1. So, I was just messing around with some practice questions in my university physics textbook and I came across a particular question which was fairly easy to answer but which I would like to take further - however, it may be too advanced for me at the moment being a first year UK university physics student.

So, the question basically was:"The wavefunction

is a superposition of 2 free-particle wavefunctions, both k1 and k2 are positive.

a) Show that this wavefunction satisfies the Schrödinger equation for a free particle of mass m.
b) Find the probability distribution function for ψ(x,t)."

So, the first question (a) was simple of course - just taking partial derivatives then a bit of algebra. And the next part (b) gave me an answer of

using the Born interpretation procedure (complex multiplied by complex conjugate gives mod. complex squared which is the probability density - or "distribution function").

So, I was wondering how one could use this result for the probability distribution function to normalise the wavefunction - as I have done for simpler complex wavefunctions before. This is where I got to:

Now, I am sure that that is incorrect somehow as it just doesn't seem to add up - however, my reasoning is that integrating the probability distribution function between -infinity and +infinity will give you an answer of 1; as the particle has to be somewhere in this one dimensional space. But then I'm not too sure about integrating the function with respect to x then t and multiplying them to give the answer 1 - as when I do this, the equations get into a horrible mess and I end up with a value of A = 0 for normalising the wavefunction; which is nonsensical.

So, I was wondering how one may go about computing such a situation - it may be too advanced for me at the moment, but I love to learn! By the way, no this is not a "homework" question - I am not the kind of person who cheats in such a way for homework answers; however, evidently I cannot prove this so believe this is a homework question if you want to, it makes no difference to me as I know it isn't!

Thanks,

x(x-y)

2.

3. Ah wait, yes! I realise that integrating the modulus of the wavefunction ψ(x,t) squared with respect to time between the limits of negative and positive infinity doesn't make any sense - so, is it just an integral of ψ(x,t)^2 with respect to x assuming that t is a constant as if the wavefunction is normalised for one value of time then it follows that it must be normalised (where the normalisation factor remains constant) for all other times? Hmmm, well, I shall try this - but not now, I need to sleep really; but perhaps not, physics is far better than sleep I find.

4. I think I have it - after a long procedure of calculus and algebra, however I am not sure if the answer is correct. It would be great if someone could corroborate - I shall post my answer and full working soon!

5. You absolutely sure its not cos of x and its derivatives??

6. So, as I said I would provide in my last post – here is the procedure for which I found an answer (it may not be correct):

We have a wavefunction ψ(x,t) of the form

and we want to normalise this function to find A, thus we want to compute this integral

which is equal to 1 because the particle must be somewhere in space (along the one dimensional line in this case). However, if we compute the modulus of the wavefunction squared then we find that we still have a term with the variable time t in it (as shown in my original post). So, I stipulate that it would be easier to find another integral in order to determine the constant A – if the wavefunction can be normalised at a specific time then it must be normalisable for all other times t, thus we have the term:

Then one can manipulate Schrodinger’s one dimensional time dependent wave equation to find an expression for the integrand above – we multiply all the terms of Schrodinger’s equation by the complex conjugate of the wavefunction over (I times hbar) giving:

Then the complex conjugate of this equation is

adding the complex and complex conjugate expression will give an alternate expression for the integrand of the integral above:

which can be simplified to

If one substitutes this into the integral then you will obtain this expression overall:

Then you can work through the algebra of this latter expression remembering that

where k1 and k2 are both positive, and also remembering the rules for exponents etc. Doing this gave me the expression:

So that

And, there is the answer for A that I obtained! If somebody could corroborate this – or show that it’s wrong – I’d very much appreciate that. Thanks!

7. Originally Posted by fiveworlds
You absolutely sure its not cos of x and its derivatives??
What do you mean? Which part?

EDIT:

In regards to my long post above - please note that I have been a fool as it actually works out to be that A = 0 (which must be wrong) as I made the mistake of thinking it was a 1 on the right hand side of the equation rather than a zero. Dammit. Back to the drawing board I suppose...

8. Your graph will start at 0 then its sinx if its 1 its cosx

9. Originally Posted by fiveworlds
Your graph will start at 0 then its sinx if its 1 its cosx its p=np
The graph of the probability distribution function:

at t = 0 will have a value of 4A^2 at the origin (x = 0) and the value of the probability distribution function will vary between 4A^2 and 0.

And what's the relevance of p=np here?!

10. P=np you are trying to convert a exponential into a polynomial did you not understand this graph is cosx and sinx dependant on how you approach the question you are trying to solve with one point you need both

11. cos x and sin x dependent? I'm fairly sure that the graph of the prob. dist. function is only cos dependent - there is no sine term in the function.

13. Originally Posted by fiveworlds
Stop early? Relative uncertainty? I really haven't got a clue about what you're talking about - I can show you how I derived the prob. dist. function if you wish and thus how it does not involve any "sin x" function.

14. Exactly where is your second function.

15. Right, ok, we know that the wavefunction is

and we want to find the probability distribution function

And remember that

thus

See? No sine function there.

16. Ill put up an extended proof im working on it at the moment you dont expect me to just give everything away. i havent been paid yet maybe ill make a bit outta lectures.

17. Originally Posted by fiveworlds
Ill put up an extended proof im working on it at the moment you dont expect me to just give everything away. i havent been paid yet maybe ill make a bit outta lectures.
Well, I shall wait and see for what you post here - I must say that, unfortunately, a lot of what you have said here hasn't made much sense; I'd like to see some mathematical support for what you have been posting. I still do not know what you mean with respect to the prob. dist. function when you say "stop early", "relative uncertainty" and other phrases used. Hmmm... In the meantime, I shall continue working on this problem every now and then - however, I have a suspicion that it is not a function which can be normalised "easily" - either that or I'm missing something fundamental to the problem!

18. I think I know why the procedure I used in post #5 didn't work now - after reviewing page 38 of "E.*Merzbacher, Quantum Mechanics, Wiley, New York, 2nd edition, 1970" I found that that mechanism only works as long as

which it doesn't for the wavefunction in this case.

19. back to work. i know what to do i might be a while.

20. Well I tried another method of letting t = 0, as if the wavefunction can be normalised at one time then it must be normalised for all other times if it is a square-integrable function. However, again, when working through the algebra - I found that A = 0 (nonsensical). Also, I found that when using a program to try to integrate the probability distribution function - it stated that the integral does not converge, leading me to suspect that this wavefunction cannot actually be normalised (or at least not in a way that I know of)!

21. I really wish I could help but I have no idea what to add even though I'm itching to! What is the jist of what you are trying to do? Get rid of those infinities and zeroes? All I know about QT that might help is that you have a limit of reocurring instances in a given system and that must include this problem so there cannot be an infinite number involved? Is there a way you can use a very large number or calculate, ah I have no idea what I'm talking about.

What does it mean to normalise a wave function? To predict it? To rationalise it? To predict a pattern?

22. Originally Posted by Quantime
I really wish I could help but I have no idea what to add even though I'm itching to! What is the jist of what you are trying to do? Get rid of those infinities and zeroes? All I know about QT that might help is that you have a limit of reocurring instances in a given system and that must include this problem so there cannot be an infinite number involved? Is there a way you can use a very large number or calculate, ah I have no idea what I'm talking about.What does it mean to normalise a wave function? To predict it? To rationalise it? To predict a pattern?
In order to normalise a wavefunction one must find the value of the arbitrary constant(s) in the expression - for example, if we have the wavefunction which has a value:

between x = 0 and x = pi, and a value of zero everywhere else, then in order to normalise this function then we need to find the value of A. So, we can use the fact that the probability density (or distribution) function is given by the equation

where ψ(x,t)* is the complex conjugate of the wavefunction, which is (in this case):

notice that ψ(x) has been treated as a separated function of x in this case - where ψ(x) = ψ(x)* = A sin(x) - this can be done for convenience as ψ(x) is non-complex.

So, if we integrate this function (which is the probability density) between 0 and pi (where the wavefunction was defined above) then we will obtain a value of A (note that the integral equals 1 - maximum probability - because the particle must be somewhere in this region):

Hope that helps your understanding of normalisation?!

23. remember how you asked what the relevance of p=np was here its the tech heisenberg was talking about. But wexstill dont have all the tech i can show how it should work.

24. Thats chipping the iceberg with me and it is very interesting thank you

If you cannot be certain of the particles position but you know it is inbetween two values, could you then say that it is a fixed value say inbetween 0 and pi plus or minus a certain number? Would the wavefunction have an upper limit of behaviour and a lower limit? Can you eliminate time from the equation? Just bouncing ideas around. Probably no idea what I'm talking about.

25. eliminate time? no you are reliant on it what im talking about is that you are limited by your computer. My advise would be take something simple like an electron in a oscilloscope and work your way up from there. dont just start in the middle the first is prob a months work

26. Originally Posted by Quantime
Thats chipping the iceberg with me and it is very interesting thank you If you cannot be certain of the particles position but you know it is inbetween two values, could you then say that it is a fixed value say inbetween 0 and pi plus or minus a certain number? Would the wavefunction have an upper limit of behaviour and a lower limit? Can you eliminate time from the equation? Just bouncing ideas around. Probably no idea what I'm talking about.
1) What is a fixed value? The constant "A" or the position of the particle? If you meant the former then, yes, in the basic sense the constant A (calculated from normalisation of the wavefunction) is a fixed value (not "plus or minus a certain number") at all times t and all positions x if the wavefunction has a value defined (which is greater than zero and is finite) within a certain region and all other regions give the wavefunction with a zero-value (essentially setting up a square well potential). And if you mean the latter then, well, the position of the particle can simply defined by the range of x = a to x = b (in this case x = 0 to x = pi).

2) I'm not entirely sure I understand the question here.

3) No, the wavefunction - in this case - is dependent upon time as a variable and so one could not eliminate it from the wavefunction equation. However, as shown by my post above, by performing the modulus squared of the wavefunction (which is the complex conjugate of the wavefunction multiplied by the wavefunction) one eliminates time t from this expression - making the normalisation process much easier!

27. exactly but the time function is for the position on your graph that the electron stops moving. then you need to simplify to your constant

28. My question is that you shouldn't have a zero or infinite in there because that is wrong somewhere, I was looking for a solution by jiggling the variables for the sake of jiggling them, from a purely hypothetical viewpoint, such as removing time for instance. I'm sure the equation would break down but isn't that the point? Would it be easier finding the problem by forcefully changing variables and constants to see what happens? Not permanently of course and not part of the proof, just for the sake of seeing what happens to the different parts of the equation. Purely lateral thinking I have no education on wave function normalisation at all I was itching to jump in the deep end.

29. i wouldnt see the point of using infinity practically you are looking at small timeframes in anything you will do thats what i did for pnp. Take a snippet of what you are looking for your list of numbers and place them as your turning points for your graph. some np problems might not ever run but in theory there is always an awnser. So maybe take your runtime as constants. you have the speed of your waveform. therefore the distance it will cover. you need the a.o.e (area of effect) of the particles magnetic field and its strength and how you can see the changes. Its like programming remember windows when it started.

30. Originally Posted by Quantime
My question is that you shouldn't have a zero or infinite in there because that is wrong somewhere, I was looking for a solution by jiggling the variables for the sake of jiggling them, from a purely hypothetical viewpoint, such as removing time for instance. I'm sure the equation would break down but isn't that the point? Would it be easier finding the problem by forcefully changing variables and constants to see what happens? Not permanently of course and not part of the proof, just for the sake of seeing what happens to the different parts of the equation. Purely lateral thinking I have no education on wave function normalisation at all I was itching to jump in the deep end.
The plus and minus infinities are necessary for limits of the integral (they are the boundaries we are integrating between) as they tell us that the particle(s) must be somewhere within this space - a one dimensional space in this case. Essentially, if we have a wavefunction of a particle ψ (x,t) which exists at any point in the region of then the probability of finding the particle in this region must be equal to 1 (i.e. we are certain that it lies somewhere in infinite one dimensional space - it cannot be outside these limits).

In order to obtain the probability density function P(x) we must use the Born Interpretation of the wavefunction which is that the probability density is equal to the modulus squared of the wavefunction which, in turn through basic complex analysis, is equal to the wavefunction multiplied by it's complex conjugate. So if we integrate this function between our limits then we will obtain an answer of 1 (for reasons described above). So, written mathematically:

this expression allows us to normalise a "simple" wavefunction.

31. could you use a smaller number than infinity ie. if you knew the particle must be within 2 metres or whatever. i dont think your particle will wind up on the other side of the universe in a small timeframe. you are treating this really generally? Is it possible to reduce the integral smaller? its a 2d space but surely particles move in 3d generally how would you handle the change from 2d to 3d? i want to know how to manage the change from the particle moving in the oscilloscope. to when the function changes to the magnetic fields of the particles to 3d space and be able to resolve the vectors. If you are able to then you know the position and velocity of a particle in an atom.

32. Originally Posted by fiveworlds
could you use a smaller number than infinity ie. if you knew the particle must be within 2 metres or whatever. i dont think your particle will wind up on the other side of the universe in a small timeframe. you are treating this really generally? Is it possible to reduce the integral smaller? its a 2d space but surely particles move in 3d generally how would you handle the change from 2d to 3d? i want to know how to manage the change from the particle moving in the oscilloscope. to when the function changes to the magnetic fields of the particles to 3d space and be able to resolve the vectors. If you are able to then you know the position and velocity of a particle in an atom.
Indeed, I am computing the integral in the original post generally as it has not been defined to exist within a finite position range - therefore, we only know that it exists somewhere along a one-dimensional line. Also, this is only a 1 spatial dimension problem - the wavefunction

only involves the variables x and t (there are no y or z dimensions in this problem) - whilst k and omega are just the wavenumber and angular wave frequency respectively, which are constant; and A is a constant to be determined.

33. no this is definately two dimensional your points always lie on a line. in 2 dimensional space. pnp is this its the math for calculating waveforms this is one part you are thinking straight line but it isnt this graph has height. connect the dots similar to any wave. you know the derivative is the rate of change.what is your height prob 1,-1 or something similar. but all waves are at least 2d. x and y axis. Need any more help look up oscilloscope on wikipedia. If you use a computer it should be easy to take into account all your variables and therefore no more uncertainty. we have the tech but we need a few programs.

34. I have received a good response from another science forum, so I shall study the information given and see if I can achieve some solution - however, the level is a bit too advanced for me at the moment; I shall be more suited to solve this problem when I come onto Quantum Mechanics 2 in the second year of my Physics course!

35. actually i understand eventually your integrated waveforms always give constants which is 1d however since waveforms dont move like this at all they are always normally moving in at least 2d space unintegrated. the only time the waveform would be 1d in reality is at absolute zero.
secondly i was talking about heisenberg uncertainty and well and it can be done. however the ones that cannot be done are the ones that cannot easily be converted from the exponential to the polynomial time eqns and verifiable lists. np hard and stuff like this does not mean p!=np this is a misconception p=np but most of the time we cant find the awnser but it does exsist. However finding the exponential eqn here ha good luck solving it. basically for those interested you need the worst exponential eqn fractals between each point in 3d basically an infinite no of points to the height of the graph. However if can write the eqn for both you get the actual path subject to some uncertainty unless you knew the vectors if you knew where the particle started ie create the atom. though mega difficult math i dont know how long this would take but a long time. You need to take deflection in magnetic fields into account so think more like a spiral i guess and prob something similar tqo its normal wave pattern. oh and ps in this case your voltage etc is basically heat a change in temperature at all and you will mess up. and yes heat is electricity. Oh i think maybe im wrong but is heat not required for catalysis. sometimes sometimes not but there is always energy loss. maybe its impossible to make the perfect system or close to. youd have to know exactly what you are doing

36. Originally Posted by x(x-y)
So, I was just messing around with some practice questions in my university physics textbook and I came across a particular question which was fairly easy to answer but which I would like to take further - however, it may be too advanced for me at the moment being a first year UK university physics student.

So, the question basically was:"The wavefunction

is a superposition of 2 free-particle wavefunctions, both k1 and k2 are positive.

a) Show that this wavefunction satisfies the Schrödinger equation for a free particle of mass m.
b) Find the probability distribution function for ψ(x,t)."

So, the first question (a) was simple of course - just taking partial derivatives then a bit of algebra. And the next part (b) gave me an answer of

using the Born interpretation procedure (complex multiplied by complex conjugate gives mod. complex squared which is the probability density - or "distribution function").

So, I was wondering how one could use this result for the probability distribution function to normalise the wavefunction - as I have done for simpler complex wavefunctions before. This is where I got to:

Now, I am sure that that is incorrect somehow as it just doesn't seem to add up - however, my reasoning is that integrating the probability distribution function between -infinity and +infinity will give you an answer of 1; as the particle has to be somewhere in this one dimensional space. But then I'm not too sure about integrating the function with respect to x then t and multiplying them to give the answer 1 - as when I do this, the equations get into a horrible mess and I end up with a value of A = 0 for normalising the wavefunction; which is nonsensical.

So, I was wondering how one may go about computing such a situation - it may be too advanced for me at the moment, but I love to learn! By the way, no this is not a "homework" question - I am not the kind of person who cheats in such a way for homework answers; however, evidently I cannot prove this so believe this is a homework question if you want to, it makes no difference to me as I know it isn't!

Thanks,

x(x-y)
If you'd like I can check out all of the math worked out here. I could use the exercise, get all those synapsyes firing. However by just looking at the wave function it seems apparent to me that it simply cannot be normalized. You do realize that not all wavefunctions can be normalized, right?

37. Originally Posted by pmb
If you'd like I can check out all of the math worked out here. I could use the exercise, get all those synapsyes firing. However by just looking at the wave function it seems apparent to me that it simply cannot be normalized. You do realize that not all wavefunctions can be normalized, right?
I realise now that this wavefunction is a plane wave and thus cannot be normalised in "normal" Hilbert space - the Hilbert space needs to be extended to rigged Hilbert space in order to include the plane wave. And I know that the mathematics I have done in the OP, and subsequently I believe, is correct - it all works out; however, indeed, I have been using the wrong methods to try to normalise the wavefunction as explained above.

38. [QUOTE=x(x-y);380329]
Originally Posted by pmb
And I know that the mathematics I have done in the OP, and subsequently I believe, is correct - it all works out; above.
I don't follow. How can it be correct if the function is not normalizable and you obtained normalization factors?

I'm being lazy today due to many days lack of sleep so please be patient with me. Otherwise I'd recheck the math in all the gorey detail from the start. I'll do if upon request of course. But your response confuses me. Please help clarify it for me?

39. i have to agree with pmb here i know very little i do this because i enjoy it and maybe in ten years ill be fairly good at it for the moment im trying my best to keep up with you because its interesting. It isnt my field of study if it was id be a bit better

40. [QUOTE=pmb;380343]
Originally Posted by x(x-y)
Originally Posted by pmb
And I know that the mathematics I have done in the OP, and subsequently I believe, is correct - it all works out; above.
I don't follow. How can it be correct if the function is not normalizable and you obtained normalization factors?I'm being lazy today due to many days lack of sleep so please be patient with me. Otherwise I'd recheck the math in all the gorey detail from the start. I'll do if upon request of course. But your response confuses me. Please help clarify it for me?
I meant that the mathematics itself is correct - i.e. I haven't computed anything incorrectly as the probability distribution function is correct:

However, I have not actually used the correct methods to extract the normalisation coefficient - i.e. the mathematics is right but the methods used are wrong as they cannot be used for this plane wavefunction.

41. Originally Posted by x(x-y)
I meant that the mathematics itself is correct - i.e. I haven't computed anything incorrectly as the probability distribution function is correct:

However, I have not actually used the correct methods to extract the normalisation coefficient - i.e. the mathematics is right but the methods used are wrong as they cannot be used for this plane wavefunction.
For a wave function to be normalizable you'd have to integrate

over the entire x-axis. However the integral will not converge and its for this reasonits not normalizable. Therefore there must be an error in post #5 since your result gave a finite normalization coefficient

42. Originally Posted by pmb
Originally Posted by x(x-y)
I meant that the mathematics itself is correct - i.e. I haven't computed anything incorrectly as the probability distribution function is correct:

However, I have not actually used the correct methods to extract the normalisation coefficient - i.e. the mathematics is right but the methods used are wrong as they cannot be used for this plane wavefunction.
For a wave function to be normalizable you'd have to integrate

over the entire x-axis. However the integral will not converge and its for this reasonits not normalizable. Therefore there must be an error in post #5 since your result gave a finite normalization coefficient
Yes, indeed - I realised that the integral does not converge and so the wavefunction (which represents a plane wave) cannot be normalised using classical Hilbert space, one must extend Hilbert space to rigged Hilbert space in order for the plane wave to be able to be normalised.

Also, if you read the "EDIT" subsection of post #6 then you will find that I realised that I had made a slight error which had lead to me assume that A has a finite value - when, in fact, A actually had a value of zero (i.e. nonsensical). So, yes, you just need to read post #6; maybe I should've just deleted that last part of post #5.

43. Originally Posted by x(x-y)
Also, if you read the "EDIT" subsection of post #6 then you will find that I realised that I had made a slight error which had lead to me assume that A has a finite value - when, in fact, A actually had a value of zero (i.e. nonsensical). So, yes, you just need to read post #6; maybe I should've just deleted that last part of post #5.
Ah ha! I see. That's the problem when one sticks his nose in to a thread which is so long. Sorry.

Question - How much experience do you have with QM, if you don't mind me asking? All I saw you refer to in the OP was a "university textbook." I can't tell from that what level it is. The reason I'm asking is that all I had to do was merely look at the wave function to know that it wasn't normalizable. That comes with time and practice. Then one does the calculation and if its finite then you know you made an error. This is based on what has come to be known as "Wheeler's First Moral Principle". I can quote the entire principle if you'd like. It's from Spacetime Physics - 2nd Ed by Taylor and Wheeler

44. Originally Posted by pmb
Ah ha! I see. That's the problem when one sticks his nose in to a thread which is so long. Sorry.