0
$\begingroup$
  1. My Understanding of uncertainty principle goes that if some particles are in same state, then their measurement of certain property (say $x$ and $p$) will be different for different particles. Plotting that probability distribution will give me a real valued function $\psi(x)^2$ and $\phi(p)^2$ where $\psi(x)$ and $\phi(p)$ is a complex function (which later we know it as wave function). These two curves are related to each other in a way that increasing or decreasing the width (ie s.d) of either curve, affects the other curve and so this is what uncertainty principal is all about.

  2. Now in double slit experiment, my understanding is that for a particle there is a quantity i.e. $\sqrt P$, whose square gives the probability $P$ and if $P_1\;\&\; P_2$ are two such property then their superimposition is $(\sqrt{P_1}+\sqrt{P_2})^2$ instead of $P_1 + P_2.$

Now my issue is I can't see $(1)\to(2)$ and $(2)\to(1)$; $(1)$ basically tells me how the particle should be distributed and $(2)$ tells me how two distribution adds up.

$\endgroup$
  • $\begingroup$ I don't require a rigorous mathematical proof, only slight intuition on why should this follow. $\endgroup$ – Manish Kumar Singh Sep 24 '15 at 21:30
2
$\begingroup$

To me it seems you are a little confused on the basics, so I think going over some of the basics will do more to help you than trying to answer your question directly. As a result, first I am going to make a few statements about the basics of quantum mechanics, then I am going to comment on your two statements, and lastly I'll come to your question.

OK, first I want to describe how I would explain what you are discussing in your statements (1) and (2), without making direct reference to what you said.

We will restrict ourselves to the case of a single particle, without spin, that lives on a line (1 spatial dimension). Classically, the state of the particle at a given time is determined by its position, $x$, and its momentum, $p$. If we know $x$ and $p$ at a given time, we can in principal use the equations of motion to compute $x$ and $p$ for all other times.

Quantum mechanically, $x$ and $p$ are not well defined values. Instead there is a "quantum state" of the system, which we can refer to as $|\Psi\rangle$. The quantum state contains all the information about the system.

$| \Psi \rangle$ is an abstract vector in a Hilbert space, in order to talk about specific observable quantities we need to "bring it down to earth" by expressing it in a particular basis (where the basis vectors typically correspond to observable quantities). It is often useful to express $|\Psi \rangle$ in the position basis, the "components" of $\Psi$ in this basis form a function $\psi(x) = \langle x | \Psi \rangle$. Other times it is useful to express $|\Psi \rangle$ in the momentum basis, in which case the "components" are $\phi(p)\equiv \langle p | \Psi \rangle$. A very important result is that $\psi(x)$ and $\phi(p)$ are related, indeed we simply change our basis to go from one representation of $|\Psi\rangle$ to the other.

The "components" of $\Psi$ in these different bases are equal to the probability amplitude $\mathcal{A}$ for a given event to occur. The probability amplitude is a complex number. The probability of an event to occur is given by the norm squared of the probability ampliude, $P=|\mathcal{A}|^2$. So, $\psi(x)$ is the probability amplitude for the particle to be at position $x$ and $\phi(p)$ is the probability amplitude for the particle to have momentum $p$.

The uncertainty principle says that we cannot measure $x$ and $p$ simultaneously. One way this is reflected in formalism of quantum mechanics is that we can talk about the quantum state by giving $\psi(x)$ OR by giving $\phi(p)$; these two functions are not independent (either one suffices to fully determine the quantum state).

This is all standard textbook stuff. Let's get to your statements.

(1) I disagree with your characterization of the uncertainty principle. The uncertainty principle is more about how measurements of position are related to measurements of momentum (in particular we can't know both arbitrarily precisely at the same time). What you call the uncertainty principle is really more a statement about how the particle does not have a definite position, which in the formalism of quantum mechanics is reflected in the fact that $\psi(x)$ need not be infinitely sharply peaked around a certain value of $x$.

I also just want to point out that plotting $\psi(x)$ does not give you the probability distribution for finding a particle at $x$. It is really $|\psi(x)|^2$ you want to plot to get that information. Since $\psi(x)$ is a complex function it contains more information than just the norm. In particular $\psi(x)$ also has a phase, which is crucial in relating $\psi(x)$ with $\phi(p)$.

(2) Even though $P=|\mathcal{A}|^2$, you have to be very careful when you write $\mathcal{A}=\sqrt{P}$. The reason is that $\mathcal{A}$ is a complex number, so the square root of $P$ does not contain all the information . You also need to specify the phase, which you cannot simply infer from $P$.

The phases are very important when you add up two probability distributions. The most extreme example of this, consider an amplitude $\mathcal{A}=\mathcal{A}_1+\mathcal{A}_2$. The surpising fact is that $\mathcal{A}$ can vanish even if neither $\mathcal{A}_1$ or $\mathcal{A}_2$ vanish. The reason is that we can have $\mathcal{A}_1=-\mathcal{A}_2$. If we were adding probabilities (or their square roots) directly, this could never happen because probabilties are always positive. But amplitudes can have any phase, in this case the two amplitudes have a relative phase of $e^{i\pi}=-1$.

An important time when you add amplitudes together is when you change the basis! As I have said, $\psi(x)$ and $\phi(p)$ are related by a change of basis formula. This formula requires you to add amplitudes in a specific way. The phases in $\psi(x)$ are crucial in correctly constricting $\phi(p)$. In other words, $|\psi(x)|^2$ does not contain all the information in the full quantum state $|\Psi\rangle$. One consequence of this is that knowing $|\psi(x)|^2$ (just the amplitude) does not suffice to tell you $\phi(p)$ or $|\phi(p)|^2$, but knowing $\psi(x)$ (amplitude + phase) does.

Lastly, let me comment on your question. I basically disagree with the way you have worded both (1) and (2), so I don't agree that (1) should imply (2) or that (2) should imply (1). I think one takeaway message, however, is that you have to be very careful about distinguishing probability amplitudes from probabilities. The fact that probability amplitudes are copmlex numbers is crucial: a lot of information is stored the relative phases of different amplitudes. It is a mistake to try to phrase everything in quamtum mechanics purely in terms of probabilities, rather than probability amplitudes.

$\endgroup$
  • $\begingroup$ Andrew: "Quantum mechanically, $x$ and $p$ are not well defined values." -- But you keep referring to $x$ and $p$ (values) in your answer nevertheless. Rather: considering a suitable state $\Psi$, both $\psi[~x~]$ and $\phi[~p~]$ have non-zero standard deviations $\sigma$. "The uncertainty principle says that we cannot measure $x$ and $p$ simultaneously." -- Referring to one trial, right. That's due to how to measure such values, i.e. due to operators $\hat x$ and $\hat p$. Consequently the relation between $\psi[~x~]$ and $\phi[~p~]$. $\endgroup$ – user12262 Sep 25 '15 at 5:40
  • $\begingroup$ Fair enough about your first comment, maybe a better way to say what I meant was that the state is not determined by numerical values of $x$ and $p$, unlike in classical mechanics. As far as the uncertainty principle (UP) goes, of course I agree with you. I wanted try to keep things relatively simple, so I didn't want to get into talking about operators and commutation relationships. The main point I wanted to make about the UP is that it's about limitations making measurements of $x$ and $p$ simultaneously, whereas the OP was under the impression that the UP is just about uncertainty in $x$. $\endgroup$ – Andrew Sep 25 '15 at 5:49
  • $\begingroup$ Andrew: "that the state is not determined by numerical values of $x$ and $p$, unlike in classical mechanics." -- Well, I don't know much about classical mechanics; but your discussion of $\psi$ and $\phi$ shows nicely that the characterization of a state can require some function (or distribution) rather than only a single value. "I didn't want to get into talking about operators and [...]" -- If you want to explain "limitations [on] making measurements" you'll need to get into "how to measure", and specificly the role of $p$ and $\hat p$, too. Unfortunately, our OP might be overwhelmed. $\endgroup$ – user12262 Sep 25 '15 at 6:11
  • $\begingroup$ The uncertainty principle is not about position and momentum only; it is about any pair of non-commuting observables. Also, every quantity does not have precise values before being measured (no matter their distributions), because the entire point of quantum mechanics is exactly that measurements determine the values of the observables after being performed. $\endgroup$ – gented Sep 25 '15 at 7:20
  • $\begingroup$ Yes I know :-) that's why I specified that I was talking about a single spineless particle in one dimension. It's good these clarifications are being made so that people reading the comments can get into the subtleties. I intended my answer to be very basic, really the only point I was trying to make is that one should distinguish carefully between probability amplitudes and probabilities, which seemed to me that it was a major source of confusion for the OP. $\endgroup$ – Andrew Sep 25 '15 at 7:26
0
$\begingroup$

Plotting that probability distribution is what my wave function $ψ(x)$ is.

No the $ψ(x)$ is a complex valued function. The probability density is given by $ψ(x)ψ(x)^* $, i.e. the complex conjugate squared with the original.

For the second point, this $P$ and its square roots are meaningless; there are no two independent probabilities; there is one quantum mechanical problem with a $ψ(x,y)$ ( $x,y$ on the screen) solution.

The $ψ(x,y)$ is a complex valued function for the solution of the problem "an electron scatters off two slits". This $ψ(x,y)$ when complex-conjugate-squared for these specific boundary conditions shows the interference pattern of the sinusoidal functions in its nature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.