Collapse of wavefunction into a forbidden eigenstate for a free particle

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