Sarahisme Posted March 12, 2006 Share Posted March 12, 2006 hey all, just having a bit of trouble with this question: what i think i have figured out so far is that, V(x) = 0 (because its a free particle) and i am also thinking that i should be using the time dependent SE? would anyone be able to me a pointer in the right direction? Cheers Sarah Link to comment Share on other sites More sharing options...

RyanJ Posted March 12, 2006 Share Posted March 12, 2006 This should help you out Cheers, Ryan Jones Link to comment Share on other sites More sharing options...

timo Posted March 12, 2006 Share Posted March 12, 2006 what i think i have figured out so far is that, V(x) = 0 (because its a free particle) and i am also thinking that i should be using the time dependent SE? "V(x)=0 because it´s a free particle" is ok. Using the time-dependent version sounds reasonable, too. After all, a time-parameter is given explicitely. would anyone be able to me a pointer in the right direction? Sure. You´re given two functions and are asked if they are solutions to a differential equation. Plug them in and see if both sides of the diff. eq. equal (and under what conditions they do so - there´s two free parameters w and k in the problem). Perhaps it helps to rewrite the sine and the cosine as cos(x) = 0.5(exp(ix) + exp(-ix)) and sin(x) = -0.5i(exp(ix)-exp(-ix)), perhaps it´s more a burden than a help - figure it out yourself. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 12, 2006 Author Share Posted March 12, 2006 ok now what i've got is that: rearraging the TDSE: since V(x) = 0 [math] \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} = i \hbar \frac{\partial \Psi}{\partial t}[/math] (1) then for [math] \Psi_I [/math] : [math] \frac{\partial^2 \Psi}{\partial x^2} = -k^2Acos(kx - \omega t) [/math] and [math] \frac{\partial \Psi}{\partial t} = \omega ASin(kx - \omega t) [/math] subbing these into 1 gives: LHS = [math] \frac{\hbar^2 k^2 A}{2m}cos(kx -\omega t) [/math] and RHS = [math] i \hbar \omega A sin(kx - \omega t) [/math] but then i am a little bit stuck... :S Link to comment Share on other sites More sharing options...

Severian Posted March 12, 2006 Share Posted March 12, 2006 So to work you need LHS=RHS for all values of x and t. Pick one x and t and see what happens, eg. x=0, t=0. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 12, 2006 Author Share Posted March 12, 2006 So to work you need LHS=RHS for all values of x and t. Pick one x and t and see what happens, eg. x=0, t=0. ok, so i do: [math] \frac{\hbar^2 k^2}{2m}cos(kx - \omega t) = it \omega sin(kx - \omega t) [/math] then pick x=0 and t=0, and so get: [math] \frac{\hbar^2 k^2}{2m}cos(0) = it \omega sin(0) [/math] which leads me too: [math] \frac{\hbar^2 k^2}{2m} = 0 [/math] but how does that show anything? i think i am little lost Link to comment Share on other sites More sharing options...

Severian Posted March 13, 2006 Share Posted March 13, 2006 You made an assumption in your first line. (If A=0 you can't divide it out.) But lets go with the assumption. You have shown that for a non-zero solution (ie with A not zero) you need [math]\frac{\hbar^2 k^2}{2m}=0[/math]. But this is a necessary condition - not a sufficient one. What happens for values of [math]kx-\omega t[/math] of [math]\pi/2[/math] and [math]\pi/4[/math]? (Remember the equations need to work for all values of x and t.) Then think about how this answers the original question. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 13, 2006 Author Share Posted March 13, 2006 You made an assumption in your first line. (If A=0 you can't divide it out.) But lets go with the assumption. You have shown that for a non-zero solution (ie with A not zero) you need [math]\frac{\hbar^2 k^2}{2m}=0[/math]. But this is a necessary condition - not a sufficient one. What happens for values of [math]kx-\omega t[/math] of [math]\pi/2[/math] and [math]\pi/4[/math]? (Remember the equations need to work for all values of x and t.) Then think about how this answers the original question. for kx-wt = pi/2 ----> itwA = 0 for kx-wt = pi/4 ----> [math] \frac{\hbar^2 k^2}{2m} A = it \omega A [/math] k, sorry, not to sound too stupid, but i am still stuck. :S Link to comment Share on other sites More sharing options...

Sarahisme Posted March 13, 2006 Author Share Posted March 13, 2006 plz delete this post, accidently posted same thing twice! Link to comment Share on other sites More sharing options...

timo Posted March 13, 2006 Share Posted March 13, 2006 for kx-wt = pi/2 ----> itwA = 0for kx-wt = pi/4 ----> [math] \frac{\hbar^2 k^2}{2m} A = it \omega A [/math] k' date=' sorry, not to sound too stupid, but i am still stuck. :S[/quote'] Well, from three values for kx-wt you already extracted the conditions k=w=0. Checking if cos(0*x - 0*t) is a solution for the Schroedinger eq. shouldn´t be too hard. Link to comment Share on other sites More sharing options...

Severian Posted March 13, 2006 Share Posted March 13, 2006 Ok, let me be more explicit. Lets assume A is not zero for a moment. The first equation [math]\frac{\hbar^2 k^2}{2m}=0[/math] implies that [math]k=0[/math] (since the other variables are just numbers). The one at pi/2 implies w=0. So putting this back in the original equation [math]\Psi_I = A \cos (kx - \omega t) = A \cos 0 = A[/math] So the equation is only a solution if it is a constant (the case A=0 is a subset of this too). This shows why the question isn't a very good one because both answers are correct. [math]\Psi_I[/math] is a solution of the Schroedinger equation if k=w=0. But, I am sure that this is not the answer intended by the question. This solution is a particle with zero momentum and zero frequency (infinite wavelength) which is non-physical. (There would be a problem with normalization too, but this is true also for a plane wave.) Think about it also in tems of eigenstates: Let's write [math]\Psi_I = \frac{A}{2}\left( \psi_+ + \psi_- \right)[/math] with [math]\psi_\pm = e^{\pm (kx-\omega t)}[/math] Putting [math]\psi_+[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = \hbar \omega[/math] so it is a valid solution as long as this holds. It is an energy eigenstate with [math]E= \frac{\hbar^2 k^2}{2m}[/math] Putting [math]\psi_-[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = - \hbar \omega[/math]. [math]\psi_-[/math] is an energy eigenstate with [math]E= -\frac{\hbar^2 k^2}{2m}[/math]. But this cannot hold if the earlier one did (unless [math]k=0[/math]). The state cannot be in two (non-degenerate) eigenstates at once. (I think that was the point of the question). Link to comment Share on other sites More sharing options...

Sarahisme Posted March 13, 2006 Author Share Posted March 13, 2006 ok thanks for that, i'm off to bed, but i'll have a look in the morning! thanks you very much once again Severian Link to comment Share on other sites More sharing options...

kenshin Posted March 13, 2006 Share Posted March 13, 2006 I thing that what I would like to point out is that the functions under considration are not well behaved.They are nither normalizable(hope so)nor their limit tends to zero as x tends to infinity.So I think they can't be a solution.But what really bothers me is that beautyfull func. like sin and cos are not solutions.Also,from eq,if tan(kx-wt) is equal to a particular contant(the original solution before putting values.),then eq has cos(kx-wt) as a solution.Contradiction??What do you people think?Or,if we trust eq, can we say that tan can have any value on no. line but that which is also against the behaviour of tan.I am so confused. .Please ponder and reply.Also excuse me for spelling mistakes if any. Link to comment Share on other sites More sharing options...

Severian Posted March 13, 2006 Share Posted March 13, 2006 They are nither normalizable(hope so)nor their limit tends to zero as x tends to infinity.So I think they can't be a solution. Neither is [math] \psi = A e^{i(kx-\omega t)}[/math] but it is a solution. Link to comment Share on other sites More sharing options...

Sarahisme Posted March 14, 2006 Author Share Posted March 14, 2006 Ok' date=' let me be more explicit. Lets assume A is not zero for a moment. The first equation [math']\frac{\hbar^2 k^2}{2m}=0[/math] implies that [math]k=0[/math] (since the other variables are just numbers). The one at pi/2 implies w=0. So putting this back in the original equation [math]\Psi_I = A \cos (kx - \omega t) = A \cos 0 = A[/math] So the equation is only a solution if it is a constant (the case A=0 is a subset of this too). This shows why the question isn't a very good one because both answers are correct. [math]\Psi_I[/math] is a solution of the Schroedinger equation if k=w=0. But, I am sure that this is not the answer intended by the question. This solution is a particle with zero momentum and zero frequency (infinite wavelength) which is non-physical. (There would be a problem with normalization too, but this is true also for a plane wave.) Think about it also in tems of eigenstates: Let's write [math]\Psi_I = \frac{A}{2}\left( \psi_+ + \psi_- \right)[/math] with [math]\psi_\pm = e^{\pm (kx-\omega t)}[/math] Putting [math]\psi_+[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = \hbar \omega[/math] so it is a valid solution as long as this holds. It is an energy eigenstate with [math]E= \frac{\hbar^2 k^2}{2m}[/math] Putting [math]\psi_-[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = - \hbar \omega[/math]. [math]\psi_-[/math] is an energy eigenstate with [math]E= -\frac{\hbar^2 k^2}{2m}[/math]. But this cannot hold if the earlier one did (unless [math]k=0[/math]). The state cannot be in two (non-degenerate) eigenstates at once. (I think that was the point of the question). hmm ok i think i begining to understand all this a bit more. So for [math] \Psi_I [/math] , it is a solution if k=w=0 or if A = 0. both are degenerate not physical cases. just to confirm... there is only those 2 cases for [math] \Psi_I [/math] ? so i did a similar thing for [math] \Psi_{II} [/math]: i got, that for [math] \Psi_{II} [/math] to satify the TDSE you need: [math] \frac{\hbar^2 k^2}{2m} Asin(kx - \omega t) = -i \hbar \omega Acos(kx - \omega t) [/math] for x = 0, t = 0, you get that w = 0 and for (kx - wt) = pi/2 you get, k = 0 so the only 2 possible cases for [math] \Psi_{II} [/math] to be a solution to teh TDSE is when either: A = 0 or when k=w=0 what do you guys think? i didnt really understand the eigenstates stuff, so yeah. do you think what i have is an acceptable answer? because it seems a bit of a degenerat kind answer...??? Link to comment Share on other sites More sharing options...

kenshin Posted March 14, 2006 Share Posted March 14, 2006 sorry repeated it. Link to comment Share on other sites More sharing options...

kenshin Posted March 14, 2006 Share Posted March 14, 2006 Neither is [math] \psi = A e^{i(kx-\omega t)}[/math'] but it is a solution. Yes you are right.I misstated it all.I say that cos and sine are not solutions as clear from the calculations done by sarahisme.Cos will be a solution iff tan has a constant value as calculated(using [math] \frac{\hbar^2 k^2 A}{2m}cos(kx -\omega t) [/math]=[math] i \hbar \omega A sin(kx - \omega t) [/math]),but niether cos,nor sine or even the function that you stated can represent a real entity as they are not well behaved.So given problem is resolved into two problems: 1)Is cos a solution of a given DE? 2)Is it a function representing a real entity? As stated,answer to both the questions is no.And hence the given functions don't represent a solution of S.E,niether physically nor mathematically.At least that's what I think. Link to comment Share on other sites More sharing options...

timo Posted March 14, 2006 Share Posted March 14, 2006 So it is resolved into two problems:1)Is cos a solution of a given DE? 2)Is it a function representing a real entity? As stated' date='answer to both the questions is no.At least that's what I think.[/quote'] @1) As Sarah showed, the first function is a solution exactly if A=0 or w=k=0. Why do you say that it isn´t? @2) You are right that it isn´t. But that´s not asked for in Sarah´s exercise. Link to comment Share on other sites More sharing options...

Severian Posted March 14, 2006 Share Posted March 14, 2006 do you think what i have is an acceptable answer? Yes, I think it is. As I said before it is not terribly clear what the question is after. Since [math]\Psi = A[/math] is a solution of the Schroedinger equation, technically the correct answer is 'yes, they are solutions, with values of k and w given by ....'. But I suspect the questioner wants you to say 'no' because [math]\Psi=A[/math] is not physical. The 'eigenstate stuff' is just terminology. By Eigenstate, I mean a solution of the equation with a definite value of w and k. So your [math]\Psi=A[/math] is an eigenstate with w=k=0. Link to comment Share on other sites More sharing options...

kenshin Posted March 14, 2006 Share Posted March 14, 2006 @1) As Sarah showed' date=' the first function is a solution exactly if A=0 or w=k=0. Why do you say that it isn´t?@2) You are right that it isn´t. But that´s not asked for in Sarah´s exercise.[/quote'] 1)Arent we forgetting that k=0 or w=0 removes one variable from the picture and hence the function changes.A=0 is certainly not an acceptable ans,at least not for me. 2)The que has been asked for a free particle.Under these conditions,the func. won't even represent something physical,what to talk of free partical? Link to comment Share on other sites More sharing options...

timo Posted March 14, 2006 Share Posted March 14, 2006 1)Arent we forgetting that k=0 or w=0 removes one variable from the picture and hence the function changes. A=0 is certainly not an acceptable ans,at least not for me. The zero function is a function, actually a pretty important one . The constant function is a function, too. Of course the function changes for different parameters (A, k and w). 2)The que has been asked for a free particle.Under these conditions,the func. won't even represent something physical,what to talk of free partical? Strictly speaking, it does not ask if the solutions are physically meaningfull - at least to my reading. It asks "are the functions solutions of a particular diff. eq" where the particular diff. eq is the Schroedinger eq. for a free particle. That might sound like nitpicking but there´s a catch in it. The functions [math] \psi_{\pm} [/math] Severian brought up are not solutions for a free particle, either. But as you might know, for differential equations there is something which I suppose is called the "solution space". For linear diff. eqs this solution space can be seen as a vector space (unless you refuse the zero-function to be a function - in that case you lack the zero-vector). In this case, the functions [math] \psi_{\pm} [/math] are a (mathematically) valid base for this vector space. You can construct all solutions of the diff. eq. from these base vectors - including the physically meaningfull ones. That´s why asking if a particular function is a solution for a 'physical' differential equation can make sense even if the function itself cannot represent a physical entity. Link to comment Share on other sites More sharing options...

kenshin Posted March 14, 2006 Share Posted March 14, 2006 The zero function is a function' date=' actually a pretty important one . The constant function is a function, too.Of course the function changes for different parameters (A, k and w). Strictly speaking, it does not ask if the solutions are physically meaningfull - at least to my reading. It asks "are the functions solutions of a particular diff. eq" where the particular diff. eq is the Schroedinger eq. for a free particle. That might sound like nitpicking but there´s a catch in it. The functions [math'] \psi_{\pm} [/math] Severian brought up are not solutions for a free particle, either. But as you might know, for differential equations there is something which I suppose is called the "solution space". For linear diff. eqs this solution space can be seen as a vector space (unless you refuse the zero-function to be a function - in that case you lack the zero-vector). In this case, the functions [math] \psi_{\pm} [/math] are a (mathematically) valid base for this vector space. You can construct all solutions of the diff. eq. from these base vectors - including the physically meaningfull ones. That´s why asking if a particular function is a solution for a 'physical' differential equation can make sense even if the function itself cannot represent a physical entity. 1)I have no problem with zero or constant function.I am just saying,that if it were the case(k=0 etc) then examinor/teacher/instructor would have used the function cos(kx) or cos(wt).Also isn't it that A=0 will mean no partical(free or bound)at all?So I am assuming that A is not = to 0 and niether are k or w. 2)secondly,we are discussing physics and here for me a solutions means one which represents a physical quantity/entity.Also you said that"You can construct all solutions of the diff. eq. from these base vectors - including the physically meaningfull ones. ".Is it possible for a non-well behaved function as well? What do you people think?Please enlighten me. Link to comment Share on other sites More sharing options...

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