Normalization of the wavefunction

[@biswarup077]

It would have been very difficult to jump start my my self learning QM using Griffith's book if had not found your videos. Thanks for making these videos.This was just what I needed.

[@rashidalali6510]

What a coincidence, I also came here because of Griffith

[@ahmed24-z2b]

and same here

[@Peter_1986]

A lot of physics books have an irritating tendency to go on forever about a million different things at the same time, so that it takes a very long time to read them. Video tutorials tend to be much more straightforward and to the point.

[@riosvm]

Best explanation of wave function normalization on KZbin.

[@sphericalchicken]

No real magic there... 1/i is equal to -i. You can see that starting with i^2 = -1. Dividing both sides by i should give you i = -1/i, so multiplying both sides by -1 gives you -i = 1/i. It's handy sometimes to move i to the numerator, so I've made that 1/i --> -i conversion.

[@mikevaldez7684]

An Excellent exposition of Griffth's much abbreviated proof in chapter 1 of the 2nd ed.

[@sphericalchicken]

It's worse than that, actually -- Psi(x) can be positive or negative, and can even by complex. The missing piece is that Psi (the wavefunction) is not itself the probability density. Instead, we treat |Psi(x)|^2, the squared absolute magnitude of Psi(x), as the probability density, and |Psi(x)|^2 is always a positive real number.

[@genekisayan6564]

hey, could you explain why we can factorize with a derivative ? I am about the question left unanswered in the video

[@vasudevankn575]

soo much thanks mahn, been searching for this for years

[@tommygaa]

at 14:34 it should be PSI* (psi star) :)

[@zenojimneuromansah8665]

You are excellent at explaining your mathematical reasoning. Your mini visual proof of e ^-ix multiplied with it's complex conjugate was very helpful. thanks

[@jacquelinebaeza7462]

I took QMI last semester and aced it. I had no idea what any of the math meant, but thanks to you know I understand what this means!! only took me 2 weeks into QMII to realize I didn't know what I was doing

[@goldenchopstick1788]

great explanation of probabilistic interpretation! I finally understood the relationship thanks.!

[@goutham94]

dear sir there is a slight mistake in the 1st term of schrodinger equation on the RHS at 6:24 Great video sir.. thank you :D

[@Andrew6James]

@Brant Carlson Could you please expand on the @18:48

[@StephenRayner]

What software are you using?

[@dannijunglejim5692]

Great explanation! Much appreciated before my exam!

[@asifhossain2863]

this will greatly help in my today's exam.... superb!!!!

[@deconfinedQPT]

at 19:08 the reason it is true that because differentiation is a linear transformation hence superposition and homogeneity is preserved, for people who wonders as to why he was able to rewrite the expression

[@executorarktanis2323]

explain more is this not simple common

[@universal69]

The reason isn't this complex

[@alexfriebe1508]

would love a video on expectation values and spherical polar coordinates

[@FlintPet]

14:48 There is a star missing on the last blue psi for those who are confused :)

[@nelsonappiagyei4713]

At 23:08 why did you not integrate from negative infinity to positive one but negative one to positive one?

[@ahmed24-z2b]

the epsi is zero at any intervak except this

[@hidden_anonymous]

Great lecture. Thank you!

[@danv8718]

Fantastic series! Thanks a lot for sharing

[@maxmiller5656]

thanks, clear and concise!

[@عبدالرحمنالعصيمي-ي6ظ]

Thank billion time for this amazing video , go ahead for such amazing explanations👍🏻😩😩😩

[@juanreyes8564]

Hi I am having trouble understanding what you said at 5:35 that it is not posible to have a function that stays non zero or goes to infinity as x goes to infinity and still have to be integrable. If you could explain that will be great thanks.

[@juanreyes8564]

I think i understand: The wave function has to be zeros at both ends -infinity and pos infinity because if its not the the integral from -infinity to infinity will not be 1 thefore the wave function is non-normalizable.

[@jonnyyy9716]

If the wave function satisfies the conditions for normalisation does it suffice to differentiate the integral of psi squared dx = 1 wrt time? And then the RHS goes to 0?

[@stijndhondt9611]

This is golden. Thanks alot!

[@RI-xt4nh]

Didn't know Eric Foreman taught quantum mechanics.

[@RICHFRVR]

Haha

[@kimyongtan3818]

Brant Carlson, May I know why 18:48 is true?

[@RaveSlave2DaGrave]

late but just take the derivative using product rule, some of the terms cancel out leaving you with the equation Brant left

[@Andrew6James]

@@RaveSlave2DaGrave Can you be more specific please?

[@TheBrazilianFury]

@@RaveSlave2DaGrave thanks!!

[@richroylance4630]

Excellent lecture...thank you.

[@dannijunglejim5692]

At around 13:30 You divide the Schrödinger equation by ih, and the way I would have computed that, the i would have stayed next to (2m), however you have placed it as if you were multiplying the equation by i. I also am unsure of what happened to the negative symbol in that step. Could you please explain the reasoning behind these steps?

[@weisun184]

This is my mom’s account but I think he multiplied by i on both side. On the left side of the shrodinger eqn you would have i^2 = -1 and hence the signs are flipped.

[@gibsonmaglasang]

thank you so much prof. carlson. i learned a lot from your lecture videos. do you have lec videos on many particle physics or quantum field? thanks :)

[@mkminerals12343]

dear,i m confused here,once you said that infinite square amplitudes are not normalizablet,right,as in dirac delta function,/////then you fit this idea to integral(summation) of infinite basis. as we know that square integrable functions converges in hilbert space in h2 space,i.e in infinite basis. so normalizable.

[@kharonofficial]

What calculus you talked when you solved the integral?

[@gforcebreakin]

sounds like main guy from that 70s show

[@lasha97]

I can't get the part when you bring up functions in 02:37 how can probability be negative? sigh represents probability of finding a particle in some point of x right? well how can it be negative?

[@magtutorial3606]

That's probability density psi(x) Not probability psi(x)2

[@sacha7958]

I'm pretty sure that the normalization constant at the end could also have been MINUS the square root of 15/16.

[@thewalesj897]

Brilliant!

[@nashtrojan]

great video.

[@compphysgeek]

I dont know who the first person was to put arrows on graphs but that person deserves some sort of punishment

[@reimalm7191]

hi, can the constant be a negative value

[@niemandwirklich]

I was asking myself the same question - mathematically yes, but would it make sense for a wave function having a negative sign? I think so, yes, but I could be wrong...

[@Dekoherence-ii8pw]

The wave function has a real and an imaginary part. Both of those parts are a wave which oscillate between positive and negative. If you multiply the wave function by -1 you just change the phase of the wave by 180 degrees (pi radians). @@niemandwirklich

[@Dekoherence-ii8pw]

If the constant is (for example) -3.5, then positive 3.5 would work just as well. The sign of the constant doesn't make a difference, because we're interested only in the square of the absolute value.

[@muhammadziaulislamarsalan1392]

thank u sir...

[@a1ang0r85]

can anyone show the step for 22:00?

[@cellerism]

That is the most simple way of expressing it. u cant get anything more simpler. Its just like x^2=x(x) or u can say x^2+x^2=x(x+x)

[@executorarktanis2323]

@@cellerism pls help is it just that he took partial dx common or something big i missed

[@AlchemistOfNirnroot]

wrt the derivative of the partial derivative of psi*, why isnt +(iV/h-bar)psi not psi star?

[@AlchemistOfNirnroot]

1st derivative****

[@professionalprocrastinator8103]

Yeah he forgot

[@Vector_world2036]

Position density

[@SamsherSinghJoon]

explain ???//

[@basilpines2201]

3:44 am and 6 hours till ny final

[@juliopchile]

d'fuck??