Wave Function of Particle in Finite Potential Well

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  • Опубліковано 19 сер 2024
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КОМЕНТАРІ • 37

  • @highkey4316
    @highkey4316 5 місяців тому

    amazing. i think a middle schooler could get a good idea about this problem from this video. thank you a million times over for this

  • @muckrakerwm.8498
    @muckrakerwm.8498 9 років тому +18

    Hello,
    You have a no-nonsense style which I find exceptional to say the least. In your video 'Particle in a Finite Potential Well' you left the final steps for the viewer to solve, which is, finding the solutions using the sine & cosine as well as the complex exponentials to determine the subsequent solutions as tan (lx) and cot (lx). Do you have a follow-up video to the above title pertaining to deriving these solutions? I am interested in how you do it.

  • @laganjeet2177
    @laganjeet2177 6 років тому

    You are a great teacher, even a kindergarten kid can understand....

  • @gerrynightingale9045
    @gerrynightingale9045 7 років тому

    As a follow-up to this lecture regarding 'Eigen States'...all the energy in the known Universe is a virtual 'Constant' . By this concept, every particle in the Universe is also in a 'constant state of psi function' w/regard to it's 'potential'.

  • @gerrynightingale9045
    @gerrynightingale9045 7 років тому

    "All of the energy and matter that existed still exist. Matter does not create energy of itself. The actions of matter enable energy to become manifest".

  • @TheLittleGopher
    @TheLittleGopher 3 роки тому

    Explained way better than my lecturer lol thanks

  • @wildglorypsn
    @wildglorypsn 8 років тому +5

    very well explained...thank you so much :)

  • @zeenaligog
    @zeenaligog 8 років тому +3

    all is: solving TISE in three region and then apply the boundary conditions for wave function and its derivative and nothing else

  • @Rykurex
    @Rykurex 9 років тому +8

    I'm still trying to learn this topic and I can't understand your steps around the 6 minute mark. On the top of the right hand side of the board you have
    H Ψ'' + (E-V_0) Ψ = 0 (This I understand)
    If you divide by -2m/(h^2), then wouldn't the Ψ'' term also become negative? Or if you just divide by 2m/(h^2), would you have two positive terms?
    I understand the steps afterwards, but if there was a + sign instead of a -, wouldn't the solutions have imaginary exponents? Or you could change the E-V_0 to V_0-E and then have the - sign on the outside, which would change the value of alpha to have (V_0 - E)
    Thanks to anybody who can help!

    • @AKLECTURES
      @AKLECTURES  9 років тому +12

      I just checked my notes and it looks like I meant to write (Uo - E) and not (E - Uo) as I have it in the lecture! So the equation on the upper right hand side should be Ψ'' - α^2(Uo - E)Ψ = 0! Sorry for the confusion Ryan! And thanks for noting my sign error! :)

    • @Fuad_
      @Fuad_ 7 років тому

      :)

    • @ninjaengineer9393
      @ninjaengineer9393 4 роки тому

      @@AKLECTURES Thank you and no worries

  • @evakld9679
    @evakld9679 3 роки тому

    THANK YOU SO SO SO MUCH

  • @whattheisfor9200
    @whattheisfor9200 4 роки тому

    your lectures are amazing

  • @sunnypala7629
    @sunnypala7629 6 років тому

    Awesomely explained sr

  • @krystofsrsen8655
    @krystofsrsen8655 8 років тому +2

    So what is the wave function in this case actually? It doesnt say anywhere on the internet.

  • @tomsawyer6831
    @tomsawyer6831 6 років тому +1

    absolutely phenomenal explanation, thank you

  • @Scientificeducation785
    @Scientificeducation785 6 років тому

    JzakAllah o khaira

  • @chandrasekharanchandrasekh4066
    @chandrasekharanchandrasekh4066 5 років тому

    outstanding class .thank you

  • @maryannann8480
    @maryannann8480 6 років тому +2

    Hi there! I know you posted this long ago, so I hope you will see the comment and be able to answer me. I am reading the physical chemistry textbook by Atkins, De Paula, (10th edition) and on page 318, they say the opposite for the C=0 and D=0. How is that possible? They use the term C (in their case A) for calculating the motion in the forward direction, while you do the opposite. Is your interpretation the same thing but in different way or was there something missed here?
    Thank you

  • @AlexS-cb4ij
    @AlexS-cb4ij 6 років тому

    Thank you!

  • @adosar7261
    @adosar7261 5 років тому

    Does any force exist in this potential well ? Both inside and outside the well the derivative of potential energy with respect to position is 0 therefore zero force. So what is the point to have 2 different potentials ?

  • @ninjaengineer9393
    @ninjaengineer9393 3 роки тому

    Can someone explain where the K^2 term has gone? This value keeps being left out?

  • @BlackFiredDragon
    @BlackFiredDragon 4 роки тому

    Why do we need the normalization fifth equation? We have 4 unknowns, not 5

  • @vorsankoy3548
    @vorsankoy3548 8 років тому

    thank you, but I wanted to know about energy eigenfunction and eigenvalues in the same situation ı mean ın fınıt box

  • @Jayvaid
    @Jayvaid 6 років тому

    when you multiplied by 2m/h^2,there is a slight error in the final equation...or may be I missed something?please clear it

    • @diegode966
      @diegode966 5 років тому

      It was adjusted by the annotation on the video.

  • @MatanMor_3ix4d
    @MatanMor_3ix4d 3 роки тому

    king

  • @inadream3574
    @inadream3574 5 років тому

    Can someone plz confirm the final answer?

  • @Jarrod_C
    @Jarrod_C 5 років тому

    how does normalizing give the 5th equation?

    • @gabor6259
      @gabor6259 5 років тому

      Integral from -inf to +inf psi_1 + psi_2 + psi_3 dx = 1.

  • @TutorMak
    @TutorMak 5 років тому

    Sir please calculate c and d...

  • @Jarrod_C
    @Jarrod_C 5 років тому

    Where are the 5 equations?

  • @manuelsojan9093
    @manuelsojan9093 5 років тому

    is this assuming E

  • @jeyaneepan
    @jeyaneepan 8 років тому +2

    how to calculate "A"

    • @mustafaturk9280
      @mustafaturk9280 5 років тому +2

      It has been 3 years but anyway..
      All you have to do is normalizing the wave function in region 2. If you do so, you will find that A=(2/L)^(1/2)