This question is a small step removed from two earlier questions about the uncertainty relation in the QM context. knzhou's answer touches on the point, and L. Motl's link-answer here is a little closer.

The passage from Wikipedia in relevant part is (italics added):

"When a state is measured, it is projected onto an eigenstate in the basis of the relevant observable. For example, if a particle's position is measured, then the state amounts to a position eigenstate. This means that the state is not a momentum eigenstate, however, but rather it can be represented as a sum of multiple momentum basis eigenstates. In other words, the momentum must be less precise. This precision may be quantified by the standard deviations..."

Well, the crucial step is missing I think. I understand the wave-mechanical explanation well, and I appreciate that different observables are associated with different eigenstates.

Can someone fill in these two points or suggest a reference for:

  1. In what sense can a position eigenstate be represented as a sum of momentum eigenstates? Maybe a homely example would do here...

  2. How does the math of a position eigenstate represented by multiple momentum eigenstates translate into the product of variances?

If the answer is too involved a reference would be great.

  • 1
    $\begingroup$ Position and momentum eigenstates are connected by Fourier transform. $\endgroup$ Commented Apr 20, 2017 at 20:49
  • $\begingroup$ @ZeroTheHero: Maybe I misinterpret "represented." You are saying they are FT pairs...then a position eigenstate corresponds to a FT that is a sum of multiple momentum e-states. Is that accurate? $\endgroup$
    – daniel
    Commented Apr 20, 2017 at 21:05
  • 2
    $\begingroup$ It is correct. $|x\rangle=\int d^3p \langle p|x\rangle|p \rangle$ or, if you are not familiar with the ket notation : $\psi_{x_0}(x)=\delta(x-x_0)=\int d^3p \widehat{\psi_{x_0}}(p) e^{ipx}$ where : $\psi_{x_0}(x)$ is the eigenfunction of the position in representation x, and $e^{-ipx}$ is eigenfunction of the momentum operator in $\endgroup$
    – StarBucK
    Commented Apr 20, 2017 at 21:23

1 Answer 1


Let me provide a slightly different derivation, which emphasize the Fourier transform aspect of the relation between the $x$ and $p$ representations.

Let $\vert p\rangle$ be a state with definite momentum. Physically, we know that such a state is a plane wave so that, in the $x$-representation: $$ \psi_p(x)=\langle x\vert p\rangle =\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\tag{1} $$ Conversely it follows that $$ \langle p\vert x\rangle := \psi_x(p)= \frac{1}{\sqrt{2\pi\hbar}}e^{-ipx/\hbar} \tag{2} $$ Somewhat informally, the $\vert p\rangle$ states are eigenstates of a hermitian operator $\hat p$ so they form a complete basis, with expansion of unity $$ 1=\int dp \vert p\rangle \langle p \vert\, . $$ Likewise. $\vert x\rangle$ are eigenstate are eigenstates of $\hat x$ so $$ 1=\int dx \vert x\rangle \langle x \vert\, . $$

Thus, to your question 1: $$ \vert x\rangle = \int dp \vert p\rangle \langle p \vert x\rangle = \int dp \vert p\rangle \frac{1}{\sqrt{2\pi\hbar}}e^{-ipx/\hbar} $$ and more generally $$ \langle p\vert\psi\rangle: = \psi(p) = \int dx \langle p\vert x\rangle \langle x \vert\psi\rangle \, =\int dx \frac{1}{\sqrt{2\pi\hbar}}e^{-ipx/\hbar} \psi(x) $$ and conversely $\psi(x)$ is the (inverse) Fourier transform of $\psi(p)$. Note that the normalization of (1) and (2) is ``symmetrical" in that both the direct and inverse Fourier transform carry a factor of $1/\sqrt{2\pi\hbar}$; this is not the usual convention of many math textbooks.

To your question 2: there are "classical" uncertainty relations related to product of variances for pairs of variables related by Fourier transforms like $p$ and $x$ (or $E$ and $t$) but of course these are not quantum in any sense. In this case they simply express that to localize a wave packet in space requires an increasingly wider distribution of the conjugate momentum. There is nothing quantum in this product of variances.

The quantum uncertainty relations, related to products of variances of non-commutating operators, is fundamentally different as it arises as a result of quantum non commutativity of operators.

  • $\begingroup$ The quantum version is a formal inequality involving operators: $\Delta A\Delta B\ge \frac{1}{2}\vert \langle [\hat A,\hat B]\rangle$ and in general the average value of the commutator will be state-dependent. In the specific case of $\hat x$ and $\hat p$, the commutator is $i\hbar$ so state-independent. The $1/2$ factor is important and the h.o. ground state is state for which the strict equality holds. The classical version is an informal relation $\Delta x\Delta p\sim \hbar$ which only makes sense for conjugate variables. $\endgroup$ Commented Apr 22, 2017 at 14:30
  • $\begingroup$ The classical result depends on the definition of width. Take $\sin(x)/x$; do you defined width as distance between first $0$'s or at width at $1/2$max? Also the time-energy version $\Delta E\Delta t$ is tricky in QM because time is not an operator. I have to go... stay well. $\endgroup$ Commented Apr 22, 2017 at 15:10
  • $\begingroup$ Writing the question out helped me answer it, so I deleted. $\endgroup$
    – daniel
    Commented Apr 23, 2017 at 9:22
  • $\begingroup$ @daniel I'm tight for time right now but will update/edit my answer within 72hrs to add a short discussion on the problem of properly defining the width in the classical context. $\endgroup$ Commented Apr 23, 2017 at 12:06
  • $\begingroup$ OK look forward to it, and I do think your answer is fine as is. $\endgroup$
    – daniel
    Commented Apr 23, 2017 at 12:27

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