# What's wrong with this derivation that $i\hbar = 0$?

Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where $p$ is the eigenvalue of $\hat{p}$. Then, we have $$[\hat{x},\hat{p}] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$$ From the above equation, denoting by $\langle\cdot\rangle$ an expectation value, we get, on the one hand $$\langle i\hbar\rangle = \langle\psi_p| i \hbar | \psi_p\rangle = i \hbar \langle \psi_p | \psi_p \rangle = i \hbar$$ and, on the other $$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle = \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$ This suggests that $i \hbar = 0$. What went wrong?

EDIT: This seemingly little curiosity, as I perceived it initially, evolved into a major problem, as can be seen from the discussion below. Quantum mechanics indeed leads to non-physical results in my opinion and not just to resolvable paradoxes, at that regarding one of its basic claims, the one involving commutators. It seems most of what could be said on this issue has already been addressed below and probably there is no need to continue the exchange here. I would suggest that we discuss it further in the chat.

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I am not sure that I am right but here's a potential explanation: You wrote $$\langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle = \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle$$ but the second term in the RHS shouldn't be that. The operator $\hat{p}$ acting on the bra should be $\hat{p}^{\dagger}$ hence you don;t get the same thing as the first term. –  bra-ket Aug 30 '11 at 7:18
Ok I was wrong since $\hat{p}$ is Hermitian. Sorry. –  bra-ket Aug 30 '11 at 7:28
ganzewoort, I think you need to stop editing - you're making a mess of your question. Your point has already been addressed by the answers. In order to preserve the value of this question for the site I'm going to revert the latest edit, but you can continue to discuss it in chat, and if you can develop a good followup question you're free to ask it. –  David Z Sep 2 '11 at 7:59
By the way, I'll mention for the record that this is not the place to make claims that quantum mechanics is an invalid theory; that falls under the "personal theories" prohibition in the FAQ. People who (think they) get unphysical results out of it are generally misapplying it. If you want to discuss this issue, we can do that, but in the chat room. –  David Z Sep 2 '11 at 8:03
–  Joe Jun 19 '12 at 13:30

Both p and x operators as operators do not have eigenvectors in the strict sense. They have distributional eigenvectors which are only defined in a bigger space of functions than the space of square-normalizable wavefunctions, and which should be thought of as only meaningful when smeared a little bit by a smooth test function.

The normalization for $\langle \psi_p | \psi_p \rangle$ is infinite, because the p-wave is extended over all space. Similarly, the normalization of the delta-function wavefunction, the x-operator eigenvector, is infinite, because the square of a delta function has infinite integral.

You could state your paradox using $|x\rangle$ states too:

$$i\hbar \langle x|x\rangle = \langle x| (\hat{x}\hat{p} - \hat{p}\hat{x})|x\rangle = x \langle x|\hat{p}|x\rangle - \langle x|\hat{p}|x\rangle x = 0$$

because $|x'\rangle$ is only defined when it is smeared a little, you need to use a seprate variable for the two occurences of x'. So write the full matrix out for this case:

$$i\hbar \langle x|y\rangle = x\langle x|\hat{p}|y\rangle - \langle x|\hat{p}|y\rangle y = (x-y)\langle x|\hat{p}|y\rangle$$

And now x and y are separate variables which can be smeared independently, as required. The p operator's matrix elements are the derivative of a delta function:

$$\langle x|\hat{p}|y\rangle = -i\hbar \delta'(x-y)$$

So what you get is

$$(x-y)\delta'(x-y)$$

And you are taking $x=y$ naively by setting the first factor to zero without noticing that the delta function factor is horribly singular, and the result is therefore ill defined without more careful evaluation. If you multiply by smooth test functions for x and y, to smear the answer out a little bit:

$$\int f(x) g(y) (x-y) \delta'(x-y) dx dy= \int f(x)g(x) dx = \int f(x) g(y) \delta(x-y)$$

Where the first identify comes from integrating by parts in x, and setting to zero all terms that vanish under the evaluation of the delta function. The result is that

$$(x-y)\delta'(x-y) = \delta(x-y)$$

And the result is not zero, it is in fact consistent with the commutation relation. This delta-function equation appears, with explanation, in the first mathematical chapter of Dirac's "The Principles of Quantum Mechanics".

It is unfortunate that formal manipulations with distributions lead to paradoxes so easiy. For a related but different paradox, consider the trace of $\hat{x}\hat{p}-\hat{p}\hat{x}$.

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As it seems that you are not completely satisfied with Ron Maimon's answer, I'll put it in a bit different way.

The problem is that in your derivation you have a hidden ambiguity. $$\langle{\psi}_p\vert \hat{x}\vert\psi_p\rangle = \infty \;\;\;\;\; \Rightarrow \;\;\;\;\; \langle[\hat{x},\hat{p}]\rangle=...=(p-p)\langleψ_p|\hat{x}|ψ_p\rangle = 0\cdot\infty = \text{any number}$$ The problem is with the functions. Eigenfunctions of both momentum operator and coordinate operator are not really functions. They do not belong to integrable functions space and thus you can not freely work with them pretending that they are. Sometimes you can, but if you do at some point you find yourself in trouble.

If you take some "proper" function and do the math, you will find no problems. Let us take e.g. $$\psi(x)=\frac1{\sqrt{\pi}}e^{-x^2/2}$$ Then $$\int_{-\infty}^{\infty} \psi(x) x\left( -i\hbar \frac{∂}{∂x} \right) \psi(x)dx = i\hbar \frac1{\pi} \int_{-\infty}^{\infty} x^2 e^{x^2}dx = i\hbar \frac1{2\sqrt{\pi}}$$ $$\int_{-\infty}^{\infty} \psi(x) \left( -i\hbar \frac{∂}{∂x} \right)x \psi(x)dx = i\hbar \frac1{\pi} \int_{-\infty}^{\infty} \left(x^2-1\right) e^{x^2}dx = i\hbar \frac1{2\sqrt{\pi}}-i\hbar$$ The difference is what you expected.

If you take $\psi_a(x)=\frac1{a}\psi(x/a)$ and note that $\lim_{a\to0}\psi_a(x) = \delta(x) = \vert x\rangle$ you will get an idea how this paradox for $\vert x \rangle$ may be solved and check that the solution is a correct way to handle $0\cdot\infty$. Similar trick may be used to resolve your paradox. Just functions which have $\psi_p$ as a limit are less convinient.

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@ganzewoort It seems that you did not get my point. There is no functions of physical nature. When you speak of functions you are in math part of the science. Thus, you should follow the math rules. In this particular case, you can not freely integrate non-integrable functions. Technically, you break well-defined value into two undefined parts. Of course, you get a paradox. It is absolutely as doing something like $1=1+\sum_1^{\infty}(1-1)=\sum_1^{\infty}1+\sum_0^{\infty}(-1)=\sum_0^{\infty}(1‌​-1)=0$. You get $1=0$ in the same way as you did $i\hbar=0$. –  Misha Aug 31 '11 at 8:16
@ganzewoor I did. Read carefully last paragraph of the answer. The limit $a\to0$ of $\langle \psi_a \vert [x,p] \vert \psi_a \rangle$ exists and correct. Your procedure mathematically incorrect (because $\langle \psi_p \vert x \vert \psi_p \rangle$ does not exists as a Lebesque integral) and should not give the right answer. –  Misha Aug 31 '11 at 13:55
@ganzewoort It is not a problem of quantum mechanics. You have very strange idea of the scientific method and have big problems with math. So, it seems that it is your problem. You have your answer and have no counterarguments, so I am nothing to do here. –  Misha Aug 31 '11 at 16:28
@ganzewoort I will enter in this controversy with the following observation: Mathematics for physics is a tool. As when you are unscrewing a screw you take a screw driver and not a hammer, when you are describing physical systems you need the appropriate well defined for the problem mathematical functions. Both answers attempt to illustrate for you this very clear point as far as the scientific methods of physics are concerned. –  anna v Aug 31 '11 at 19:05
@ganzewoort , where did I invent new rules? You are giving me a reference to a chemistry paper? YES, chemists are famous for their rigorous attitude towards Qm! I really hope you are joking. MOST papers use a lot of naive ways of using QM because its much quicker, but when people face inconsistencies (like your example), they do a more rigorous calculation. you are mistaking ad hoc/quick-and-dirty way of doing QM with a rigorous one! –  Heidar Aug 31 '11 at 21:44

i think the error is that the representations of operators $\hat{x}$ and $\hat{p}$ when the state wavefunction is expanded in the basis of operator $\hat{p}$ are not the ones given in the question but the "opposite", i.e

in $\psi_p$ the operator representations are respectively:

$$\hat{x} = -i \hbar \frac {\partial} {\partial p}$$ $$\hat{p} = p$$

so:

$$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle \ne \langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$

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I think you are saying that when we have an eigenvalue for the momentum representation, and that is when we can replace with one real number the quantities in the equations, the space representation is not diagonal, and vice verso. –  anna v Aug 6 at 5:16
@annav, yes exactly, it is a completely dual sitiuation (at least for $\hat{x}$ and $\hat{p}$ operators in simple cases). This is what the OP missed, expanded the wavefunction in the momentum representation and left the x in the position representation –  Nikos M. Aug 6 at 10:57
@annav, the fact remains that both $x$ and $p$ cannot be simultaneously diagonalised (in other words they dont commute), that is the whole point, the OP by using incompatible representations where each $x$ and $p$ is diagonal (in its own representation) re-derived classical mechanics –  Nikos M. Aug 6 at 11:08
I think yours is the conceptually simplest answer. –  anna v Aug 6 at 11:30
@annav, thanx! i'm surprised myself :) –  Nikos M. Aug 6 at 12:05