# Is it possible to derive HUP from SE?

1. Apr 1, 2009

### sokrates

Is it possible to derive Heisenberg's Uncertainty principle just using Schrodinger Equation?

If yes, is there any source that you can refer me to?

2. Apr 1, 2009

### quZz

Uncertainty principle can be derived without SE =) just from the commutation relations of coordinate and momentum operators

3. Apr 1, 2009

### sokrates

But the question is can it be derived using SE only!

Because SE is deterministic, in the sense that, the time evolution of a state evolves deterministically ...

And since uncertainty relation requires an external (environmental) input to the system, a one particle Hamiltonian cannot capture that.

So if you have a one-particle SE, you should not be able to derive HUP in principle...

So the point is, is it possible?

4. Apr 1, 2009

### clem

The SE comes from saying that x is the Fourier transform variable for p.
Then the HEP is just the relation between widths for any Fourier transform.

5. Apr 1, 2009

### sokrates

I believe SE comes from itself ( it is fundamental )

Both historically (the way Schrodinger derived it)
and pedagogically.

So at least adopting the Schrodinger picture, we must take SE to be fundamental.

I think I still don't see the it clearly - the way you can justify HEP using SE.

I think it's added on top.

6. Apr 1, 2009

### JustinLevy

What do you consider the Schrodinger equation?
Please write it out given a single particle in an arbitrary potential V. If you are choosing a basis (such as the position basis), make that explicit. I think the problem here may be that you are not aware you are choosing a basis, and that you cannot have a state that is simultaneously an eigen state of momentum and position... this comes from the experimental fact that hbar is non-zero.

If what you consider the Schrodinger equation has a term like $$-\frac{\hbar^2}{2m}\nabla^2$$ then you are already using what clem commented about above.

7. Apr 1, 2009

### sokrates

$$(i \hbar \frac{\partial}{\partial t}+qV) \Psi = \frac{1}{2m}(\sigma \cdot (\overline{p}+q\overline{A}))^2 \Psi$$

is what I consider the Schrödinger equation ( Derived from Dirac equation to "naturally" include spin.)

A= vector potential
V= scalar potential
sigma = Pauli spin matrices in vector form

So how do I exactly derive the uncertainty principle ==== starting from this equation ====

8. Apr 1, 2009

### StatusX

The Schrodinger equation has to do with dynamics, while the uncertainty of an observable is a kinematic property. In other words, to know the uncertainty of some observable, we only need to know the state, not how it evolves. So the Schrodinger equation has nothing to do with the HUP.

9. Apr 1, 2009

### isabelle

The problem is that without commutation relations, V, A, an p are operators that you don't know anything about. In order to get the structure of QM we need the fundamental commutation relation [x,p] = i.

10. Apr 1, 2009

### sokrates

This verifies my initial conjecture then.

Indeed, it is added on top.

StatusX - it is very clarifying thank you

11. Apr 1, 2009

### Ben Niehoff

From the time-varying wavefunction we can define the probability current. From the expectation that the "velocity" of the probability current ought to be equal to p/m, we can compare to the S. equation and then deduce the form of the momentum operator, p.

Once we have p, we can evaluate the commutator [p,x]. Using this, we can construct the HUP.

Actually, come to think of it, I don't think you need the S. equation in there at all. Just the probability current.

In fact, one can "derive" the Schrodinger equation by starting from the probability current, identifying the operators p and x, and then rewriting the Hamiltonian in terms of these operators. Voila!

12. Apr 1, 2009

### JustinLevy

If you leave this discussion considering that it is "added on top", then I feel you may have missed some of the point here.

An equation by itself is meaningless. As isabelle tried to explain to you, without more information, you can't specify what those operators are based on any classical concepts.

Indeed, saying "states evolve according to the Schrodinger equation" (or that the Hamiltonian describes how the wavefunction, not the density etc, evolves in time ... or yet other ways of wording this) is only one of the postulates of non-relativistic quantum mechanics. You need ALL the postulates to really have a functioning theory. You can derive the HUP from these postulates.

So yes, you are correct that Schrodinger's equation is not sufficient ... but not for the reason you seem to think.

13. Apr 1, 2009

### sokrates

An equation by itself is not meaningless.

Schrödinger equation predicted the observed spectra of Hydrogen atom (firstly) and many others...

Just by itself. No commutators, no Hilbert space.

This is the first point.

Secondly, I asked whether Schrödinger equation was sufficient to derive Heisenberg's Uncertainy Principle ('by itself') or not and StatusX was right on the spot in his/her answer.

I think I am pretty sure that I have the answer to the question I asked. But thanks for your concern.

14. Apr 1, 2009

### Staff: Mentor

Here's an outline of one way to get the HUP using only properties of the Schrödinger wave function. You can find the detailed steps in textbooks and probably on the Web if you look hard enough. I don't know if this meets your criterion of "just using the Schrödinger equation".

1. Plane-wave free-particle solutions of the SE, with a precise momentum, have the form

$$\Psi = A e^{i(px - Et)/\hbar}$$

These solutions extend to infinity in both directions in x and hence are not normalizable.

2. To get normalizable free-particle solutions, you have to superpose (add) waves with different momenta and amplitudes, by integrating:

$$\Psi = \frac{1}{\sqrt{2 \pi \hbar}} \int^{+\infty}_{-\infty} {A(p) e^{i(px - Et)/\hbar} dp}$$

This is basically a Fourier integral, whose properties are well-known from the theory of Fourier analysis.

3. Define the uncertainties in terms of expectation values of x and p:

$$\Delta x = \sqrt{<x^2> - <x>^2}$$

$$\Delta p = \sqrt{<p^2> - <p>^2}$$

The product $\Delta x \Delta p$ depends on the form of the amplitude function A(p) in the integral shown above. It is possible to show, from the properties of Fourier integrals, that the smallest possible value of this product, for any A(p), is $\hbar / 2$. This gives the Heisenberg Uncertainty Principle:

$$\Delta x \Delta p \ge \frac{\hbar}{2}$$

15. Apr 1, 2009

### sokrates

This definitely looks promising. Thank you very much for taking the time to outline it.

But it surpassed my knowledge at the moment, I'll probe further if I run into difficulties.

I am okay up to step 3 (excluded) and I can't follow the reasoning given in the end.

16. Apr 2, 2009

### JustinLevy

No, that is incorrect.
Look at what you wrote and claimed was the Schrodinger equation above. You cannot even use that to calculate anything until you can specify the operators in some basis. Once you've taken momentum to be $$i\hbar\nabla$$ in the position basis, you have already used some of the essential pieces that lead to the HUP.

I'm not arguing against StatusX's statements, but based on your statements you seem to be taking away only partial information, and/or wrong interpretation, from this discussion. I'm trying to help, but I seem to be failing to communicate clearly.

This is essentially the answer given to you previously by several people. Since you already understand steps 1 and 2: Notice #1 is written in the position basis. How do you know that #1 is an eigen state of the momentum operator? How do you know that in #2, a fourier transform relates the momentum and position basis?

If you can understand these, then you see the features that leads to the HUP.

Also notice, that the way the outline was written, step three also relies on the measurement postulate (so you can calculate expectation values). ALL the postulates of a theory are important to make predictions.

It seems you are taking some things as "given" so that you can view the Schrodinger equation as standing on its own. The problem here is that these "givens" you are glossing over actually contain important information related to this problem. If you consider these 'givens' as wrapped up in the Schrodinger equation as a single package, then yes, you can derive the HUP from 'just' the Schrodinger equation ... but I would strongly suggest it is better to be explicit about these givens.

17. Apr 2, 2009

### sokrates

I am not claiming that Schrodinger Equation. You can find that formula in textbooks if you don't recognize it.

When Schrodinger invented his formula, as I tried to explain to you, there was no established "quantum-mechanical formulation" yet. Without operators, bases (all this spectacular terminology you persistently bring up) the prediction was evident. It's a simple equation that has a strong correlation with reality.

If you want to drag this debate onto a philosophical frame, I am not the right person. One could discuss what an equation "means" forever... But for me, (and I believe for many other working scientists) it has a clear-cut definition. And Schrodinger Equation is one prime example. It stood by itself 80 years ago, and is still considered fundamental.

I have taken the part that I need. Thank you for your concern though.

Frankly, I tried to say "I don't understand the details so let me get back to you". And no, it was not the answer given by several people. The response you are mentioning started from the SE, solved it and presented further arguments. Apart from the conceptual issue StatusX brought up, nobody before gave a clear pathway from SE to HUP.

18. Apr 2, 2009

### jensa

The OP seems to get a little irritated so it is probably not a good idea to post a reply but I must say that I agree with what JustinLevy said. The Schrödinger equation as a differential equation alone does not have any physical meaning. Physics only comes in when you attach meaning to the wave-function you are solving it for. So in order to calculate something of value you must first specify: How do I obtain physical observables from this wavefunction

It is in this step the HUP pops up. If you have not made any statements about how to obtain observables from your wavefunction you obviosly cannot make any statement about observables, regardless if you have an equation for the wavefunction.

I think the point JustinLevy makes is very relevant to your original question and not just some attempt at opening a philosophical debate.

Last edited: Apr 2, 2009
19. Apr 2, 2009

### sokrates

Thanks for the message.

My response is :No I don't think so. ( I think I have said enough to convey what I really mean by "meaningful" equations in my previous posts. I believe it's clear enough if you don't take it out of its context)

and I am not going to pollute this thread with any more non-technical aspects of the problem.

Last edited: Apr 2, 2009
20. Apr 2, 2009

### vanesch

Staff Emeritus
Just my 2 cents. I agree fully with what StatusX wrote, in that the HUP is part of the kinematical description, while the SE is what defines the dynamics. However, I suppose people have different opinions on this, depending on how they got heuristically to a suggestion of what a wavefunction is, what the SE is and so on: there are different ways to suggest how to do so, and different textbooks have different approaches.

But in general, you first set up your Hilbert space by defining the spectra of a complete set of commuting observables (for instance, X, Y and Z with spectra (x,y,z) all real 3-tuples for a featureless point particle). You then introduce their conjugate observables by *requiring* the HUP: [x,px] = i hbar etc... which turns out to define their eigenstates.
And from there on, you try to guess a sensible hamiltonian (by analogy with a classical system, or by inspiration, or whatever), which fixes the time evolution of your states through the SE.

In this approach, the HUP is a fundamental part of the definition of the kinematics, even before we think of any dynamics and hence a SE.

However, as others suggested here, several textbooks take a more suggestive approach by first suggesting that a particle in uniform motion must correspond to something like a plane wave, and then try to deduce how one could get something that is "position" and "momentum" from that wave, and how we could then set up a first order equation that takes these suggestions, and gives the original plane wave as a solution.

21. Apr 3, 2009

I would like to add that HUP does not really need to be *required* with this approach. One of the postulates of QM is that observables are represented by Hermitian operators which then define the eigenstates. The observables quite naturally do not need to commute with one another and if you define the momentum operator as a generator of translations, i.e. an operator which takes position eigenstate $$|x\rangle[/itex] to $|x+dx\rangle$, the position and momentum operators will naturally satisfy the HUP. In a similar way the Hamiltonian can be thought of as the generator of time-translations which, if you follow this road, automatically leads you to the Schrödinger equation (with a as of yet undetermined Hamiltonian). So with this approach the Schrödinger equation really does not take a fundamental role. Yes. My intention with the previous post was to emphasize that to say anything meaningful one must make these "interpretations" as to what the wave-function represents (probability amplitude) and what we would expect to be velocity (or momentum). I.e. how to get physical observables. It is this stage which will give you the HUP principle. To me it seems like the approach to start off by defining Hilbert space and operators (observables) simply formalizes this (vague) suggestive approach. Last edited: Apr 3, 2009 22. Apr 3, 2009 ### vanesch Staff Emeritus Indeed. You can define the conjugate operator as being the generator of translations of the original operator (which will lead you then to the commutation relations), or you can start from the classical conjugation relations (the Poisson brackets) which then lead you to require the right commutation relations. Both are mathematically equivalent. Indeed, the SE is nothing else but the differential expression that the Hamiltonian is the generator of time translations. 23. Apr 3, 2009 ### JustinLevy I am not challanging that form of the Schoridinger eq. The point is, you cannot derive the measurements you claim from that equation alone. For that equation contained several operators, and to put it into a useful form you would need to write what those operators are in some basis, which you need additional information/equations to do. ... as soon as you have written the momentum operator as $$i\hbar\nabla$$, you have incorporated details intimately related to your question. Historically, Schrodinger didn't even know what the "wave" he was solving for was. Several papers later, he was incorrectly suggesting it could be the electron density. It was Max Born that soon figured out that the appropriate choice was to consider $$\Psi^*\Psi$$ a probability density. I assume your question wasn't meant to be interpreted as regarding the consequences of the original historical Schrodinger equation and ideas. The useful concepts are instead the Schrodinger equation as it is understood in modern quantum mechanics. The form of Schrodinger's equation you are probably thinking of, is not the one you wrote, but the one written in the position basis. Whether or not you were aware of this choice of basis does not change the fact that a choice was made. And even more importantly, how one knew what the momentum operator was in this basis gave imporant additional information. This is not a philosophical issue. As you can see from discussion here, there are many ways to build up quantum mechanics. I don't want to get into all of that, so let's just look at two views, that have been used so far in this discussion, of what the SE encompasses. If you consider Schrodinger's equation for a single particle with no internal degrees of freedom, and in an arbitrary scalar potential to be: $$i\hbar \frac{\partial}{\partial t} \Psi = \hat{H} \Psi$$ where $$\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})$$ then it seems quite reasonable to say the SE is not enough to lead you to the HUP. (As StatusX mentioned earlier, because the SE gives you dynamics of the state, but HUP is related to the state itself ... not the dynamics.) If instead, you consider Schodinger's equations for a single particle with no internal degrees of freedom and in an arbitrary scalar potential to be: $$i\hbar \frac{\partial}{\partial t} \Psi(x,t) = H \Psi(x,t)$$ where $$H = -\frac{\hbar^2}{2m}\nabla^2 + V(x)$$ then notice the additional information you have encoded into the equation. If you consider that first term the KE, then we can extract from the SE what the momentum operator is. It is this extracted information (nothing else from the SE is really needed for the remaining steps), along with postulates on what Psi and what a measurement made on Phi is, allows us to calculate $\Delta x$ and [itex]\Delta p$$ for a state. It is possible to obtain the commutator [x,p] from the previous information and then prove the HUP. If you consider this as "from the SE alone", then so be it ... but realize it was the implicit givens that you are considering lumped in with the SE that allows this derivation. It was not the equation of the dynamics that led to it, it was those other postulates / implicit givens that you seem to insist on sweeping away as philosophical fodder.

It is not philosophical fodder, but actually the crux of the issue in responding to your question.

Last edited: Apr 3, 2009
24. Apr 3, 2009

### xlines

I believe it is possible. Should potential be in form such that SE is of a Sturm - Liuville 's type, on physical grounds we introduce such boundary conditions that will give you a spectrum of square integrable functions that form a base in Hilbert space IIRC. Since HUP is always valid in Hilbert's space, I think this is the way to proceed . I think S-L theory is the connection between Schroedinger's and Heisenberg's formulation of QM. If I may note, there is not much of a physical content to this : you will not find out something deep about HUP, but you'll stuck with question "Why is H. space 'good enough' to model QM?"

25. Apr 3, 2009

### xlines

Very descriptive answer and precise insight. Well done!