Collapse of wavefunction into a forbidden eigenstate for a free particle

[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
$$P(x)=|\psi(x)|^2=\frac{1}{\Delta x} \chi_{(x_0-\Delta x/s,x_0+\Delta x/2)}.$$