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