1
$\begingroup$

In the Wikipedia article it says "the propagator is a function that specifies the probability amplitude for a particle to travel from one place to another in a given period of time".

Let us write the propagator going from state $A$ to state $B$ as $K_t(A,B)$.

According to the definition this amplitude should correspond in some sense to the probability $P(A|B)$. But this can't be so since: $$\sum_AP(A|B) = 1$$ but $$\sum_A |K_t(A,B)|^2 \neq 1$$ instead: $$\sum_A K_t(A,B) =1$$

However we do have $\sum\limits_B K_t(A,B)\Psi_0(B) = \Psi_t(A)$ which corresponds to the probabilities $P(A) = \sum\limits_B P(A|B)P(B)$. But we have $|\sum\limits_B K_t(A,B)\Psi(B)|^2 = |\Psi_0(A)|^2$. Hence the amplitude $K_t(A,B)$ can't be associated with any probability as the equation cannot be factored. Wave functions, of course, satisfy $\sum\limits_A|\Psi_t(A)|^2=1$

Hence, what right have we to call $K_t(A,B)$ a "probability amplitude" when it's absolute square does not correspond to a probability?

It is acceptable to call $K_t(A,B)$ a "time evolution operator" as we can interpret it as a matrix $K(t)^{AB}$ where $A$ and $B$ can be thought of as indices of a matrix.

Is this just being semantically picky? Or is there a fundamental difference between the complex amplitude given by a wave function $\Psi$ and the complex "amplitude" given by the propagator $K$?Because it seems to me that $K$ does not satisfy the definition of amplitude given by a complex number whose modulus squared represented a probability.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
2
$\begingroup$

According to the statement that you quote from Wikipedia, the propagator is a conditional probability amplitude: Given a particle in one place, it tells us the probability amplitude for the particle in another place. In other words, to get a probability amplitude, one needs to multiply the propagator with another probability amplitude and integrate. This is exactly what you are doing at one point: $$ \sum\limits_B K_t(A,B)\Psi(B) = \Psi_0(A) . $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ True but in probability theory the sum of conditional probabilities still adds up to 1. But I take your point perhaps a better name for it would be "conditional amplitude". Personally I think it the terminology of calling both objects amplitudes can lead to confusion but that's just my point of view. I think it's because the Shrodinger wave function picture and "sum over histories" pictures both use the term "amplitude" to mean different things. $\endgroup$
    – user84158
    Jun 28, 2022 at 4:42
  • $\begingroup$ For example if you follow the link to "probability amplitude" it describes something normalised to 1. So this is inconsistent in Wikipedia. $\endgroup$
    – user84158
    Jun 28, 2022 at 4:50
0
$\begingroup$

If $\Psi(B)$ is a delta-function, then you get back to your initial equations. Thus, the problem is really with how one normalizes the continuum states - not specifically with the propagator. E.g., this problem would not arise, if you work in k-space.

Improve this answer
$\endgroup$
3
  • $\begingroup$ $\Psi(B)$ cannot be a delta function since $\int |\delta(B)|^2 dB \neq 1$ . This is a common trap which I've also fallen into from time to time! If you work in k space you get the same problems. $\endgroup$
    – user84158
    Jun 28, 2022 at 14:58
  • $\begingroup$ @zooby this is what my answer says - normalizing continuum states is problematic: you need an eigenfunction of $\hat{x}$. On the other hand, if you consider the same free particle propagator in momentum space, your initial wave function is just $\Psi(B)=\delta_{B,B_0}$, so that $\Psi(A)=\sum_BK(A,B)\Psi(B)=K(A,B_0)$. But, if you wish semantic difference, you can simply put in bold the probability amplitude for a particle to travel from one place to another rather than just the probability amplitude: in the same way as you don't confuse probability amplitude with probability. $\endgroup$
    – Roger V.
    Jun 28, 2022 at 15:07
  • $\begingroup$ Don't quite know what you mean since you can easily normalise $\Psi_0(B)=e^{-B^2}$ $\endgroup$
    – user84158
    Jun 28, 2022 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.