Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I will denote operators with hats. Suppose we got an operator of the form $i[\hat p, \tan^{-1}(e^{\hat x})]$ and we want to calculate the amplitude for a transition from a state $|p_i\rangle$ to the same state $|p_i\rangle$, as might happen in elastic scattering problems. Below are two evaluations which look like they are correct but take to two different results. Denoting $\hat f=\tan^{-1}(e^{\hat x})$,

1) \begin{equation} \begin{array}{lcl} \langle p_i|i[\hat p,\hat f]|p_i\rangle&=& i\langle p_i|(\hat p \hat f-\hat f\hat p)|p_i\rangle =i\langle p_i|(p_i \hat f-\hat fp_i)|p_i\rangle\\ &=&i\,p_i\langle p_i|(\hat f-\hat f)|p_i\rangle=0 \end{array} \end{equation}

2) \begin{equation} \begin{array}{lcl} \langle p_i|i[\hat p, \hat f]|p_i\rangle&=&i \langle p_i|\left(-i\hbar\frac{\partial \hat f}{\partial x}\right)|p_i\rangle =\hbar\langle p_i| \frac{e^{-\hat x}}{1+e^{-2\hat x}}|p_i\rangle\\ &=&\hbar\langle p_i| \int dx |x\rangle \langle x| \frac{e^{-\hat x}}{1+e^{-2\hat x}}|p_i\rangle =\hbar\int dx \frac{e^{- x}}{1+e^{-2 x}} \langle p_i| x\rangle \langle x |p_i\rangle\\ &=&\hbar\int dx \frac{e^{- x}}{1+e^{- 2x}} \frac{e^{\frac{i}{\hbar}p_i x}}{\sqrt{2\pi \hbar}}\frac{e^{-\frac{i}{\hbar}p_i x}}{\sqrt{2\pi \hbar}} =\frac{1}{2\pi}\int_{-\infty}^{+\infty} dx \frac{e^{- x}}{1+e^{- 2x}}\\ &=&\frac{1}{2\pi}\,\frac{\pi}{2}=\frac{1}{4} \end{array} \end{equation} The integral in the evaluation 2) can be found in Jeffrey, Eq. (11), Sec. 15.3.1 .

I chose the form of $\hat f$ to bring the wanted results, but the question is general. Many other simpler forms can be employed to bring similar discordant results.

Moreover, the results may be generalized saying that: from the first evaluation the result is always zero, from the second evaluation the result is always $\propto f\big|_{x=+\infty}-f\big|_{x=-\infty}$.

What do I do wrong?

share|cite|improve this question
Excuse me, about the first line in your 2), $ \frac{ \partial{\hat{f}}}{\partial x} = \frac{ \partial{ArcTan(\hat{x}) }}{\partial x} = \frac{1}{1+\hat{x}^2} $ , I cannot get $e^{-\hat{x}}$ etc – user26143 May 2 '14 at 12:04
Sorry, there was a typo. I meant $ArcTan(e^{x})$. I corrected now. Thanks. – Wizzerad May 2 '14 at 12:24
One can also more elegantly generalize this discordant result as follows. Say we have a potential $V(\hat x)$ for which we want to calculate the amplitude $\langle p_i | V(\hat x) | p_i \rangle$. I define $G(\hat x)=\int_x V(\hat x)$. Then I have $\langle p_i | V(\hat x) | p_i \rangle = \langle p_i | \frac{\partial}{\partial x}G(\hat x) | p_i \rangle=(-ih)^{-1}\langle p_i | [\hat p, G(\hat x) ]| p_i \rangle=0$. This looks pretty astonishing. – Wizzerad May 2 '14 at 13:39

The problem is ill posed from scratch because $|p_i\rangle$ is not an element of the Hilbert space and, a fortiori, it does not belong to the domain of $\hat{p}$. The same problem arises when considering $\hat{f}|p_i\rangle$.

As a matter of fact, rigorously speaking $\langle p_i|i[\hat p,\hat f]|p_i\rangle$ does not exist. From the mathematical point of view the problem stops here since ex falso quodlibet.

However something can be said with a suitable interpretation of $\langle p_i|i[\hat p,\hat f]|p_i\rangle$. Naively but straightforwardly one can give this interpretation, omitting inessential signs and constants, $$\langle p_i|[\hat p,\hat f]|p_i\rangle= i\int_{\mathbb R} e^{-ip_ix} \left(\frac{d}{dx} f(x) e^{ip_ix}\right) dx - i\int_{\mathbb R} e^{-ip_ix} f(x)\frac{d}{dx} e^{ip_ix} dx\:. \quad [1]$$ This integral can be computed and produces the result (2). The result (1) cannot be obtained with this straightforward interpretation of the formalism, because it relies upon the formal self-adjointness of $i\frac{d}{dx}$, that is the identity: $$i \int_{\mathbb R} \psi(x) \frac{d}{dx} g(x) dx = i\int_{\mathbb R} \left( \frac{d}{dx}\psi(x)\right) g(x) dx \quad [2]$$ This identity, in fact holds for some classes of functions $\psi,g$ (in some cases also if $\psi, g$ do not belong to $L^2(\mathbb R)$). Comparing with the first term in the right hand side of [1], we have that it should be, $$g(x) = f(x) e^{ip_ix} \quad \psi(x) = e^{-ip_ix}\:.$$ However these functions have been fixed just to make false [2]!

share|cite|improve this answer
As to your first paragraph: Naturally $|p_i\rangle$ belongs to the quantum space of $\hat p$. I just define it so by saying $\hat p|p_i\rangle=p_i|p_i\rangle$. As to your second paragraph: Such a scalar product can be defined since ket/bra vectors and operators belong to the same ($\hat x$,$\hat p$) quantum space. However, we are right here discussing how to calculate it, so its value is still undefined as far as this post and its answers are concerned. – Wizzerad May 2 '14 at 20:38
No, it does not belng to the Hilbert space of the theory, $|p\rangle$ is just a formal object, it is not a $L^2(\mathbb R)$ element. This is, after all, the source of all this sort of problems, to use formal arguments, and believe that they are rigorous procedures, where they do no make any sense. – Valter Moretti May 2 '14 at 21:44
If you want to give a sense to all those formal procedures you must adopt the Gelfand formalism of rigged Hilbert spaces. But once again, even in that approach the object we are discussing turns out to be ambiguous. There is no answer to the question because, actually there is no question at all. It is just a wrong way to handle the formalism. – Valter Moretti May 2 '14 at 21:49
I think I have only now understood your answer, after many months! You meant that states that are not L^2 aren't physical and taking their expectation value with any operator does not make any sense. The Hilbert space to consider in quantum mechanics is made only of physical states, which are the ones that have norm=1. States with infinite norm cannot be taken as physical states and must be used only for expanding physical states. Right? This sounds reasonable, indeed. Thanks. – Wizzerad Feb 13 '15 at 19:40
It is a subtle issue. However, directly in position representation $<p|V|p> = \int_V V(x) dx$. If $V(x) = dG/dx$ for some $G$ which vanishes for large $|x|$ (notice that this is nothing but another way to say that $\int_V V(x) dx=0$), then the final result is $0$. However it depends on the shape of $V$... – Valter Moretti Feb 17 '15 at 10:54

It is not an answer directly to your case, but some relevant observations. Nevertheless it is too long to post it into comment.

We consider a similar situation, (I omit the hats)

$$\langle p | [x,p] | p \rangle \tag{1} $$

There are two ways calculating expectation of (1).

The first, $$ \langle p | xp -px | p \rangle = p ( \langle p|x|p \rangle - \langle p|x|p \rangle) =0 \tag{2} $$

The second, $$\langle p | i | p \rangle = i \delta(0) = i \infty \tag{3} $$

There is an inconsistency. The problem is, $$\langle p |x|p \rangle = \int \int dx dx' \langle p|x \rangle \langle x| x| x '\rangle \langle x'| p\rangle = \int x dx = \infty -\infty \tag{4} $$, which is ill-defined, unless we pick up principal value.

We may make a more conservative calculation in the line of the first approach,

$$ \lim_{p'\rightarrow p} \langle p | [x,p] | p' \rangle = \lim_{p'\rightarrow p} (p' - p) \langle p | x | p' \rangle = i \lim_{p'\rightarrow p} (p' - p) \frac{ \partial}{\partial p} \delta(p-p') = i \lim_{p'\rightarrow p} \delta (p - p') = i \delta(0) \tag{5} $$

Everything is consistent now.

I think the situation is similar for $f=ArcTan(e^x)$, though I did not work out the integral in $\lim p' \rightarrow p $ approach.

share|cite|improve this answer
Thanks. I passed by this example before formulating my question. I must admit I solved it similarly but slightly differently, though. I did: 1) $\lim_{p'\to p}\langle p'|[x,p]|p\rangle=\lim_{p'\to p}\langle p'|(xp -px)|p\rangle=0$ 2) $\lim_{p'\to p}\langle p'|[x,p]|p\rangle=i\hbar \lim_{p'\to p}\langle p'|p\rangle=i\hbar\lim_{p'\to p}\, 0=0$ The result is however different from yours, though they are again consistent with each other. The problem is that this approach does not save me for more complicated operators, like $x^2$ already. That is how I came to what above. – Wizzerad May 2 '14 at 14:44
I think from the "normalization" convention, $\langle p' | p \rangle = \delta (p-p')$, so you have to take the delta function. – user26143 May 2 '14 at 14:47
When I say $\lim_{p'\to p}$ I suppose $p'\neq p$. That is how the limit makes sense to me in the first place. The limit of the Dirac Delta $\lim_{x\to0}\delta(x)$ is zero. Of course the Delta is not continuous, so its limit does not coincide with its value, but still. Do I do it wrong? So, yes, we could write the last passage as you suggested $\lim_{p'\to p}\delta(p-p')=0$, but the result would be the same. – Wizzerad May 2 '14 at 14:51
As a simple-minded physicist, I adopt $\lim_{x\rightarrow 0} \delta(x) = \delta(0) = \infty$. Some mathematician may be able to answer this point, sorry – user26143 May 2 '14 at 14:53
Mm, ok. Here we disagree. The limit is thought as the limiting value of the function approaching the desired point which might be different from the function at that point. To me it looks like you just replace $\lim_{x\to x'}f(x)=f(x')$ which is the condition for continuity. That is not our case, since the Delta is not continuous. If I took your definition of limit, then I would not know what is the limit of the step function at 0 (it might be 0 or 1) or the limit of the function 1/x at 0 (it might be -$\infty$ or +$\infty$). In those cases, you must define the left and right limits – Wizzerad May 2 '14 at 15:03

(1) Looks wrong simply because you did not apply the operators onto the state $|p\rangle$ correctly. Operators act from right to left, so you should get: $$ \left(\hat{p}\hat{f}-\hat{f}\hat{p}\right)|p\rangle=\hat{p}\hat{f}|p\rangle-\hat{f}\hat{p}|p\rangle =\hat{p}\left(\hat{f}|p\rangle\right)-\hat{f}\left(\hat{p}|p\rangle\right) \tag{1} $$ because $\hat{f}$ needs to act on $|p\rangle$ before you act $\hat{p}$ on it. The first term on the left of the second line (and neglecting constants) is really $$ \hat{p}\left(\hat{f}|p\rangle\right)=\frac{\partial}{\partial x}\left( f|p\rangle\right)=\frac{\partial f}{\partial x}|p\rangle+fp|p\rangle\neq pf|p\rangle\tag{2} $$ which is not what you have supposed.

The (2) looks to be correct because it is using the quantum analog of the Poisson bracket to define the commutator: $$ [\hat{p},\,\hat{f}]=i\hbar\left(\frac{\partial\hat{p}}{\partial\hat{x}}\,\frac{\partial\hat{f}}{\partial\hat{p}}-\frac{\partial\hat{p}}{\partial\hat{p}}\frac{\partial\hat{f}}{\partial\hat{x}}\right)=-i\hbar\frac{\partial\hat{f}}{\partial x}\tag{3} $$ because $\partial\hat{f}/\partial\hat{p}=0$.


The proper definition of the momentum operator is $$ \hat{p}=-i\hbar\frac{\partial}{\partial x} $$ Thus, by applying the missing constants to (2) and subtracting the $fp|p\rangle$ term from (1), we get $$ -i\hbar\frac{\partial f}{\partial x} $$ which is the same thing as equation (3). Thus, there is no paradox because you were not correctly applying your operators.


The problem OP has is in the confusion about operator orders. When operators act on bras, we must take the (Hermitian) adjoint of the operators: $$ \langle p|\hat p\hat f|p\rangle=\left(\langle p|\hat f^\dagger\hat p^\dagger\right)|p\rangle\neq \left(\langle p|\hat p\right)\hat f $$ when using the middle term, a similar equation to my Equation (2) above will be obtained (i.e., ket replaced by a bra).

share|cite|improve this answer

I think I found an answer that, at least for me, is satisfactory.

Transition amplitudes from a state $|a\rangle$ to the same state $|a\rangle$ make full sense if I am dealing with a bound state, whose wave-function is squared integrable. All problems I pointed out won't happen with bound (squared integrable) states. On the other hand, if I deal with a free state, whose quantum number is continuous, I'll have to insert a generally different state for the final state. I have pointed out scattering problems, but in scattering problems the scattered state $|p_ f\rangle$ is set to be in general different with respect to the initial state, although their energy is sometimes the same.

One may be also tempted to consider $\langle p_i|V(\hat x)|p_i \rangle$ as expectation value of the operator $V(\hat x)$ onto the state $|p_i \rangle$. However, that looks wrong again if the state is not squared integrable, like $|p_i \rangle$. In fact, let us suppose we want to get the expectation value of the operator $\hat p$ onto the state $|p_i \rangle$. We naturally want the outcome to be $p_i$. This will not happen if we write the expectation value as $$ \langle p_i|\hat p|p_i \rangle=p_i\delta(0)=+\infty $$ since the states are not squared integrable. The best we can get is by inserting a final different state and get the distribution $$ \mathcal{D}=\langle p|\hat p|p_i \rangle=p\delta(p-p_i) $$ so that if integrated we get the wanted result: $$ <p>=\int \mathcal{D} \,d\!p=p_i~, $$ where $<p>$ means here "mean value for p".

In my opinion, when dealing with density matrices and traces in (x,p) quantum space, one must also do it with care, considering what above.

In general, now I do not see anymore the need of considering elements of the type $\langle p_i|V(\hat x)|p_i \rangle$, whatever the potential is, if $|p_i\rangle$ is not squared integrable. In perturbation theory, which is what I mostly need, the first term of the Dyson series reads $\langle p_f|p_i \rangle$ and is set to zero, since we suppose $p_f \neq p_i$. If we don't suppose that, everything blows up to infinity at the zero order and that is what we do not want.


To emphasize what I mean, let me reconsider my equations with squared integrable states: \begin{equation} \langle p_i|x\rangle = \frac{e^{-\frac{i}{\hbar}p_i x}}{\sqrt{V}}\;,\; \langle x|p_i\rangle = \frac{e^{\frac{i}{\hbar}p_i x}}{\sqrt{V}}\;, \end{equation} where $V$ is the integration 'Volume' in 1D (someone might want to call it $V^{1/3}$). Equation (1) reads as it is, while Eq. (2) now reads: \begin{equation} \begin{array}{lcl} \langle p_i|i[\hat p, \hat f]|p_i\rangle&=& \hbar\int dx \frac{e^{- x}}{1+e^{- 2x}} \frac{e^{\frac{i}{\hbar}p_i x}}{\sqrt{V}}\frac{e^{-\frac{i}{\hbar}p_i x}}{\sqrt{V}} =\frac{\hbar}{V}\,\int_{-\infty}^{+\infty} dx \frac{e^{- x}}{1+e^{- 2x}}\\ &=&\frac{\hbar}{V}\,\frac{\pi}{2}=0 ~,\qquad (1.1) \end{array} \end{equation} and the discrepancy has gone. States in (1.1) are probably a good approximation of a Gaussian state with large spatial width, which is what we would get in usual experiments. I therefore expect that $\approx 0$ is the result we would find. As a matter of fact, a rigorous calculation with Gaussian states (which are squared integrable) would be interesting. We could send the spatial width to large values to see how the expectation value behaves. I'll see if I can do it in the following days.

share|cite|improve this answer
The Dyson series reminds me when dealing with S matrix in scattering, we write $S=1+iT$ to exclude the $p_i=p_f$ .. – user26143 May 3 '14 at 10:18
Yes, with Gaussian states as defined in Sakurai, Eq. 1.7.35, all comes to be ∝1/d≈0, where d is the spatial width which is here supposed to be >>1 (so to approximate a plane wave). This is so since the function under integral goes to 0 fast as x increases. So the result for the evaluation of the element in discussion here is 0 as suggested. This is rigorous and not much is left to be discussed, I think. I see that I have probably raised some bad mood and I am sorry for that. I will try to post some happier questions next time. :-) Thanks to everybody who joined the discussion – Wizzerad May 4 '14 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.