Respected sir, For a particle in a finite well with no potential on it is considered as free particle.. classically a free particles will retain its state with no change in energy.. but here.. we have different energy states even for a particle inside a potential well where it is supposed as free.. from where these energy states come??
This particle is said to be bound not free, as walls of the well are having infinite potential. Classically, like a rope tied at two ends, its middle can oscillate freely but not the nodes. Thanks
at 16:30 why psi(x)= C sin(lx) +D cos(lx) instead psi(x)= Ceikx+De-ikx. from whete sin(lx) came? The real part of Ceikx is C cos(lx). then why we have written C sin(lx). Thank you.
Thanks for your comment! The wavefunction for bound states in the finite square well is often written as ψ(x)=Csin(lx)+Dcos(lx), which comes from solving the Schrödinger equation in regions with constant potential. While Ceikx+De−ikx represents the wavefunction in terms of complex exponentials, cos(kx) and sin(kx) are the real and imaginary parts of these exponentials. So, the sine/cosine form is equivalent and more suitable for this case. Thanks
Alright, thank you very much for the reply. However, will the boundary conditions change if we have 5 regions instead of 3. Such as -infinity/0/-Vo/0/infinity? Is this situation possible?
You know L is proportional to E+V by supposition, where V is negative. So if V becomes more than E, E+V becomes negative and as a result L will become complex. Thanks
It's actually the constraint that in Region II, we're having bound states as the particle is bound to some potential. In case, when E is greater than V, it will no more be a bound state but a scattering state. Thanks
sir please can you elaborate on how we have 'l' as positive. Not getting quite clear how it was converted from complex to real. thankyou sir you are doing great job
Thanks for the appreciation and your interest in my lectures. Would you please mention the location of the video where you have this confusion. This will help me in addressing your question more effectively.
@@SAYPhysics Well if that is the case then for region 1 and 3 we have to take separate case(E>0),if they are unbound state how can we take them under bound state case?
@@shibaneethakur5035 Dear, we have a negative potential here and we want to calculate BSs. The geometry is such that we divide it in 3 different regions. At 2 regions (boundaries of potential) we have scattering behavior of the incident wf, while at 1 region (as it experiences the potential), the wf is trapped in. Thanks
@@SAYPhysics But the negative potential I only for the nucleus,not for outside region. Then for negative potential I.e E0,that's the completely different case
In an infinite square well, the energy levels are higher because the potential walls are infinitely high, meaning the particle is fully confined with no chance of escaping. In a finite square well, the energy levels are lower because the potential walls are finite, allowing for the possibility of tunneling, so the particle is less confined. Therefore, for the same quantum state, the energy is greater in an infinite square well compared to a finite square well. Thanks
@@SAYPhysics thanks Sir for your response but what i meant ..why we did m-not write the solution of the equation in regoin I as Aexp (ikx)+Bexp(-ikx) ..... Thanks in advance
Inside the well for bound states we write i in the solution. When out of potential, these are either an exponentially decaying or scattering solutions, so no need to write i in it. Thanks
Respected sir,
For a particle in a finite well with no potential on it is considered as free particle.. classically a free particles will retain its state with no change in energy.. but here.. we have different energy states even for a particle inside a potential well where it is supposed as free.. from where these energy states come??
This particle is said to be bound not free, as walls of the well are having infinite potential. Classically, like a rope tied at two ends, its middle can oscillate freely but not the nodes. Thanks
at 16:30 why psi(x)= C sin(lx) +D cos(lx) instead psi(x)= Ceikx+De-ikx. from whete sin(lx) came?
The real part of Ceikx is C cos(lx). then why we have written C sin(lx).
Thank you.
Thanks for your comment! The wavefunction for bound states in the finite square well is often written as ψ(x)=Csin(lx)+Dcos(lx), which comes from solving the Schrödinger equation in regions with constant potential. While Ceikx+De−ikx represents the wavefunction in terms of complex exponentials, cos(kx) and sin(kx) are the real and imaginary parts of these exponentials. So, the sine/cosine form is equivalent and more suitable for this case. Thanks
Vmin= -Vo and to obtain a bound state at region II : Energy (EVmin= -Vo , so E-Vmin >0 . L becomes real and positive
Thanks for your input 🙂
Thanks from India..Sir you are doing some really great job!
I'm pleased to hear from you Swapnil. I'll appreciate if you could inform your fellows and friends to benefit from it. Thanks
(Psi) should vanish at + and - infinity. However, in this particular situation x>a and x
Psi doesn't become zero at +-a, rather it exponentially decays on the boundary which cause Psi to vanish at infinity at both ends. Thanks
Alright, thank you very much for the reply. However, will the boundary conditions change if we have 5 regions instead of 3. Such as -infinity/0/-Vo/0/infinity? Is this situation possible?
Yes, as boundaries changes, so the BCs. Thanks
Best method of teaching
Thanks for the appreciation dear ☺️
You made a mistake in region 2 , Schrodinger equation is written wrong that .Due to which there is confusion in e+v to be positive when E>vmin
I rechecked p-79 DJ Griffiths 2nd ed. It's right. Thanks for your concern....
At 13:20 , how by making E>v , made l positive? I got confused, Can you explain me please.
You know L is proportional to E+V by supposition, where V is negative. So if V becomes more than E, E+V becomes negative and as a result L will become complex. Thanks
sir in region 2 , since V0 is a real positive quantity and E is negative but less then V0 ..so their sum will be positive real quantity?
It's actually the constraint that in Region II, we're having bound states as the particle is bound to some potential. In case, when E is greater than V, it will no more be a bound state but a scattering state. Thanks
appreciated sir 👍
sir please can you elaborate on how we have 'l' as positive. Not getting quite clear how it was converted from complex to real.
thankyou sir you are doing great job
Thanks for the appreciation and your interest in my lectures. Would you please mention the location of the video where you have this confusion. This will help me in addressing your question more effectively.
Bound state is confined to a single region,how we can take the equations in different region for E
Yes, you're right. But we'll have to find in all three regions. BTW, Region I and III gives scattering states. Thanks
@@SAYPhysics
Well if that is the case then for region 1 and 3 we have to take separate case(E>0),if they are unbound state how can we take them under bound state case?
@@shibaneethakur5035 Dear, we have a negative potential here and we want to calculate BSs. The geometry is such that we divide it in 3 different regions. At 2 regions (boundaries of potential) we have scattering behavior of the incident wf, while at 1 region (as it experiences the potential), the wf is trapped
in. Thanks
@@SAYPhysics
But the negative potential I only for the nucleus,not for outside region.
Then for negative potential I.e E0,that's the completely different case
@@SAYPhysics
I meant under E
Thanks for the all lecture sir .....
Sir why in scatternig states E>0
Welcome. When E0 (positive), there's no restriction on the particle movement and we call it as free or scattered. Thanks
Thanks sir..
Sir you are doing a great job...
You're welcome.
Respected sir what is the energy difference between finite and infinite square well which one has greather energy ?
In an infinite square well, the energy levels are higher because the potential walls are infinitely high, meaning the particle is fully confined with no chance of escaping. In a finite square well, the energy levels are lower because the potential walls are finite, allowing for the possibility of tunneling, so the particle is less confined. Therefore, for the same quantum state, the energy is greater in an infinite square well compared to a finite square well. Thanks
@@SAYPhysics Thanks sir this was mcqs in our quantum paper on 27 august 2024
Where it was in QM paper? Thanks
I am a student of physics Dept University of Buner .
Sir app ni khamari university mi seminar bi attend ki thi kham bhi us seminar mi thi.
@SaharNasib right dear. Good to know about you dear 😊
Sir , why the solution of Schrodinger equation in region 1 we donot have expikx I mean imaginary i in exp why on k==exp kx
Because we know that in this region V=0, so there will be no bound states but only scattering states. Thanks
@@SAYPhysics thanks Sir for your response but what i meant ..why we did m-not write the solution of the equation in regoin I as Aexp (ikx)+Bexp(-ikx) ..... Thanks in advance
Inside the well for bound states we write i in the solution. When out of potential, these are either an exponentially decaying or scattering solutions, so no need to write i in it. Thanks
Dear sir!
When x is - infinity and we take a decaying function why?
Psi should vanish at plus minus infinity. Thanks
Good
Thanks
This video is not full
Yes, watch the remaining two parts (L17.2 and L17.3) as well to complete the lecture. Thanks
Mashallah sir g
Thanks