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Quantum mechanical observables of a system are represented by self - adjoint operators in a separable complex Hilbert space $\mathcal{H}$. Now I understand a lot of operators employed in quantum mechanics are unbounded operators, in nutshell these operators cannot be defined for all vectors in $\mathcal{H}$. For example according to "Stone - von Neumann", the canonical commutation relation $[P, Q] =-i\hbar I$ has no solution for $P$ and $Q$ bounded ! My basic question is :

  • If the state of our system $\psi$ is for example not in the domain of $P$ (because $P$ is unbounded), i.e., if $P\cdot \psi$ does not, mathematically, make sense, what does this mean ? Does it mean we cannot extract any information about $P$ when the system is in state $\psi$ ?
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3 Answers 3

It doesn't really mean anything bad at all, in spite of the confused answers one sometimes receives about this matter. The big point is: there is no very direct physical meaning to applying an operator $H$ or $Q$, even if it is an observable, to a wave function $\psi$, even if $\psi$ is in the domain of $H$ or $Q$. (See also Operator versus Linear Transformation, http://physics.stackexchange.com/a/18933/6432 .

In my opinion, the only operator with very direct physical meaning is the time evolution operator $e^{iHt}$ for some $t$, or for all $t$, and as you seem to realise, by the Stone-von Neumann theorem, this exists even for discontinuous $\psi$ even for unbounded $H$.

But, also this is my opinion, the exact definition of Hilbert space is not that important and many mathematical physicists who worry (too much) about whether discontinuous wave functions, which are obviously outside the domain of $H$, are physical or not, manage to formulate Quantum Mechanics just fine on a dense subspace of a Hilbert space. Other, with the opposite worry, again, in my opinion, worrying too much about something which isn't really that bad, talk about rigged Hilbert spaces or nuclear spaces in order to somehow include infinite norm states and exclude non-differentiable wave functions....see Sudbery, Quantum Mechanics and the Particles of Nature. for this.

There is a mathematical reason for thinking that the exact choice of what domain or what space, whether the Hilbert space, or a subspace of smooth wavefunctions inside the Hilbert Space, or the Schwartz space of rapidly decreasing and smooth functions, or an extension of the Hilbert space to include some dual objects such as distributions, to think of as the domain of these operators, is....unimportant because no matter what space you choose, you get the same physics, and that reason is a theorem of Wilfrid Schmid, Henryk Hecht, and Dragan Milicic, or at least somebody or other, which says that if you have a semi-simple Lie group operating on the space, (if the QM is going to made relativistic you eventually have to assume the Lorentz group acts) and if the representation has a finite composition series (this excludes quantum fields), then the algebraic structure of that rep. is independent of which space you consider (within broad limits). Earlier versions of such results were proved by Nelson, Garding, and Harish-Chandra, and gave a very pleasant surprise to Hermann Weyl and everyone else involved at the time...

Now very concretely, even if $\psi$ is not in the domain of $H$, or any other observable $Q$, it is still true that the Hilbert space has a Hilbert basis of eigenstates of $H$ and hence even if $\psi$ is horrible and discontinuous and everything bad, it still holds that

$$\psi = \lim_{n\rightarrow \infty}\sum_{i=0}^{i=n} c_i v_i$$

where $v_i$ is the normalised eigenstate of $H$ (or $Q$) with eigenvalue, say, $\lambda_i$, and $c_i$ are complex numbers, the so-called Fourier--Bessel coefficients of $\psi$, and the convergence is not pointwise but in the L$^2$ norm. Now notice: each finite sum $$\sum_{i=0}^{i=n} c_i v_i$$ is an analytic function, if $H$ is hypo-elliptic, as is often true, e.g., the harmonic oscillator, and is at any rate smooth and in the domain of $H$.

And it is still true, by the axioms of QM, that the probability that $H$ (or $Q$) will, if measured, take the value of $\lambda_i$, is $\vert c_i \vert ^2$ whether or not $\psi$ is in the domain of $H$ (or $Q$).

Pedagogically, there is this widespread confusion that an observable, since it is an operator, ought to be applied to a function since it is an operator, but this is just a naive confusion. If anything should be applied to the wave function as an operator, it is the exponential of $iH$, which is always bounded. To repeat: just because $H$ is an operator, and $\psi$ is a function, doesn't mean you should apply $H$ to $\psi$. Although when you can, that may be a useful shortcut, it is not necessary to ever do it, and the axioms of QM, when stated carefully, never ask you to apply $H$ to $\psi$. What they ask you to do is, for the unitary time evolution to apply the exponential of $iH$, and for the Born rule probabilities, expand $\psi$ to get its Fourier--Bessel coefficients. The sloppy way of thinking, which one often sees, works fine for many simple QM problems, but leads to people asking precisely this OP, precisely since it is sloppy. The careful axiomatisation states things the way I formulated them.

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Thank you both ( specially Joseph F. Johnson ) for your contributions! I`m trying to adjust the physicists way of thinking and doing maths as I try to learn physics ! –  Serifo Blade Jan 17 '12 at 0:57
    
Why do you repeat my answer and comments, and at the same time call them confused? Do you just not like the presentation? You aren't saying anything technically wrong regarding the physics (so no downvote), but you are imputing misunderstandings to me that just don't exist. Your answer might also be pedagogically better, I don't know, but it is annoying to be accused of not understanding this stuff, when I just told you the unitary operator is fine, and went into detail about the limiting process you are using. –  Ron Maimon Jan 17 '12 at 4:54
    
If you are willing to accept my edits, which clarify some of your intended meaning, I intended to remove my negative comment in the answer...but I am waiting to see if yu just revert my edits...you see, I have a lot of experience on Wikipedia as an editor... –  joseph f. johnson Jan 17 '12 at 5:17
    
@joseph: I reviewed them, they are fine, they didn't change the meaning, but they seem very minor to me. They aren't changing what I said, but perhaps they make it more clear, I don't know. You changed "mean energy" to "expected energy" and "not relizable with a finite energy reservoir" to "not realizable". I don't know why this makes such a big difference, but if you like it better, why would I revert it? –  Ron Maimon Jan 17 '12 at 5:25
    
I changed your vague, confused, and possibly (for all I know) false statement about not realizabe with finite energy to the precise and true « could never arise from a state with finite average energy by time evolution ». In fact, since time evolution preserves expectations of operators which commute with the Hamiltonian, this is a truism, and puts the whole question on the same footing as saying that a state with energy expectation of 7 calories could never arise by time evolution from a state with energy expectation of 7.3 calories. If the universe has an infinite number of degrees of f –  joseph f. johnson Jan 19 '12 at 2:57

It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.

An example for P is the state

$$\psi(p) = {1\over\sqrt{p^2+1}} $$

This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since

$$ \langle \psi |P P|\psi\rangle =\infty$$

means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.

An analogous $\psi$ for x would be defined by replacing $p$ above by $x$: $$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized, even though zero is its average position. But it has an infinite variance in position--- if you measure the position again and again, you will not have an average deviation. This wavefunction is infinitely spread out, so it is not really useful for describing a particle which is somewhere. It is an idealization, like a plane-wave state.

These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.

See also here: Regularisation of infinite-dimensional determinants

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If it is not an eigenstate, it makes no sense to say it "has X energy" even if you plug in "infinity" for X. Basically, this answer is completely confused. A sum of eigenstates where the sum converges in L$^2$ norm but not pointwise is going to be outside the domain of $H$ but it still has finite probabilities of having a finite energy if observed. –  joseph f. johnson Jan 16 '12 at 2:47
    
I should have stated that even more strongly: such a state will have finite norm, and will, with certainty, have finite energy if its energy is measured, even if it is way outside the domain of $H$ –  joseph f. johnson Jan 16 '12 at 3:23
    
The answer is correct if you understand "infinite energy" to mean "infinite mean energy", as should have been obvious without saying. A state which has finite energy with certain probability does not necessarily have finite mean. Whether you choose to consider the state physical is up to you. –  Ron Maimon Jan 16 '12 at 4:15
    
A) consider editing your post to fix this. Suggestion: the normal word is « expectation », it is really incorrect to use the word « has » before a measurement has been made. B) It's still wrong: the comments about localisation are also undefined, vague, and useless since 1) particles do not have to be somewhere, and 2) QM may as well study wave-like phenomena too. The OP was not about QFT... and C) the vague remarks about not realisable by time evolution are either extremely confused or wrong, too. –  joseph f. johnson Jan 16 '12 at 4:41
    
@josephf.johnson: I agree. I will make the edit. The question was "what does it mean if p|psi> is not normalizable?" The answer is that <psi|pp|psi> is infinite. It does not mean anything too terrible beyond infinite mean energy, and it does answer the question. "Not realizable by time evolution" is absolutely correct--- you cannot make an infinite mean energy state, you can only make an approximated regulated version. –  Ron Maimon Jan 16 '12 at 7:41

Faced with some of the confusing aspects of the « bad » wave functions, such as those which are discontinuous or whose expectations are undefined, some people might wish to adopt the attitude that there is something unphysical about an unbounded operator. This is the opposite point of view to the more common one that the unbounded operators which represent observables are physical but the « bad » wave functions, i.e., those which are normalisable but are not in the domain of the observable, are unphysical.

There are physical reasons against considering unbounded observables, except for the Hamiltonian itself (which, as often remarked, is a case apart from other observables). The analysis of the measuring process by Wigner, and by Araki and Yanase, famously shows that unless an observable $Q$ commutes with the Hamiltonian, no measurement apparatus of finite size can implement exact measurements of the observable. But since observables are supposed to be connected with measurements, this could suggest that such a $Q$ is not exactly physical. AT least this could justify the attitude that the exact properties of $Q$ are not physical.

One could replace $Q$ by a truncated and smoothed or cut-off version, which might avoid this, but then it would be a bounded operator, and it would wreck all the usual calculations of the textbooks making them unmanageable. But it would mean that these « bad » wave functions were no longer so bad: they would have finite expectations and so on.

This was not exactly your question, but it is somewhat relevant to some of the more confusing aspects of unbounded observables, discontinuous wave functions, etc. This line of thought leads to formulating Quantum Mechanics in terms of algebras of bounded operators only, as done by Irving Segal, Streater, Wightman, and many others, and has led to interesting treatments of Quantum Field Theory and Statistical Mechanics by Araki, van Hove, Ruelle, and many others.

These confusing aspects can also be avoided in other ways, which have already been mentioned.

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Thank you for this post, indeed you have captured some of my thoughts ! –  Serifo Blade Jan 21 '12 at 2:56

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