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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.

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    $\begingroup$ Position and momentum eigenstates are connected by Fourier transform. $\endgroup$ – ZeroTheHero Apr 20 '17 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 Apr 20 '17 at 21:05
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    $\begingroup$ 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 Apr 20 '17 at 21:23
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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.

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  • $\begingroup$ 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. $\endgroup$ – ZeroTheHero Apr 22 '17 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. $\endgroup$ – ZeroTheHero Apr 22 '17 at 15:10
  • $\begingroup$ Writing the question out helped me answer it, so I deleted. $\endgroup$ – daniel Apr 23 '17 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. $\endgroup$ – ZeroTheHero Apr 23 '17 at 12:06
  • $\begingroup$ OK look forward to it, and I do think your answer is fine as is. $\endgroup$ – daniel Apr 23 '17 at 12:27
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I will work with space in 1 dimension here.

First, the wavefunctions $| \psi \rangle$ live in an abstract Hilbert space. To have access to information on our system we have to choose a base of this Hilbert space and represent the function on it.

I can take the position basis. If I decompose this abstract vector on it, I will have $$| \psi \rangle = \int dx \psi(x) |x\rangle$$

As you can see it is just a summation using an integral on vector basis.

I could also decompose this abstract vector on the momentum basis. I thus have :

$$| \psi \rangle = \int dx \widehat{\psi}(p) |p\rangle$$

These two basis are in fact orthogonal ones so $\langle p|p'\rangle=\delta(p-p')$ for example.

Now, $|x\rangle$ and $|p\rangle$ are in fact eigenvectors of operators $\widehat{x}$ and $\widehat{p}$ that I will detail.

The momentum basis vectors are defined as the eigenstates of the $\widehat{p}$ operator.

To know how this operator works we have to explain how he works in a given basis. The momentum in the position basis works like this :

$$\langle x|\widehat{p}=-ih \partial_x $$

The $\langle x|$ at the left of the operator mean that I work in position representation. So it will tell us how it acts in practice in the basis of the position.

So, to give an example, if I apply the momentum operator on a wave function, it act in position representation as :

$$ \langle x|\widehat{p} |\psi\rangle = -ih \partial_x \psi(x)$$ In momentum representation, the momentum operator just works simply like this :

$$ \langle p|\widehat{p}| \psi \rangle = p*\widehat{\psi}(p)$$

Now lets talk about the position operator :

The position operator is denoted $\widehat{x}$.

Its action in position representation is simply :

$$\langle x | \widehat{x} | \psi \rangle = x*\psi(x)$$

Now lets find the eigenvectors of $\widehat{p}$ and $\widehat{x}$.

I choose to work in position basis.

The eigenvectors of $\widehat{x}$ verifies the following equation : $\widehat{x}|x\rangle=\lambda |x\rangle$. Thus, if I project on $|x' \rangle$ I have, because the basis is orthogonal :

$$\langle x'|\widehat{x}|x\rangle=\lambda \delta(x-x')$$ To simplify, I decide to note the eigenvalue of the $| x\rangle$ eigenvector $x$. It is just a notation.

Thus : $\widehat{x}|x\rangle=x |x\rangle$

I will now find the eigenvector of $\widehat{p}$ operator.

$$\langle x|\widehat{p}|p\rangle=-ih \partial_x \psi_p(x)$$

Where $\psi_p(x)=\langle x | p \rangle$ is the expression of the momentum eigenspace on the x basis. So if you want we have : $|p\rangle = \int dx \psi_p(x) |x\rangle$.

So as I find the eigenvector I have the equation :

$$\langle x|\widehat{p}|p\rangle=-ih \partial_x \psi_p(x)=p*\psi_p(x)$$ where p is the eigenvalue associated to $|p\rangle$.

The solution of this 1st order diff equation is just :

$$\psi_p(x)=C*e^{i\frac{p}{h}x}$$

And we find the constant $C$ by normalisation (in practice we work in a finite box so $C=\frac{1}{\sqrt{L}}$

And as $TF(\delta(x-x_0))=e^{i\frac{p}{h}x_0}$ we find the TF relation between them.

Well I hope I didn't made too much mistakes its my first answer on this website. I was probably a little too long.

As a book that explains all this in detail I strongly advise you Quantum Mechanics from Cohen Tanoudji. It is an easy to read books that explains things very well and it is very complete. There are of course chapters explaining with much more details everything I said here. I learned all the Quantum Mechanics with it.

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  • $\begingroup$ You refer to the Fourier transform as "FFT". However, FFT is an acronym for the Fast Fourier Transform, which is a particular numerical algorithm that computes the discrete Fourier transform (DFT). Neither the DFT nor the FFT have anything to do with this answer. $\endgroup$ – DanielSank Apr 21 '17 at 6:16
  • $\begingroup$ Pardon me but how does "the two basis are orthogonal" imply "$\langle p\vert p'\rangle=\delta(p-p')$". Aren't $\vert p\rangle$ and $\vert p'\rangle$ in the same basis or am I misreading something here? $\endgroup$ – ZeroTheHero Apr 21 '17 at 6:55
  • $\begingroup$ Yes you are right, I edited. I always write FFT when I just think about a TF I don't know why. @ZeroTheHero : $|p\rangle$ and $|p'\rangle$ are elements of a same basis but they are not the same vectors. If I take the canonical basis of $\mathbb{R^3}$ : $(1,0,0)$ and $(0,1,0)$ are two elements of this basis but the vectors are different. $\endgroup$ – StarBucK Apr 21 '17 at 9:15

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