[Astrokel]

Why should wave function be continous without any abrupt change in value at any point? My teacher said the wave function must be always positive or always negative because it cannot have a sudden change from positive to negative value. Why is that so? Thanks!

[tamim83]

The wavefuction has to be a "smooth fuction".  The reason is that the wavefuction should be differentable, or you have to be able to take a derivative of it and not get infinity .  This has to do with measurements; to find the momentum you apply the momentum operator (p=idx/dpsi) and you should get an answer that is not infinity. 

[pantone159]

There isn't any reason that the wavefunction can't change from + to - though, as long as the change is continuous.  E.g. the two lobes of a p orbital have opposite signs.

[Astrokel]

tamim83: althought i do not understand the answer much(im in grade 12), still thanks for the answer!

pantone159: I have attached 2 graphs, and the question was which of the following wave functions are not possible for a particle found in a certain region of space? why?

Left graph - Not a wave function because 1 value of x gives 2 values of psi as wave functions must be a continous function.

Right graph: Not a wave function because of the abrupt change in value from +v1 to 1-ve

Both are my teacher's answers

I have the following questions:

1) Wave function have to be a one-to-one function since it does not allow 1 x value gives 1 psi value.
2) 1 x-value corresponds to only 1 psi value because psi is a unique mathematical quantity?
3) You said that the as long as it is continious it does not matter of the change from + to - Why is the right graph not a wave function then? It is continous.

Thanks for any help given and sorry for the hand drawn graphs =D cookies for both of you 

[Valdorod]

It looks to me like there is a confusion between a function and a wave function.

you can "generally" make the following statement:

all wavefunctions are functions, but not all functions are wavefunctions.

Your teacher is correct in saying that a "wavefunction" cannot change from positive to negative.

We need to keep in mind that the probability of finding the electron is related to the square of psi.  the probability can be either positive or zero, the electron is somewhere or it is not there (zero probability) but the probability cannot be negative.  Psi can be all negative or all positive, since we are relating the square value to the probability, but as your teacher said it cannot be both.

Not all functions are solutions to the shroedinger equation, thus wavefunctions have limitations that many functions do not have.  Take a look at the following diagram which shows the first 3 eigenfunctions for the particle in a box.  You can see that you can have sinusoidal waves as a wavefunction, but all values are positive.  the case on the right is not allowed.



Valdo

[Borek]

Your teacher is correct in saying that a "wavefunction" cannot change from positive to negative.

Before you will spread more nonsense, take a look at the particle in the box wavefunction. In general

ψ_n = sqrt(2/L) sin(nπx/L)

where L is width of the box, π is π, x is coordinate, n is quantum number describing energy level. Obviously EACH wavefunction for n>1 IS negative is some interval (or even several separate intervals), that's just sin function. And you simply don't understand the plot you are looking at - vertical axis shows energy levels, but each wavefunction plot is separate and vertical line means zero.

I suppose when you refer to wavefunction you mean already square of the wavefunction, which - as a square - is positive everywhere. But wavefunction can be negative. Unfortunately both you and Astrokel's teacher teach wrong.

[tamim83]

Your teacher is correct in saying that a "wavefunction" cannot change from positive to negative.

We need to keep in mind that the probability of finding the electron is related to the square of psi.  the probability can be either positive or zero, the electron is somewhere or it is not there (zero probability) but the probability cannot be negative.  Psi can be all negative or all positive, since we are relating the square value to the probability, but as your teacher said it cannot be both.

Not all functions are solutions to the shroedinger equation, thus wavefunctions have limitations that many functions do not have.  Take a look at the following diagram which shows the first 3 eigenfunctions for the particle in a box.  You can see that you can have sinusoidal waves as a wavefunction, but all values are positive.  the case on the right is not allowed.

Wavefunctions can have + and - amplitude (or they can change sign) as long as the change in curvature is not abrupt (no pointy wavefunctions).  A great example is atomic orbitals (like a p orbital) which can have changes in sign.

As for the pictures the energies must be positive (En=n2h2/8mL2) but the wavefunction can have + and - amplitude as they do in both pictures.  The only problem with the picture on the right is that the enrgy (if your y axis is energy) is equal to zero.  The n quantum number cannot equal zero so E cannot equal zero for this problem (this can be explained using the Uncertainty principle). 

The wavefunctions are plotted superimposed on the energy levels and the energy levels do not reflect the sign of the wavefunction. 

[Astrokel]

Thanks for all the replies! Im struggling and its rather confusing, however it is part of my syllabus.

tamim83,

1) Wave function is not determined by changes in sign in amplitude but rather energy and it has to be >0 Is this right?

2) I hava attached another graph. Is the wavefunction pointy as you have described? My teacher said it is a wave function as he defined abrupt change in value as + to - and not pointy.

Thanks for being patience! 

[Astrokel]

P.S: the y-axis is psi.

[Valdorod]

I stand corrected,

Obviously I had a brain fart and did not distinguish between the wave function and the probability distribution.  You are correct in that the wave function can be sinusoidal especially when E is greater than the potential energy, while the probability cannot be negative due to the square of the wave function.

I am by no means the only one who occasionally missteps in this forum, and I am by no means ashamed to admit it when such is the case.

http://www.chemicalforums.com/index.php?topic=26828.msg101052#msg101052

http://www.chemicalforums.com/index.php?topic=26433.msg99514#msg99514

http://www.chemicalforums.com/index.php?topic=25100.msg94882#msg94882

http://www.chemicalforums.com/index.php?topic=24319.msg92298#msg92298

Unfortunately both you and Astrokel's teacher teach wrong.

As far as what and how I teach or not, I would welcome you to please sit in on one of my lectures before making such a generalized statement.  There are not a log of things that I am able to do, but those that I do, I am very good at.

Of course from Poland to Texas is not exactly a drive around the corner.  Nonetheless, if you are ever in the area I would gladly offer you a place to stay, and would gladly show you around West Texas (there is a lot more to the dessert than people realize)

On the other hand, siting in on one of my classes may be dangerous to you or my students.  This summer half the class did not pass an exam that was exactly as their practice one, which was accessible to them well in advance.  With such a group you might be led to murder, or suicide - I surely have banged my head on the wall a couple of times this summer.

Cheers

Valdo

[pantone159]

pantone159: I have attached 2 graphs, and the question was which of the following wave functions are not possible for a particle found in a certain region of space? why?

Left graph - Not a wave function because 1 value of x gives 2 values of psi as wave functions must be a continous function.

Right graph: Not a wave function because of the abrupt change in value from +v1 to 1-ve

The left graph isn't a function at all.  Continuity isn't really the problem, it isn't a valid function for anything.

There is nothing wrong with the right graph.  (Except that you used a mouse-based paint program to make line drawings, but I'll give you a pass on that.  )

[Borek]

I stand corrected

I am glad to hear that. You may not remember that, but few months ago we have long discussion on chemistry of ammonium chloride, you never admitted you were wrong there, although several people explained to you that you are and why. I must confess that I am still perceiving you through this discussion, could be I am wrong - as you know, I accasionally am.

Interestingly, I was today invited to visit Maine. I am not yet starting to plan my trip - I will wait till there will be no need for visa - but combination of Texas, Maine and Washington (my old friend lives in Seattle) and trips between seems to be a good starting combination to visit almost everything

[pantone159]

combination of Texas, Maine and Washington (my old friend lives in Seattle) and trips between seems to be a good starting combination to visit almost everything

Quite a distance to cover, that.   

[Borek]

Friend of mine (other one) spent back in eighties 2 years on post-doc at SUNY in Buffalo. After that he took his family to the car and they rode to Pacific and back. It took them about a month.

[Astrokel]

The left graph isn't a function at all.  Continuity isn't really the problem, it isn't a valid function for anything.

There is nothing wrong with the right graph.  (Except that you used a mouse-based paint program to make line drawings, but I'll give you a pass on that.  )

Hey pantone, that means as Valdorod said, all wavefunctions are functions but not all functions are wavefunctions. Thats to say, a function has to pass the vertical line test first before determining if its a wavefunction, right? What about the third graph i have posted? It's a valid function, however it has a pointy peak, which is considered as abrupt change in values? Thank you!! (for the help and of course, my passing) 

well, the problem in my physics A level is because of the introduction of new syllabus, which included and emphasize on modern physics topics which previously wasn't. My teacher was trained in civil engineering, and he admitted all this(semiconductor, schrodinger, uncertainty) was new to him. He had never learnt before in his life until he has to teach the course now. So all this is new to me and my teacher, thanks for the helps everyone!