I Collapse of wavefunction into a forbidden eigenstate for a free particle

[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.

 

[Nugatory]

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.

We are using the Born rule here. It says that the probability of finding the momentum between $p$ and $(p+dp)$ is $dp|\langle p\mid\phi\rangle|^2$ (where $\phi(p)=\langle p\mid\phi\rangle$ is the wave function in the momentum representation) and that’s what I’m integrating to find the probability that the momentum is between $p_0-\epsilon$ and $p_0+\epsilon$.

 

[vanhees71]

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

If you have a list of books, you shouldn't blindly "go through" them but check them out in the library and see, which one helps you best. I'm really surprised that it seems to be uncommon nowadays to consult several books, when one has a problem in understanding. For me that's the normal way to study.

 

[vanhees71]

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.

I don't know, how often we have to repeat the obvious! Pure states are represented by square-integrable functions. The distributions that occur as "generalized eigenfunctions" of operators with continuous spectras are distributions, i.e., generalized functions and not square-integrable functions and thus don't represent states. I don't know, how to explain this differently than I did already several times before.

 

[Kashmir]

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.

Understood it. We've to deal with the delta function a little bit more carefully as you've shown. Thank you for writing it. :)

 

[Kashmir]

We are using the Born rule here. It says that the probability of finding the momentum between $p$ and $(p+dp)$ is $dp|\langle p\mid\phi\rangle|^2$ (where $\phi(p)=\langle p\mid\phi\rangle$ is the wave function in the momentum representation) and that’s what I’m integrating to find the probability that the momentum is between $p_0-\epsilon$ and $p_0+\epsilon$.

got it. Thank you so much. :)

 

[vanhees71]

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.

One must emphasize however, that this calculation is invalid, because your integral formally isn't 1 but the undefined expression $\delta(0)$. You cannot square the $\delta$ distribution as if it were a function.

Rather you have to "smear" the "position eigenfunction" first, before you can square it. Taking up your example you can describe the situation that you have measured the position of the particle to be in the interval $(x_0-\Delta x/s,x_0+\Delta x/2)$ by
$$\psi(x)=\int_{x_0-\Delta x/2}^{x_0+\Delta x/2} \delta(x-x_0) = \frac{1}{\sqrt{\Delta x}} \chi_{(x_0-\Delta x/2,x_0+\Delta x/2)}(x),$$
where the characteristic function of the said interval is defined as
$$
\chi_{(x_0-\Delta x/s,x_0+\Delta x/2)}=\begin{cases} 1 &\text{if} \quad x \in (x_0-\Delta x/2,x_0+\Delta x/2), \\ 0 & \text{if} \quad x \notin (x_0-\Delta x/2,x_0+\Delta x/2). \end{cases}$$
Then the probability distribution to find the particle at position $x$ is given by
$$P(x)=|\psi(x)|^2=\frac{1}{\Delta x} \chi_{(x_0-\Delta x/s,x_0+\Delta x/2)}.$$