4

Here are the two equations I'm concerned with: $$\Psi = \sqrt{\frac2a}\sin\left(n\frac{\pi x}a\right)$$ $$E = n^{2}\frac{\hbar^2π^2}{2ma^2}.$$ If we have a ball with mass 1 kg, confined in a 1 m infinite potential well, and its kinetic energy is 1 J, then the corresponding $n$ will be very high. As a result we get a very frequently oscillating probability density function. Which essentially says that the probability of finding the ball is pretty much the same everywhere for human eyes/senses.

But why is the probability of finding the ball the same everywhere?

In reality, we would find (by seeing) the ball at one particular place at a particular time.

So what is the probability function telling me? Which probability is equal everywhere (because the probability of finding ball changes with time if it has some energy)?

Buzz
  • 16,031
Rohit Shekhawat
  • 1,305
  • 1
    That expression is for a box of lenght $a$. The position of a ball with 1 J kinetic energy will not remain at rest, every time you look at the ball, it will be in a different position. – Mauricio Apr 19 '23 at 12:10
  • 1
    You can also calculate its mean position and use https://en.wikipedia.org/wiki/Ehrenfest_theorem you will recover the classical behaviour. – Mauricio Apr 19 '23 at 12:13
  • 2
    Such a ball will simply move with a velocity of 0.7m/s. There would be nothing to see. Heisenberg wrote about this around 1927, or so. He used the more interesting case of Rydberg atoms and how one would recover something along the lines of "planetary motion" from that. Mott worked out the math for the "motion" of alpha particles in a cloud chamber in 1929. He showed how multiple weak observations reduce a quantum state to a classical trajectory. – FlatterMann Apr 19 '23 at 12:15
  • 1
    Related: Can quantum mechanics model a classical particle in a box?, in which I discuss the time-evolution of a Gaussian wave packet in a box.. – Michael Seifert Apr 20 '23 at 18:41
  • @MichaelSeifert Thanks, I was looking for exactly that. – Rohit Shekhawat Apr 22 '23 at 06:13

5 Answers5

6

Which essentially says that the probability of finding the ball is pretty much same everywhere for human eyes/senses.

But why is the probability of finding the ball same everywhere?

Firstly, because we are talking about probability, and then because we are talking about quantum mechanics:

  • The probabilistic aspect means that we have the same probability to find a ball in any part of the well. That is, every time we perform our experiment, we find the ball in a different place, but the frequency with which it is found in different regions of equal size (e.g., in the right and left halves of the well) is the same
  • Quantum mechanics means that our measurement is not only probabilistic, but that we cannot perform the measurement twice on the same system, since a measurement collapses the wave function and changes the system state (at least in the Copenhagen interpretation.) Thus, we necessarily work with an ensemble - a large set of wells with balls, all prepared according to the same protocol, and each measured only once. The probability then predicts the averages over the ensemble.

Finally, what might be still confusing here is thinking of a macroscopic object as a quantum one. That is, why a real ball does not behave as a quantum particle? The question has been asked many times in different versions in this community. Here are a few links:
Do double slit experiment with big balls
Will tennis ball produce same interference pattern in double slit experiment if everything was scaled up?
Do physicists distinguish things that can be explained through the movement of matter as opposed to some unseen or non-material forces?

Roger V.
  • 58,522
  • Thanks a lot for the answer. Im concerned about the lack of time factor in probability function here. Can quantum mechanics not tell me where the ball will be after 't' seconds if it was at 'x' position initially, with 'v' velocity? Because classic mechanics seems to do this easily. Why doesn't quantum mechanics reduce to newtonian mechanics in this case? Just tell me if, this is a primitive quantum mechanical analysis ? i.e I will learn more(better) quantum mechanics that will answer my questions. – Rohit Shekhawat Apr 19 '23 at 15:46
  • 1
    @RohitShekhawat in quantum mechanics one can predict what the probability distribution will be at time $t$. Operators of momentum/velocity and position do not commute, so one can neither measure them simultaneously at time $t=0$ nor predict the exact position at a later time. This is known as Heisenberg uncertainty principle. What textbook do you use? – Roger V. Apr 19 '23 at 15:56
  • Currently Im not learning from a book, but I have David J Griffith's quantum mechanics book which will be using after getting a little better hold of the concepts :) – Rohit Shekhawat Apr 19 '23 at 16:32
6

The equation you gave for $\Psi$ is an eigenstate of the system. This means that is has one single energy and also that the probability of finding a particle at a certain position doesn't change in time. The probability density is the square of the wavefunction $|\Psi|^2$ and for eigenstates it doesn't oscillate in time, in fact it is constant. Realistic states are often a sum of multiple of these eigenstates. The probability functions of these states is not constant. The wavefunction can move from left to right and change shape and do all sorts of stuff. Below I showed an eigenstate, $\sin x$, a sum of two eigenstates and finally a cleverly constructed state that is localized in a single spot. This last state is how the wavefunction of your ball would look like: the bigger an object, the more localized your object will become. If your ball has velocity, this peak will move at constant velocity from left to right and bounce when it reaches the edge of the potential. If you want I could make a gif of this.

enter image description here

Note: the scale and normalization of these functions is completely wrong, you should just look at the shape.

AccidentalTaylorExpansion
  • 17,187
  • 4
    Hooooooly Physishit man! That's exactly what I needed to read. Thanks a looot. So, we do not actually have the knowledge of 'exact' energy of the ball, and its actually in some range. And the wave functions corresponding to all those energies combine to form the localised 'wave packet'. Am I right till here? – Rohit Shekhawat Apr 19 '23 at 15:57
  • 1
    Also, which 'n' values are actually part of the 'combined' wave function of the ball? Im mean if n=1000 for 1Joule energy, then do I take all wave functions corresponding to 1 to 1000 'n' value? Or is it like 900 to 1100 'n' value? – Rohit Shekhawat Apr 19 '23 at 16:00
  • 1
    As an example, you can make a state that is 90% in state $n=1$ and 10% in state $n=5$. The combined state would then be a sum $\Psi(x)=0.9\psi_1(x)+0.1\psi_5(x)=0.9\cdot\sqrt{2/a}\sin(1\pi x/a)+0.1\cdot\sqrt{2/a}\sin(5\pi x/a)$. The factors 0.9 and 0.1 should actually be $\sqrt{0.9}$ and $\sqrt{0.1}$, but that is not important. You would say then say this state has energy $E=0.9E_1+0.1E_5$. In the localized wavepacket all the coefficients are the same. So it contains the states $n=1, n=3, n=5$ etc. all with equal probability. – AccidentalTaylorExpansion Apr 19 '23 at 17:20
  • 1
    For the realistic case of such a system see e.g. "Quantum-mechanical motion of Gaussian wave packets on elliptical and hyperbolic Kepler orbits in the Coulomb field" S. D. Boris, S. Brandt, H. D. Dahmen, T. Stroh, and M. L. Larsen Phys. Rev. A 48, 2574 – Published 1 October 1993 and many other publications about Rydberg atoms. This has been demonstrated both theoretically and experimentally. – FlatterMann Apr 19 '23 at 18:19
  • 1
    @FlatterMann evolution of an explicit wave packet made of hydrogenic states is also considered in my answer here on Phys.SE. – Ruslan Apr 20 '23 at 09:31
  • @AccidentalTaylorExpansion By which combinations of 'E' and the corresponding 'n' wave functions, can we get localised 'wave packet'(as your picture shows) which would bounce between the extreme points of our potential well(like a ball between walls)? Can you maybe tell such equation(i.e wave packet bouncing back and forth as a function of time) for 1D potential well which I can use in Desmos? – Rohit Shekhawat Apr 20 '23 at 11:18
  • 1
    @RohitShekhawat Are you familiar with complex numbers? If not, this will be hard to do for you because time dependent solutions involve complex numbers. To satisfy your curiosity you can search for animations with search terms like "schrodinger equation wave packet" or "quantum harmonic oscillator". The last one is a quantum spring, not a particle in a box, but it provides very nice animations. – AccidentalTaylorExpansion Apr 20 '23 at 12:30
  • Ok thanks a lot. I haven't studied time dependent schrodinger wave equations. I will come back later after studying it :) – Rohit Shekhawat Apr 20 '23 at 17:41
2

It would indeed have a homogeneous probability distribution, if one could somehow manage to put the ball into a pure quantum state with a fixed $n$. However, under the conditions you gave, the energy spacing between neighboring levels is about $\delta E \sim 10^{-10}$eV. Given that for an optical photon $\hbar \omega \sim 1$eV, you cannot just "find (by seeing)" anything without seriously messing everything up/entangling billions of different eigenstates together.

Once you do mess/entangle everything up "by seeing", you end up with a random gaussian probability distribution (very narrow) for the ball position corresponding to classical physics.

John
  • 3,481
  • 1
    But we can predict what we would "see" with those 1eV interactions: it would be a slowly moving classical ball. – FlatterMann Apr 19 '23 at 12:19
1

The probability of finding the ball is not the same everywhere. The frequency of oscillation is very high, but the amplitude is exceedingly tiny. It wiggles a tiny amount but overall moves at the classically expected speed.

This is the "classical limit" of quantum physics: quantum physics gives the same results as Newtonian physics at large (i.e. non-microscopic) scales.

Rich006
  • 689
  • 4
    The probability of finding the ball anywhere is a broad distribution for the first measurement. For every (rapid) consecutive measurement afterwards the ball will be highly localized near that first measurement and the trajectory can be predicted with classical physics. – FlatterMann Apr 19 '23 at 12:23
  • @FlatterMann Wow. That is fantaaaaastic. Can you please take a look at my comments on AccidentalTaylorExpansion's answer? I have asked something very similar to your comment here :) – Rohit Shekhawat Apr 19 '23 at 16:03
  • 1
    @RohitShekhawat You can read Mott's 1929 paper about the derivation of alpha-particle tracks from wave mechanics. That's where this argument came from. I didn't invent it. – FlatterMann Apr 19 '23 at 18:15
1

A clear explanation is provided by AccidentalTaylorExpansion's answer. The wave packet( not an energy eigenstate wave function, but a linear combination of some energy eigenstates) moves as if it is a classical particle. This is a time-dependent problem, you cannot get the answer by considering the energy eigenstates by the time-independent Schrodinger equation $H\psi = E\psi$ only, you have to solve the time-dependent Schrodinger equation $i\hbar\frac{d}{dt}\psi=H\psi$. Although you solve TDSE by expanding it with energy eigenstates, they are two different equations. You can consult quantum mechanics textbook by Cohen-Tannoudji about the movement of a wave packet and the information it provides.

Besides, one point can be added. You can find that the possibility that the particle is not located at its classical position is never zero. You will never get a moving delta function. But that concerns nobody because nothing will happen if a ball in a 1D infinite potential well has no specific position. This ball is constrained in a well! It can harm nothing!

A wave function $|\psi \rangle$(, a "state" equivalently,) is an abstract object with a lot of information. You can represent it in the position basis, get a wave function in the position basis $\psi(x)\equiv\langle x|\psi\rangle$, and calculate the possibility of finding it at the position range $[a,b]$ using Born's rule $P([a,b])=\int_{a}^{b}|\psi(x)|^{2}dx$. But you can do these just because the position basis is complete in this problem, not because the observable position $X$ is "physical" in this problem. In classical mechanics, the position $X$ is always a physical observable. You use the pair $(x,p)$ to label a state in Hamiltonian mechanics. However, quantum mechanics tells that this experience is wrong, the position is not necessary when labeling a state.

TOAA
  • 192