta(x-x'')}_{\text{This is our }f(x'')} \delta(x'' - x') =\delta(x-x')$$ as expected.

You have to be careful with delta functions, because they are not proper functions, and you can get yourself into a lot of trouble by naively pretending that they are. As a general rule, delta functions are defined by how they behave inside an integral. Your mistake was when you said

$\delta(x-x'')\delta(x''-x)=0$ if $x\neq x'$ and $\delta(x'-x'')\delta(x''-x)=\infty$ (for certain $x''$ in the region from $-\infty$ to $\infty$) if $x=x'$; this means that $\delta(x-x'') \delta(x''-x) = \delta(x'-x)$.

You are using reasoning that would be applicable if the delta function were actually a function (i.e. comparing function values for various different values of the inputs) but that does not work here. You must stick to the formal defining property

$$\int dx f(x) \delta (x-a) = f(a)$$

J. Murray
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  • Thanks, J. Murray! Your explanation did save me a lot of time. – Volodymyr Mar 24 '20 at 16:51
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    @Volodymyr If you feel that your question has been answered, then you should accept one of the posted answers. Otherwise, you should post a comment with any remaining issues that you're having. – J. Murray Mar 24 '20 at 16:59
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I would say that thinking of the delta function as a function that takes value $0$ at certain points and ''$\infty$'' when the argument is $0$ can be a bit tricky. My advise is that you blindly us the definition of a delta function namely that $$\int \delta(x-y)f(y) dy = f(x)$$ and do not try to reason to it. So if you want to show that $$ \int \delta(x-x'')\delta(x''-x') dx''$$ is just $\delta(x'-x)$ just use the definition. Because what a delta function really ''is'' is just the properties of it defined by its definition.

By the way. I think your mistake is the equality $$\int \delta(x-x'')\delta(x''-x')dx'' = \int \delta(x'-x) dx''$$. If you think about this in with the definition you will see why. It should be equal to $\delta(x-x')$.

Oliver Thim
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  • What you write is correct. But, strictly speaking, you are not using the definition of the delta function because you cannot say that $\delta(x''-x') = f(x'')$ is a function. So the integral, as is, is not defined. In order to prove it properly you need to multiply by a test function $f$ and integrate over $x'$ as well. – MannyC Mar 24 '20 at 17:39