Evolution of the wavefunction

[dyn]

Hi .

For a system such as an infinite well or a harmonic oscillator if the energy is measured and it returns a value , say E₁ corresponding to the ket | 1 > then this evolves according to exp( -iE₁t/ħ) | 1 >.
So this means that for any time >0 a measurement of the system will always give the value E₁ ?
If I'm right so far , what does it mean to say the wavefunction evolves according to the Schrodinger equation ? As the system always stays in the state | 1 > ?
Thanks

 

[Nugatory]

So this means that for any time >0 a measurement of the system will always give the value E₁ ?

Yes. This isn't surprising; it's just conservation of energy and an isolated classical system will behave the same way. (Note that the solution only takes on this particular form if the Hamiltonian is independent of time. A time-varying Hamiltonian doesn't necessarily conserve energy.).

If I'm right so far , what does it mean to say the wavefunction evolves according to the Schrodinger equation ? As the system always stays in the state | 1 > ?

The wave function is changing because of the complex exponential factor. That's not so interesting for this particular state because only the phase is changing and that doesn't change the result of an energy measurement.

But suppose you had measured the position of the particle instead? Then the post-measurement wave function will be a superposition of many different energy eigenstates, and as time passes their respective phases will change relative to one another. This won't change the expectation value of the energy, but the expectation value of the position observable will change - the wave function is describing a moving wave packet with constant energy. There's a good writeup at https://ocw.mit.edu/courses/nuclear...ng-2012/lecture-notes/MIT22_02S12_lec_ch6.pdf - section 6.2 is the part we need.

 

[dyn]

Thanks for the reply. You even guessed my next question about position !
On a related note - in a hypothetical situation why can't a particle be placed in an infinite well at a specific position in a stationary state ? This would imply it had zero energy and Δx and Δp would be zero. I know that this can't happen but in practical terms why can't a particle be placed at rest in a definite position ?

 

[ZapperZ]

Thanks for the reply. You even guessed my next question about position !
On a related note - in a hypothetical situation why can't a particle be placed in an infinite well at a specific position in a stationary state ? This would imply it had zero energy and Δx and Δp would be zero. I know that this can't happen but in practical terms why can't a particle be placed at rest in a definite position ?

What is supposed to hold it there?

In practice, the onLy way you can do this is to narrow down the width of the infinite well to approach zero. But what happen to the energy eigen values when you do this?

Zz.

 

[Vanadium 50]

I know that this can't happen but in practical terms why can't a particle be placed at rest in a definite position ?

What is supposed to hold it there?

Even classically - what's supposed to hold it there?

 

[dyn]

Classically if a particle is placed at rest and no external forces act on it then surely it remains at rest ?

 

[PeterDonis]

in a hypothetical situation why can't a particle be placed in an infinite well at a specific position in a stationary state ?

Because there is no such state. There is no state that is at a specific position and is also a stationary state (i.e., a state that does not change with time).

 

[dyn]

in the classical world can I theoretically place a particle at rest at a definite position ?

 

[PeterDonis]

in the classical world can I theoretically place a particle at rest at a definite position ?

In the case where there are no forces acting, yes.

 

[Vanadium 50]

if a particle is placed at rest and no external forces act on it then surely it remains at rest ?

OK, but that requires it to be exactly at rest with exactly zero net force on it. If either is not true (always the case in real life), your object will roll around the potential well. So even classically, this doesn't happen.

 

[PeterDonis]

that requires it to be exactly at rest with exactly zero net force on it.

You're correct that this is not possible in a practical sense; but it is still true that such states exist, theoretically, in classical mechanics; whereas, as I noted before, no such states exist, even theoretically, in quantum mechanics.

 

[Vanadium 50]

That is correct - in QM no such states exist. Classically, it is a set of measure zero.

 

[dyn]

So in QM can I place a particle at a definite position with no force acting on it ?

 

[Vanadium 50]

Yes, but you would have infinite uncertainty on its momentum, so you would not know where it would be in the future.

 

[dyn]

So I can place a particle at a definite position. Its hard to understand why I can't place it there at rest

 

[ZapperZ]

So I can place a particle at a definite position. Its hard to understand why I can't place it there at rest

Because this isn’t classical mechanics. The “rules” are different. That’s the whole point!

Zz.

 

[bhobba]

So I can place a particle at a definite position. Its hard to understand why I can't place it there at rest

The Heisenberg Uncertainty principle - know the position exactly and its momentum (if you were to measure it) is totally unknown and will scoot off elsewhere - heuristically that is - quantum particles do not scoot or anything like that - unless observed what they are doing the theory doesn't say - so your guess is as good as mine.

Now if you want to understand why QM - that is a very very deep question - it requires a thread of its own - but the following may help:
https://www.scottaaronson.com/democritus/lec9.html

Thanks
Bill

 

[dyn]

The Heisenberg Uncertainty principle - know the position exactly and its momentum (if you were to measure it) is totally unknown and will scoot off elsewhere - heuristically that is - quantum particles do not scoot or anything like that - unless observed what they are doing the theory doesn't say - so your guess is as good as mine.

That's what I was trying to understand. because surely I could place it at rest in a definite position. So for a split second I would know its position and momentum but after that its gone and I could only measure its position or momentum. Also , if I place it at rest ; is that the same as measuring its momentum ? QM is very confusing !

 

[PeterDonis]

surely I could place it at rest in a definite position. So for a split second I would know its position and momentum

No.

You can place it in a definite position. You cannot place it in a definite position at rest, because there is no such state in quantum mechanics. If you know its position, you do not know its momentum, even for a split second.

if I place it at rest

You can't. See above.

is that the same as measuring its momentum ?

No, placing it at a particular position is certainly not the same as measuring its momentum.

QM is very confusing !

Yes, which means you need to be extra, extra careful in how you state things, in order to avoid confusing yourself even further.

 

[bhobba]

That's what I was trying to understand. because surely I could place it at rest in a definite position. So for a split second I would know its position and momentum

That is a problem with your reasoning - placing it there means you know its position to the accuracy of the width of the well - it says nothing about momentum - in QM you can have things like tunneling etc:
https://en.wikipedia.org/wiki/Quantum_tunnelling

In fact this is the standard particle in an infinite well problem:
https://en.wikipedia.org/wiki/Particle_in_a_box

Doesn't matter how small the width of the well is for something like an electron - which as far as we know has no size - all that is going to happen is it will do crazy things - but we know its position to the accuracy of the width of the well. It certainly is not at rest in the normal sense.

but after that its gone and I could only measure its position or momentum. Also , if I place it at rest ; is that the same as measuring its momentum ? QM is very confusing !

QM is confusing - the rules of QM forbid you placing a particle in a very small width container and it being at rest at the same time - you know position to good accuracy but as to momentum and other stuff - its complicated. And if you continuously observe its position, you have the so called Zeno effect:
https://en.wikipedia.org/wiki/Quantum_Zeno_effect

Thanks
Bill

 

[Nugatory]

That's what I was trying to understand. because surely I could place it at rest in a definite position.

Imagine how you would place a small classical object at rest: you'd start by pinching it between your finger and thumb to stop it from moving. That's equivalent to reducing the width of the well to zero, raising the question that ZapperZ asks in post #4 above: what does this mean for the energy eigenfunctions?

That word "surely" would work if this were a classical particle, but it's not and you cannot place the particle at rest at a definite position.

 

[dyn]

Thanks guys !

 

[PeroK]

Thanks guys !

It's also worth noting that the HUP is a statistical law. It says nothing about any specific measurements but about the expected statistical variance of those measurements for a particle in a given state. In principle, therefore, you could analyse your problem as follows:

1) Localise a particle to a very small region. This puts the particle in a "localised" state. Let's call this state $\psi$.
2) Measure its momentum. This represents a measurement of momentum for a particle in state $\psi$. The HUP says nothing about what value of momentum you will get; nor how precisely you can measure momentum.
3) Repeat the process. Put the particle back in state $\psi$ and measure its momentum. This time you get a different value of momentum (it probably won't be the same as the first time).
4) Repeat the experiment again. This time you get another momentum measurement
5) And so on.

After a large number of experiments you can calculate the average value of $p$ and the standard deviation $\sigma_p$ for all the momentum measurements for a particle in state $\psi$. The process of localising the particle would have resulted in some small variance in the measured position of position $x$. Call this $\sigma_x$. The HUP says that:

$\sigma_x \sigma_p \ge \frac{\hbar}{2}$

Now, just in case you want $\sigma_x = 0$, then there are distributions with infinite standard deviation, so the HUP is not violated then either. And, in fact, states with $\sigma_x = 0$ are not physically possible for a reason not directly connected with the HUP: the states are not normalisable.

In one sense, you have done what you wanted to do. You created a highly localised state and then measured the particle's momentum with high precision. But, what you found was that when you repeated the experiment the momentum took on a wide range of values. This meant that before you actually measured the momentum (in each case) you had no accurate prediction of what momentum you would get. This is another way to describe "not knowing the momentum of a particle". You did not know what momentum measurement to expect for a particle in state $\psi$.

Finally, after you have measured the momentum, the particle is no longer in the state $\psi$. It is now in the state $\phi$, say, which has a well-defined momentum (to within the limits of your experimental error). In the state $\phi$, therefore, we have a small $\sigma_p$. The HUP says that $\sigma_x$ must be large for state $\phi$. That implies that an infinite well of arbitrary small width must be physically impossible, which it is. If you imagine a very large potential well (there is no such thing an truly infinite well). As you reduce the width of the well, the expected energy of the particle increases without limit. At some point, therefore, the particle must have enough energy to tunnel out of the well and, when you look for it, it may be in the well (localised), or it may have tunnelled out and (unlike the classical particle) escaped your attempts to confine it!

When you analyse things like this, it doesn't seem quite so inexplicable. To me, at least.

 

[PeroK]

PS Some calculations never hurt. In the infinite square well of width $a$, the ground state (lowest energy) has:

$E = \frac{\pi^2 \hbar^2}{2ma^2}$

Note: as you confine a particle, then its minimum energy goes up.

Since there is no potential energy inside the well, a measurement of momentum must return $\pm \frac{\pi \hbar}{a}$. And, it returns each with equal (50-50) probability.

So, in another sense, you do know the momentum of the particle - it has only one possible value - except, you don't know which direction it will have!

The mathematical point is that equal probability measurements of $\pm \frac{\pi \hbar}{a}$ give a standard deviation of:

$\sigma_p = \frac{\pi \hbar}{a}$

If $a$ is small then you, loosely speaking, "know where you will find the particle is before you measure it". But, it could be anywhere in the well; and, in fact, it turns out that we have:

$\sigma_x = \frac{a}{\sqrt{12}}\sqrt{1 - \frac{6}{\pi^2}} \approx \frac{a}{2\pi}(1.1)$

This is what it means, mathematically, for the particle to be localised within a well of width $a$. All the measurements for $x$ are within $a$ of each other and they have a small standard deviation, proportional to $a$.

We can then see that the HUP holds for the ground state of the infinite well:

$\sigma_x \sigma_p \approx \frac{\hbar}{2}(1.1) > \frac{\hbar}{2}$

Finally, note that we really knew "everything" about the particle: its energy, approximately its position and exactly its momentum. The only thing we didn't know before we measured it was the direction of the momentum: plus or minus. In this case, that's what the HUP boiled down to. All this loose talk about "not knowing the position and momentum at the same time" wasn't really appropriate for the infinite well. Any analysis based on a woolly idea of what this really meant was futile. All we didn't know really was the direction of the momentum. And that's what the HUP was telling us in this case.

 

[PeterDonis]

the HUP is a statistical law

If you take it to be about measurement results, yes. But you can also take it to be about states: the HUP puts limits on how "close" any state can be to being a simultaneous eigenstate of non-commuting observables.

 

[PeroK]

If you take it to be about measurement results, yes. But you can also take it to be about states: the HUP puts limits on how "close" any state can be to being a simultaneous eigenstate of non-commuting observables.

Yes, in many ways I prefer that view myself: that the HUP is about states and not measurements. But, the issue at hand was what happens if you do try to measure things.