Application of Heisenberg Uncertainty Principle

[Raghav Gupta]

Homework Statement

A parallel beam of electrons traveling in x- direction falls on a slit of width d.
If after passing the slit, an electron acquires momentum p_y in the y direction then for a majority of electrons passing through the slit
A. |p_y|d ≈ h
B. A. |p_y|d > h
C. |p_y|d < h
D. |p_y|d >> h

Homework Equations

Δx Δp = h/4π

The Attempt at a Solution

I only know Hiesenberg uncertainty equation
How to deal here with inequalities?

 

[gleem]

First the Heisenberg uncertainty relation should be written ΔxΔp ≥ h/4π . What does this expression imply?

 

[Raghav Gupta]

This implies that we have uncertainty in position and momentum calculation of an electron.
How to apply that in this question ?

 

[gleem]

When the electron passes through the slit of width d what are you doing? Using the Heisenberg Uncertainly relation what statement can you make about the lateral momentum

 

[Raghav Gupta]

Δp >= h/4πd
Is that correct?

 

[gleem]

Yes, and what does this suggest?

 

[Raghav Gupta]

Sorry, I think it should be Δp>= h/4πΔd ?
There should be uncertainty involved.

 

[gleem]

No. you are localizing the electron in a space of width d which is the significance of Δx.

 

[Raghav Gupta]

Yes, and what does this suggest?

That uncertainty in momentum is greater?

 

[gleem]

Yes, or the uncertainty is not less than h/4πd.

I have to do something so I will be away for a few hours sorry.

 

[Raghav Gupta]

Yes, or the uncertainty is not less than h/4πd.

I have to do something so I will be away for a few hours sorry.

How does that relate to options?

Think you will reply after some hours.

 

[gleem]

You should be able to exclude two immediately>

 

[Raghav Gupta]

Yes, or the uncertainty is not less than h/4πd.

I have to do something so I will be away for a few hours sorry.

Do that mean certainty is less then h/4πd ?
Which is p_y<h/4πd ?
Why absolute value is being considered of momentum in options and why 4π term is neglected?
This is not true , if p_yd< h/4π
Then |p_y|d< h ?

 

[gleem]

The expression of the uncertainty principle was original suggested by Heisenberg to be ΔpΔx ≥ h. It was later formally derived and found to be ΔpΔx ≥ h/4π.

It seems to me that the two forms are mixed here. First even is they are mixed A and C are not consistent with either one. But if Δp ≥ h/4πd you cannot say Δp > h/d let alone Δp >> h/d. So to make any sense out of the choices you have to use Heisenberg's original formulation ΔpΔx ≥ h.

The absolute value is used because momentum and position are vector quantities and only the magnitudes of these quantities are of interest.

 

[Raghav Gupta]

So option C is correct, by original formulation also and also by modern formulation?
If p_yd < h/4π
Then 4πp_yd < h
Which definitely implies p_yd <h
So no matter what formulation we take ,we only arrive at this answer ?

 

[gleem]

If Δpd ≥ h/4π then 4π Δpd ≥ h. But this does not imply universally that Δpd < h.

 

[Raghav Gupta]

Do that mean certainty is less then h/4πd ?
Which is p_y<h/4πd ?

Is this correct?

 

[gleem]

No. you reversed the inequality sign.

 

[Raghav Gupta]

But I am taking certainty.
When we remove delta sign , inequality changes?

 

[gleem]

It is my understanding that the uncertainty relationship makes no statement about the actual momentum of a particle.

 

[Raghav Gupta]

So why in options delta sign is not there, if Hiesenberg uncertainty principle not tells actual momentum?

 

[gleem]

The Heisenberg Uncertainty relations states that if a particle is localized to a space of length Δx then the momentum cannot be determined to an accuracy greater than Δp_x such that Δx Δp_x ≥ h/4π . It make no statement of the actual value of p_x. The problem suggests that the electron in passing through the slit receive an impulse of Δp but I believe that interpretation is not warranted.

 

[Raghav Gupta]

So, then the answer is |p_y| d ≥ h by original Hiesenberg formulation ? ( If that is true then the question is directly formula based).
But in options it is not there.
Options are
1.|p_y| d > h
2.|p_y| d < h
3.|p_y| d ≈h
4.|p_y| d >> h
And I see options 1,3,4 bear close resemblance to the answer.
But the problem maker has given solution 2. ,Why I don't know?
Totally opposite?

 

[gleem]

I am puzzled too. what is the origin of this question?

 

[Raghav Gupta]

I am puzzled too. what is the origin of this question?

It was asked in an institute entrance exam.
I did 3. option in post 23
The answer key of that exam is saying option 2. Is correct
But one can challenge them.
Are you sure that most appropriate option is 1. ?

Also if you reconsider the question. It might be a tricky one.
Electrons are traveling in x direction and acquiring momentum in y direction which is perpendicular to it.
So something might be different?

 

[gleem]

1 seems the most correct to me. 2 is nonsense. In your OP their " correct answer" was shown to be the third choice C. You didn't mix up the answer order did you?

 

[Raghav Gupta]

1 seems the most correct to me. 2 is nonsense. In your OP their " correct answer" was shown to be the third choice C. You didn't mix up the answer order did you?

No, I didn't mix it up. For the easiness of not scrolling to first page for viewing options I posted it once again in post 23 but making new bullet points and new order.
You may have not seen this in post 25

Also if you reconsider the question. It might be a tricky one.
Electrons are traveling in x direction and acquiring momentum in y direction which is perpendicular to it.
So something might be different?

 

[gleem]

I used Δx Δp_x ≥ h but meant Δy Δp_y≥ h, the slit width is d in the y direction if the electron path is x. I don't see any trickery go on.

 

[Raghav Gupta]

Thanks Sir.

 

[BvU]

If you challenge them, you can refer to Fowler here

 

[Raghav Gupta]

If you challenge them, you can refer to Fowler here

Fowler is suggesting option 3 in post 25. It would be great if it is correct as I chose that option,
but @gleem Sir is saying option 1 is correct.
What option should I challenge?
It's getting confusing.

 

[BvU]

It's not called the uncertainty relation for nothing, then .

The outcomes aren't all that irreconcilable: if the exercise asks what's true for the majority of the electrons, then this carefully selected collection of electrons will indeed satisfy what you call 2 (<, formerly option C -- as you see, you yourself also contribute a little bit to the confusion ).

And we are into interpretation issues if we defend option 3 ($\approx$, formerly A:) which as Fowler works out, is true for the whole lot -- which is an even greater majority. The other two options (> and >>) can be ruled out.

If I were you, I wouldn't spend too much energy on this: it's simply not a very good question. Fowler (and you, and me too, and many others with us) would pick $\approx$ without hesitation. (Unless your challenging means the difference between pass or fail, in which case showing a genuine interest and a sensible defence of your choice may make a good impression and change the balance in you favour)

I would also like to point out that I strongly oppose the notion suggested in post #22:

The problem suggests that the electron in passing through the slit receive an impulse of Δp but I believe that interpretation is not warranted.

The question statement does not -- and does not have to -- suggest that at all. This whole slit business is about the wave character of the electrons: they really, intrinsically and fundamentally exhibit wave behaviour, which means that after a slit there is a diffraction pattern, even if before the slit the beam is exactly parallel (meaning infinitely wide by the same Heisenberg relation!). And the majority does end up in the central maximum.

 

[Raghav Gupta]

The outcomes aren't all that irreconcilable: if the exercise asks what's true for the majority of the electrons, then this carefully selected collection of electrons will indeed satisfy what you call 2 (<, formerly option C -- as you see, you yourself also contribute a little bit to the confusion ).

Anyways I am not challenging the problem makers, as we cannot give our evidence but only options we think correct. They might have another interpretation.
Anyways a last question,
Why this 2 or formerly C option is true for selected collection of electrons?

 

[Raghav Gupta]

Also I had a question that here in options does |p_y| is same as Δp_y?

This is an interesting video. I think question and video is same. The problematic thing were the options.

 

[BvU]

$|p_y|$ is not the same as $\Delta p_y$ but it's subtle and our language use is fuzzy (more confusion lurking!).
In principle $|p_y|$ is the expectation value of $\sqrt{p_y^2}$ and $\Delta p_y$ is the square root of the expectation value of ${p_y^2}$. Not very helpful if you aren't deep into quantum mechanics already.

Better to look at the situation described:

The statement $|p_y| \;d\ < \ h \$ can be considered true for the electrons that end up in the central maximum, i.e. the majority. That central maximum has a width $\Delta p_y \approx h/d$ which is the Heisenberg uncertainty relation (see Fowler).


I hope a real expert (e.g. @Orodruin) agrees somewhat -- or perhaps puts us right.
(I'm just an experimental physicist, so out on a limb -- but I like it )
---

 

[Raghav Gupta]

And we are into interpretation issues if we defend option 3 ($\approx$, formerly A:) which as Fowler works out, is true for the whole lot -- which is an even greater majority. The other two options (> and >>) can be ruled out.

The statement $|p_y| \;d\ < \ h \$ can be considered true for the electrons that end up in the central maximum, i.e. the majority. That central maximum has a width $\Delta p_y \approx h/d$ which is the Heisenberg uncertainty relation (see Fowler).
---

Why these two statements are contradicting made by you.
In first one you are saying approx statement is for majority electrons
And in second you are saying (< ) statement is for majority?

 

[BvU]

I don't find them contradictory; why do you ?

 

[Raghav Gupta]

Means in first quote of you in post 36 you are saying,
|p_y|d ≈ h is true for majority electrons
and in second quote of you in post 36 you are saying,
|p_y|d < h is true for majority electrons?

 

[BvU]

I see what you mean: in #32 $\approx$ I should have used the $\Delta$ instead of the | | for more clarity . (I did in #36).
Case can be made that both are true (difference between $\Delta$ and | | isn't that big), so no contradiction.

 

[Raghav Gupta]

Okay got it. That Δ and | | signs were creating confusion but not now.
I wonder, why tagging @Orodruin is not catching his attention even when I see him recently on some other thread.

 

[Orodruin]

I wonder, why tagging @Orodruin is not catching his attention even when I see him recently on some other thread.

I see it. I just do not have time to read through the thread on my breaks at work.

 

[Raghav Gupta]

Twist in a tale then.
I think Hiesenberg uncertainty principle is irrelevant here after seeing this video.


Net is a wonderful thing to explore.
Got all of question and solution in this. But still why Hiesenberg gives diff. Answer and this method a different one?
About video, it is an Indian accent one but language is English. It is an approximately 1 minute video.

 

[BvU]

Total hogwash ! Re-read Fowler to understand why; he mentions it explicitly.

 

[Raghav Gupta]

Total hogwash ! Re-read Fowler to understand why; he mentions it explicitly.

But Fowler is mainly talking about uncertainty.
Here in video, the guy is saying
By De- Broglie relationship ( as Fowler also says)
λ = h/p --- 1)
Then for diffraction, λ≈d ,
But we don't want diffraction , so d >> λ
Therefore d >> h/p
So pd >> h
So what is the hogwash here?

 

[BvU]

Spelling it out (quoting Michael Fowler, Virginia univ):

we know from experiment that this is not what happens—a single slit diffraction pattern builds up, of angular width $\ \theta \sim \lambda /w$ , where the electron’s de Broglie wavelength $λ$ is given by $p_x \cong h/\lambda$ (there is a negligible contribution to $λ$ from the y-momentum). The consequent uncertainty in $p_y$ is

$$Δp_y/p_x \cong \theta \cong \lambda/w$$

Putting in $p_x = h/\lambda$ , we find immediately that

$$\Delta p_y = h/λ$$

I can't put it into words any better than that

--

 

[Raghav Gupta]

Spelling it out (quoting Michael Fowler, Virginia univ):

I can't put it into words any better than that

--

That is correct,
Δp_y=h/λ
Then how the term d will be introduced?

 

[Raghav Gupta]

That is what I have written in post 44,
λ = h/p
Manipulating,
p = h/λ

 

[BvU]

Connect the dots: Fowler's w is your d.

 

[Raghav Gupta]

Connect the dots: Fowler's w is your d.

Yeah, got it from that
Δp_yd ≈ h , thanks.
But I should admit
2 mistakes
First from the answer key of our paper
And then from the video solution.

 

[BvU]

I'm pretty convinced the answer $\ |p_y|\;d \ < \ h\$ is actually correct