Collapse of wavefunction into a forbidden eigenstate for a free particle

[Kashmir]

It does not correspond to a single measurement of $p$ nor does it represent a state of a particle. The (pure) states of a particle can only be represented by a wave function that is square integrable, i.e., which can be normalized to 1:
$$\int_{\mathbb{R}} \mathrm{d} p |\varphi(p)|^2=1.$$
There is no proper square-integrable eigenfunction of the momentum operator $\hat{p}$, but it has only a continuous spectrum $p \in \mathbb{R}$. The corresponding "eigenfunctions" of such "generalized eigenvalues" are "generalized functions" or "distributions". They are not square-integrable.

That's easy to see in the position representation, where $\hat{p}=-\mathrm{i} \hbar \partial_x$. The eigenvalue equation reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solutions are
$$u_p(x)=N \exp(\mathrm{i} p x/\hbar).$$
That's for sure not square integrable, because $|u_p(x)|^2=|N|^2=\text{const}$. Nevertheless all $p \in \mathbb{R}$ are "generalized eigenvalues" of the momentum operator and the eigenfunctions can be "normalized" to a Dirac-$\delta$ distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} u_p^*(x) u_{p'}(x)=\delta(p-p').$$
As is known from the theory of Fourier transformations for that we simply have to set the normalization constant to
$$N=\frac{1}{\sqrt{2 \pi \hbar}} \; \Rightarrow \; u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
They build a complete set of orthonormal functions in the sense that you can represent a dense subset of square integrable functions as a Fourier transform,
$$\psi(x)=\langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p)$$
and in the other way
$$\tilde{\psi}(p) = \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x).$$
It's also clear that the transformation from the "position-space wave function" to the "momentum-space wave function" is a unitary transformation since
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x) = \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}_1^*(p) \tilde{\psi}_2(p).$$

Thank you so much. I really appreciate that you took time to write it.

I completely understand that delta functions cannot represent any realizable state.

What i was saying is that there is also another problem with relating a delta wavefunction with a single value measurement $p$ since $\delta^{2}(p-p') d p '\neq 1$ at $p'=p$ (i.e even if somehow delta function represented a single value measurement, it fails to give a probability of $1$ as shown above)

 

[PeroK]

Thank you so much. I really appreciate that you took time to write it.

I completely understand that delta functions cannot represent any realizable state.

What i was saying is that there is also another problem with relating a delta wavefunction with a single value measurement $p$ since $\delta^{2}(p-p') d p '\neq 1$ at $p'=p$ (i.e even if somehow delta function represented a single value measurement, it fails to give a probability of $1$ as shown above)

The condition that the generalised eigenfunctions statisfy is Dirac Orthonormality. The eigenfunction $p(p')$ associated with eigenvalue $p$ (in momentum space) is:
$$p(p') = \delta(p - p')$$Now:
$$\langle p_1| p_2 \rangle = \int dp' p_1(p')^*p_2(p') = \int dp'\delta(p_1 - p') \delta(p_2 - p') = \delta(p_1 - p_2)$$In particular, if $p_1 \ne p_2$, then:
$$\langle p_1| p_2 \rangle = 0$$The delta function itself is not square integrable.

 

[atyy]

For the free particle in QM, the energy and momentum eigenstates are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite energy or momentum.

What happens during measurement of say momentum or energy ?

So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle). Then we would in principle collapse the wave function to some eigenstate, but in this case we know that this is not possible (physically realizable).

So what happens during measurement and in what state is the particle after we've measured it's momentum?

Thank you

One can ask a similar question about a sharp position measurement. There is some comment on that in section 2.3.2 of https://arxiv.org/abs/0706.3526.The rule that the wave function collapses into an eigenstate of the measurement is not the most general rule for collapse, and cannot apply for position measurements, since the position eigenstate is not an allowed quantum state. There are more general rules for collapse associated with mroe general measurements called POVMs.

 

[throw]

For the free particle in QM, the energy and momentum eigenstates are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite energy or momentum.

This is just one possible interpretation of continuous spectrum eigenfunctions - it is not the only possible interpretation one can give to continuous spectrum eigenfunctions:

There are two ways out of this difficulty. The first is to give up the concept of absolute probabilities when dealing with wave functions such as (2.13) or (2.16) which are not square integrable. Instead, $|\Psi(\mathbf{r}, t)|^2 d \mathbf{r}$ is then interpreted as the relative probability of finding the particle at time t in a volume element $d\mathbf{r}$ centred about $\mathbf{r}$, so that the ratio $|\Psi(\mathbf{r}_1, t)|^2/|\Psi(\mathbf{r}_2, t)|^2$ gives the probability of finding the particle within a volume element centred around $\mathbf{r} = \mathbf{r}_1$, compared with that of finding it within the same volume element at $\mathbf{r} = \mathbf{r}_2$.

For the particular case of the plane wave $(2.16)$, we see that $|\Psi|^2 = |A|^2$, so that there is an equal chance of finding the particle at any point. The plane wave $(2.16)$ therefore describes the idealised situation of a free particle having a perfectly well-defined momentum, but which is completely delocalised.

This suggests a second way out of the difficulty, which is to give up the requirement that the free particle should have a precisely defined momentum, and to superpose plane waves corresponding to different momenta to form a localised wave packet, which can be normalised to unity.

- Bransden and Joachain, 'Quantum Mechanics'

There are plenty of books which take the alternative approach, for example Landau and Lifshitz 'Quantum Mechanics', and from this perspective many of the statements in this thread are simply indefensible: the origin of the problem really boils down to the difference between discrete and continuous probability distributions, and interpreting the meaning of measuring a single value v.s. theoretical probabilities for finding single values (we don't say single outcomes associated to general continuous probability distributions don't 'exist', only in a mathematical approach to QM is this done for some reason...)

 

[PeroK]

There are plenty of books which take the alternative approach, for example Landau and Lifshitz 'Quantum Mechanics', and from this perspective many of the statements in this thread are simply indefensible:

Are you saying that the OP, who has just started learning QM, should give up McIntyre and study Landau and Lifshitz instead?

 

[vanhees71]

Well, it doesn't hurt to consult other textbooks, if the so far used textbook is obviously not concise enough to help with a very good question as the one asked in #1. Obviously McIntire is a bit too sloppy, if it comes to unbound self-adjoint operators with a continuous spectrum. Instead of Landau and Lifshitz I'd rather suggest to read Messiah. This book in my opinion finds a quite good compromise between mathematical rigidity and the more pragmatic physicists' approach. For a more modern treatment in terms of the "rigged Hilbert space" also Ballentine is very good choice.

 

[andresB]

Checking my copy of Messiah, I find a readable treatment of projector for the continuous spectrum, pages 260 to 270. And a discussion of protective measurements, discrete and continuous cases, is given on pages 297 and 298.

 

[PeroK]

Well, it doesn't hurt to consult other textbooks, ...

It might hurt fundamentally in the sense that the OP may never really learn QM, but just stagger from one textbook to another. It's fine for an expert in the field to compare and contrast McIntyre, L&L, Messiah and Ballentine; but, for a novice that might not work at all.

Personally, I would stick with one textbook for a serious timescale - 4-6 chapters or so. I would focus on digesting the basics. Alternative and subtler analyses can come later.

 

[andresB]

It might hurt fundamentally in the sense that the OP may never really learn QM, but just stagger from one textbook to another. It's fine for an expert in the field to compare and contrast McIntyre, L&L, Messiah and Ballentine; but, for a novice that might not work at all.

Personally, I would stick with one textbook for a serious timescale - 4-6 chapters or so. I would focus on digesting the basics. Alternative and subtler analyses can come later.

While having too many sources to read from can cause some sort of analysis paralysis, the concept of learning from just one book is alien to me.

 

[vanhees71]

Hm, I always consulted other textbooks, if I couldn't answer a question with help of the main textbook I've been studying. For me that's the usual situation when learning a new subject.

 

[PeroK]

While having too many sources to read from can cause some sort of analysis paralysis, the concept of learning from just one book is alien to me.

I'm sceptical of that as an education policy below PhD/post-doctorate level. If you were teaching undergraduate E&M, you really would work from several textbooks concurrently? What about high-school geometry? I can't see it! I suggest it's bravado on your part!

It's different to suggest a choice of textbooks, but not multiple text-books concurrently. Not for the average student. And definitely not for me.

I'm happy to be corrected by those who have taught at university level, but if I was teaching I would stick to one textbook per course. And definitely not present conflicting presentations of the same material at an introductory level.

 

[vanhees71]

That's very strange to me. In our lectures we always got a list of books we can use when studying the material presented in the lectures. As I said above, I never studied a subject with using only a single book. Usually I worked along the content following the notes I took during the lecture and then consulted several textbooks when I couldn't understand something from the notes. In addition, of course, we had problem sessions and could ask the tutors and/or the professor. Last but not least we also worked together in a group of students to solve the problem sets.

 

[Kashmir]

To say that we know the exact value of the momentum is equivalent to saying that the wave function in momentum space is a delta function.

Why do you say that the wave function will be a delta function?

Suppose for arguments sake I knew the exact value of momentum to be $p_0$ .

Let the state corresponding to it be $|\psi\rangle$ . Since $\langle p \mid \psi\rangle$ gives the probability amplitude of measuring momentum around $p$ ,then it follows
$\langle p \mid \psi\rangle=0 \quad ;p \neq p_{0}$ and

$\left\langle p_{0} \mid \psi\right\rangle=1$

which means that the wave function in momentum space is
$\varphi(p)=\left\{\begin{array}{ll}0 & p \neq p_{0} \\ 1 & p=p_{0}\end{array}\right.$which is not a delta function.

 

[PeroK]

$\left\langle p_{0} \mid \psi\right\rangle=1$ , which means that the wave function in momentum space is
$\varphi(p)=\left\{\begin{array}{ll}0 & p \neq p_{0} \\ 1 & p=p_{0}\end{array}\right.$

That's not correct. The inner product of two functions is:
$$\left\langle p_{0} \mid \psi\right\rangle=\int dp' \ p_0(p')^* \psi(p')$$And that integral is zero for your eigenfunction. Note that in terms of square integrable functions, the function you suggest is equivalent to the zero function: $\varphi \equiv 0$

Instead if $p_0(p') = \delta(p_0 - p')$, then:
$$\left\langle p_{0} \mid \psi\right\rangle=\int dp' \ \delta(p_0 - p') \psi(p') = \psi(p_0)$$

 

[vanhees71]

But such a function doesn't exist! A "wave function" describing the situation that the momentum of the particle takes the exact momentum value $p_0$ in the momentum representation representation must be $\langle p|p_0 \rangle=\delta(p-p_0)$, i.e., a $\delta$ distribution. You cannot even take its square! So it's not describing a proper (pure) state of the particle. As already stressed above, you can have arbitrarily well determined values of the momentum but never a preparation in a state where momentum takes an exactly single value.

An example are the Gaussian wave functions:
$$\tilde{\psi}(p)=\frac{1}{\sqrt{\Delta p} (2 \pi)^{1/4}} \exp \left (-\frac{(p-p_0)^2}{4 \Delta p^2} \right ).$$
This describes the situation, where the momentum takes the average value $p_0$ with a standard deviation of $\Delta p$. The probability distribution for measuring the momentum is
$$P(p)=|\tilde{\psi}(p)|^2 = \frac{1}{\sqrt{2 \pi} \Delta p} \exp \left (-\frac{(p-p_0)^2}{2 \Delta p^2} \right ).$$
Indeed, that's the Gaussian distribution with average $\langle p \rangle=p_0$ and $\langle p^2 \rangle-\langle p \rangle^2=\Delta p^2$.

You can make $\Delta p>0$ as small as possible, i.e., making the probability distribution peaking as narrowly around $p_0$ as you like, but never set $\Delta p=0$. Rather the Gaussian goes to the $\delta$-distribution in the limit $\Delta p \rightarrow 0$.

 

[Kashmir]

A "wave function" describing the situation that the momentum of the particle takes the exact momentum value p0 in the momentum representation representation must be ⟨p|p0⟩=δ(p−p0), i.e., a δ distribution.

This is what I'm not able to understand.
Can you please tell me how do you conclude that "if"the momentum of the particle takes the exact momentum value $p_0$ in the momentum representation representation must be a δ distribution?

 

[Nugatory]

This is what I'm not able to understand.
Can you please tell me how do you conclude that "if"the momentum of the particle takes the exact momentum value $p_0$ in the momentum representation representation must be a δ distribution?

If the momentum takes on the exact value $p_0$, the the probability of finding any value except $p_0$ will be zero. This implies that the value of $\phi(p)$ is zero for all $p$ except $p=p_0$.

We also have $\int_{-\infty}^{\infty}\phi^*(p’)\phi(p’)dp’=1$ because the probabilities of all possible momentum results have to add to 100%.

The delta “function” is the only thing that can be zero everywhere except at a single point yet still give us a non-zero value when integrated across that point.

 

[vanhees71]

The problem is that the "eigenvalues" of the momentum operator are the complete continuous set of real numbers $\mathbb{R}$. For such "eigenvalues" there is no proper normalizable "eigenvector".

The most simple way to see it, in my opinion, is to solve the eigenvalue problem in the position representation of wave mechanics. There the operator describing momentum is (I restrict myself to 1D motion; it's straight forward to generalize to motion in 3D):
$$\hat{p}=-\mathrm{i} \hbar \partial_x.$$
The eigenvalue equation to solve is
$$\hat{p} u_p(x)=-\mathrm{i} \hbar \partial_x u_p(x) = p u_p(x).$$
The solution is obviously
$$u_p(x)=A \exp(\mathrm{i} p x/\hbar).$$
Here $A$ is an integration constant. Obviously $|u_p(x)|^2=|A|^2=\text{const}$ is not integrable, i.e., you cannot normlize $u_p$ to 1 and thus it's not a proper wave function, that describes a position-probability distribution.

The only thing you can do is to "normalize it to a $\delta$ distribution". Indeed
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) u_{p'}(x)=\int_{\mathbb{R}} \mathrm{d} x |A|^2 \exp [\mathrm{i} x (p'-p)/\hbar]=2 \pi \hbar |A|^2 \delta(p-p').$$
So we set $A=1/\sqrt{2 \pi \hbar}$, i.e., we have
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
In momentum representation you thus get a $\delta$-distribution as a "wave function" if you want to describe the situation that $p$ takes with certainty the value $p_0$:
$$\psi(x)=\langle u_p|u_{p_0} \rangle=\delta(p-p_0).$$
Of course you can't square it and take this square as a probability distribution. Thus there is no proper wave function describing this situation.

 

[PeroK]

The delta “function” is the only thing that can be zero everywhere except at a single point yet still give us a non-zero value when integrated across that point.

It's also possible to use a "sufficiently" peaked Gaussian instead of the Dirac Delta function. This ties in with the whole idea of infinitely precise measurements being impossible. This is also sometimes a useful way of justifying (by regular calculus) Dirac Delta identities. If it holds for Gaussians in the limit as the "width" tends to zero, then it holds for Delta functions.

E.g. the defining integral identity for the Delta function, that for any regular function $f$:
$$\int \delta(x)f(x) = f(0)$$

 

[Nugatory]

It's also possible to use a "sufficiently" peaked Gaussian instead of the Dirac Delta function.

Of course, but a sufficiently peaked Gaussian is not zero at exactly one point, so isn’t the pdf for an exact value.

(I’m not arguing here, just annotating. This entire thread exists because intro textbooks are understandably prone to gloss over the mathematical subtleties of continuous-spectrum observables).

 

[vanhees71]

Well, usually the textbooks discuss the Dirac $\delta$ distribution with sufficient detail. Obviously the textbook the OP reads doesn't do this with properly. Again, I can only recommend to read another textbook on this subject.

 

[throw]

While Messiah has a separate section devoted to discussing various approaches to the continuous spectrum, he does not seem to discuss the approach I referenced, which is also the approach Dirac takes in his QM book.

The continuous spectrum is implicit in just about everything one does in QFT (whether this is made explicit or not is again another question), so it's definitely worth being very careful about it.

There are arguments in QFT that the only thing we can measure precisely are the eigenvalues of things associated to free particles like their momenta, yet in this thread people are saying this is the one thing we can't measure; there is a very clear contradiction here, so it's worth being aware this is a topic with multiple perspectives not just one that's automatically correct without serious justification.

 

[vanhees71]

Another good discussion about the continuous spectrum of the position and momentum operators can be found in J. J. Sakurai, Modern Quantum Mechanics. This book also starts with the "Stern-Gerlach experiment first" and "bras+kets first" approach. Maybe it's a good additional read for the OP, because his textbook (McIntire) also uses this approach.

 

[Kashmir]

If the momentum takes on the exact value $p_0$, the the probability of finding any value except $p_0$ will be zero. This implies that the value of $\phi(p)$ is zero for all $p$ except $p=p_0$.

We also have $\int_{-\infty}^{\infty}\phi^*(p’)\phi(p’)dp’=1$ because the probabilities of all possible momentum results have to add to 100%.

The delta “function” is the only thing that can be zero everywhere except at a single point yet still give us a non-zero value when integrated across that point.

Thank you. Got it!

But the wavefunction doesn't give the correct probability of $1$ at the value of $p = p_0$ since
\begin{array}{l}
\left|\phi\left(p_{0}\right)\right|^{2} d p \\
=\delta^{2}(0) d p \\
\neq 1
\end{array}##

 

[Kashmir]

Another good discussion about the continuous spectrum of the position and momentum operators can be found in J. J. Sakurai, Modern Quantum Mechanics. This book also starts with the "Stern-Gerlach experiment first" and "bras+kets first" approach. Maybe it's a good additional read for the OP, because his textbook (McIntire) also uses this approach.

Thank you. I'll look into it. Yes the approach is similar but Sakurai is higher level.

 

[PeroK]

There are arguments in QFT that the only thing we can measure precisely are the eigenvalues of things associated to free particles like their momenta, yet in this thread people are saying this is the one thing we can't measure; there is a very clear contradiction here,

I don't accept that at all. And I'm not sure what you achieve by telling the rest of us that we are contradicting ourselves?

 

[PeroK]

Thank you. I'll look into it. Yes the approach is similar but Sakurai is higher level.

You've got McIntyre, Messiah, L&L and Ballentine to get through already, before you add Sakurai to the list

 

[PeroK]

Thank you. Got it!

But the wavefunction doesn't give the correct probability of $1$ at the value of $p = p_0$ since
\begin{array}{l}
\left|\phi\left(p_{0}\right)\right|^{2} d p \\
=\delta^{2}(0) d p \\
\neq 1
\end{array}##

That's not true for any wave-function. The modulus squared of a wave-function is a probability density function, and not a probability. You can generate the probability that the particle is found in a small region around $x_0$: $x_0 - \frac{\Delta x} 2 < x < x_0 + \frac {\Delta x}{2}$ as:
$$p(x_0) \approx |\psi(x_0)|^2 \Delta x$$The limit of this gives a probability density function:
$$p(x) = |\psi(x)|^2$$And, in general, no probability density function can be zero everywhere except one point, unless you get into Dirac delta functions etc.

 

[Kashmir]

That's not true for any wave-function. The modulus squared of a wave-function is a probability density function, and not a probability. You can generate the probability that the particle is found in a small region around $x_0$: $x_0 - \frac{\Delta x} 2 < x < x_0 + \frac {\Delta x}{2}$ as:
$$p(x_0) \approx |\psi(x_0)|^2 \Delta x$$The limit of this gives a probability density function:
$$p(x) = |\psi(x)|^2$$And, in general, no probability density function can be zero everywhere except one point, unless you get into Dirac delta functions etc.

I understand that "The modulus squared of a wave-function is a probability density function, and not a probability" that's why I wrote $|\phi(p_0)|^{2} d p$ which should give me the probability of getting a momentum around $p_0$. Since we are discussing a hypothetical single value measurement then $|\phi(p_0)|^{2} d p$ should equal $1$ but since the wavefunction is a delta function then
$\begin{array}{l} \left|\phi\left(p_{0}\right)\right|^{2} d p \\ =\delta^{2}(0) d p \\ \neq 1 \end{array}$

 

[Kashmir]

You've got McIntyre, Messiah, L&L and Ballentine to get through already, before you add Sakurai to the list

I'll first finish McIntyre then I'll go to those. For the doubts, I'll post it here. You and other people like vanhees and peter are very helpful :)

 

[PeroK]

=\delta^{2}(0) d p \\

\neq 1

1) The Dirac Delta isn't a "normal" function. In particular, $\delta^{2}(0)$ is not a number.

2) $dp$ is not a number either.

3) What you're doing is mathetically invalid. You need to handle the Delta function with care.

 

[Nugatory]

But the wavefunction doesn't give the correct probability of 1 at the value of p=p0

Sure it does: $\phi(p)=\delta(p-p_0)$, $\int_{p_0-\epsilon}^{p_0+\epsilon}\phi^*(p’)\phi(p’)dp’=1$ for arbitrarily small $\epsilon$.

 

[andresB]

I'm sceptical of that as an education policy below PhD/post-doctorate level. If you were teaching undergraduate E&M, you really would work from several textbooks concurrently? What about high-school geometry? I can't see it! I suggest it's bravado on your part!

It's different to suggest a choice of textbooks, but not multiple textbook concurrently. Not for the average student. And definitely not for me.

I'm happy to be corrected by those who have taught at university level, but if I was teaching I would stick to one textbook per course. And definitely not present conflicting presentations of the same material at an introductory level.

Well, the only courses I have had the opportunity to teach are university-level introductory physics, so my opinions are gathered from my days as an undergraduate student. Internet being too expensive in those days and physical books being a rarity, we studied with whatever we could find, sometimes not even complete books but copies of chapters or sections. I studied QM with both Griffith and Messiah, because I got some photocopies of those texts. It was annoying to only have one book, what If I don't understand something from that book? I was stuck. So, I've always assumed that reading at least two books was the preferred course of action.Btw, reading from more than one book doesn't mean going full depth into both of them. There is always a primary source, and the others are to complement, to fill the gaps.

 

[Kashmir]

Sure it does: $\phi(p)=\delta(p-p_0)$, $\int_{p_0-\epsilon}^{p_0+\epsilon}\phi^*(p’)\phi(p’)dp’=1$ for arbitrarily small $\epsilon$.

Townsend pg 194
"It is natural to identify
$d x|\langle x \mid \psi\rangle|^{2}$
with the probability of finding the particle between $x$ and $x+d x$ if a measurement of position is carried out, as first suggested by M. Born."So we can't use the born interpretation of probability for momentum here? If we use it we'll get wrong result. But if we do it the way you've done i.e $\int_{p_0-\epsilon}^{p_0+\epsilon}\phi^*(p’)\phi(p’)dp’=1$ it leads to a correct result.

 

[PeroK]

Townsend pg 194
"It is natural to identify
$d x|\langle x \mid \psi\rangle|^{2}$
with the probability of finding the particle between $x$ and $x+d x$ if a measurement of position is carried out, as first suggested by M. Born."

First, Townsend is using $dx$ here as a small inverval. He really ought to use $\Delta x$. This statistical interpretation of states is generally true. If we represent the state in position space, then:
$$\langle x \mid \psi \rangle = \int \delta(x - x') \psi(x') dx' = \psi(x)$$This is because the eigenstate of the position operator is a delta function. I.e. $\langle x \mid \ \leftrightarrow \ \delta(x - x')$

Note that if we interpret $|\psi(x)|^2$ as a probability density function, then the probability of finding the particle between $x$ and $x + \Delta x$, assuming $\Delta x$ is small, is:
$$\int_x^{x + \Delta x}|\psi(x)|^2 dx \approx |\psi(x)|^2 \Delta x$$Now, if we take the state of the system (theoretically) to be in an eigenstate of position. So, that $\psi(x') = \delta(x_0 - x')$. I.e. the result of a precise measurement of $x_0$. Then we have:
$$\langle x \mid \psi \rangle = \int \delta(x - x') \delta(x_0 - x') dx' = \delta(x - x_0)$$
And, if we consider a small interval centred on $x_0$, then the probability of finding the particle between $x_0 - \frac{\Delta x}{2}$ and $x_0 + \frac{\Delta x}{2}$ is:
$$\int_{x_0 - \frac{\Delta x}{2}}^{x_0 + \frac{\Delta x}{2}}\delta(x - x_0)^2 dx = 1$$Which is just the same as the Born interpretation, except we have taken a little mathematical care to deal with the delta function.

And, to see that result, once we have chosen $\Delta x$, we can approximate the delta function by a Gaussian that is narrower than $\Delta x$ and see that the integral tends to $1$ as the width of the Gaussian tends to zero.