A question related to de Broglie wavelength and the Uncertainty Principle

[Rishabh Narula]

"Now, if an electron has a definite momentum p,
(i.e.del p = 0), by the de Broglie relation, it
has a definite wavelength.A wave of definite
(single) wavelength extends all over space.
By Born’s probability interpretation this
means that the electron is not localised in
any finite region of space. That is, its
position uncertainty is infinite
(del x --> infinity ), which is consistent with
the uncertainty principle."
-high school physics book from India,NCERT
Physics 12th part 2.
Why does a wave of definite wavelength extend
all over the space?
Or why does it continue/goes on forever
just because it has one value for wavelength?
And why should then the opposite be true that
a wave packet consisting of different wavelengths
doesn't continue/goes on forever?

i.e I'm asking what does the opposite case
that is these lines mean-
"The wave packet description of
an electron. The wave packet corresponds to a
spread of wavelength around some central
wavelength (and hence by de Broglie relation,
a spread in momentum)."

Please explain simply if you can,my
understanding of this stuff is a beginner
level as of now.

 

[PeroK]

Try:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

The short answer is that a wave-packet is only localised when you superimpose an infinite number of waves with wavelengths over a continuous spectrum. In some ways this is an amazing mathematical trick.

It doesn't become localised just because you add a few waves together of different wavelengths.

 

[Rishabh Narula]

Try:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

The short answer is that a wave-packet is only localised when you superimpose an infinite number of waves with wavelengths over a continuous spectrum. In some ways this is an amazing mathematical trick.

It doesn't become localised just because you add a few waves together of different wavelengths.

thanks,but i still don't really get it.Is there an internet video or pictures explaining this?could you link those too.will go through the pdf and reply further too thanks.

 

[PeroK]

thanks,but i still don't really get it.Is there an internet video or pictures explaining this?could you link those too.will go through the pdf and reply further too thanks.

This is serious undergraduate level physics. If you cannot follow an undergraduate level text, then there is no simpler explanation. You just have to accept that a mathematical process called a Fourier transform does the trick and move on.

Note that the diagrams on page 16 of the jpf couldn't be clearer on this point.

 

[Dr_Nate]

One should be highly suspicious if they have a physical quantity that is infinite. Limits to infinity are generally okay, but not actual undefined quantities like infinity.

"That is, its position uncertainty is infinite (del x --> infinity ), which is consistent with the uncertainty principle." -high school physics book from India,NCERT Physics 12th part 2.

This statement from your book is not true. It can be seen to be not true at least two ways.

1) The Fourier transform of a delta function is an infinite wave. The integral of the probability of a single infinite wave is infinity so it is not normalizable, and thus it isn't a legitimate wave function in quantum mechanics.

2) Heisenberg's uncertainty principal deals with the standard deviations of position and momentum of a wave function: $\Delta x \Delta p \ge \frac{\hbar}{2}$. From OP's textbook, we are given $\Delta x = \infty$ and $\Delta p = 0$. Depending on who you ask, if you multiply these two together you will either get zero or something undefined. Either way, it won't satisfy the inequality.

 

[bahamagreen]

See if this helps...

"Heisenberg's uncertainty principle for conjugate pairs of observables follows directly from the fact that those observables are essentially the Fourier transforms of each other."
Mathpages

 

[vanhees71]

This is not true in general but for position and momentum only. Of course it's a very intuitive example. If you have the Fourier transformation of a narrow-peaked function in momentum space to the function in position space,
$$f(x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{f}(p),$$
you'll see that $f$ is a broad distribution. If on the other hand $\tilde{f}(p)$ is a broad distribution, $f(x)$ will be a narrow-peaked function.

Quantitatively the uncertainty relation for an arbitrary pair of observable $A$ and $B$, represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ reads
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
Here $\Delta A$ and $\Delta B$ are the standard deviations of the quantities calculated with the state $\hat{\rho}$ the system under consideration is prepared in:
$$\langle \hat{X} \rangle:=\mathrm{Tr}(\hat{\rho} \hat{X}),$$
with the standard deviation defined as
$$\Delta A^2 = \langle \hat{A}^2 \rangle - \langle \hat{A} \rangle^2.$$
Then $[\hat{A},\hat{B}]$ is the commutator of the two operators.

For position and momentum you have
$$[\hat{x},\hat{p}]=\frac{1}{2} \mathrm{i} \hbar$$
and thus the uncertainty relation
$$\Delta x \Delta p \geq \frac{1}{2} \hbar.$$