ChemSiddiqui Posted October 18, 2009 Share Posted October 18, 2009 Hi everyone, I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes; [math] \frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)[/math] Now we have to operate the [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] onto function [math] sin(\theta)sin(\phi)[/math]. The operator can expressed into [math] cos^2(\theta)-sin^2(\theta) [/math] after diffentiation. Now my question is that can I simply multiply this by [math] sin(\theta)sin(\phi)[/math] or will this be wrong? It would have been much easier if the expression were [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] i.e. to diffentiatie the function?! I'll be grateful for your help! Link to comment Share on other sites More sharing options...

toastywombel Posted October 18, 2009 Share Posted October 18, 2009 Hi everyone, I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes; [math] \frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)[/math] Now we have to operate the [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] onto function [math] sin(\theta)sin(\phi)[/math]. The operator can expressed into [math] cos^2(\theta)-sin^2(\theta) [/math] after diffentiation. Now my question is that can I simply multiply this by [math] sin(\theta)sin(\phi)[/math] or will this be wrong? It would have been much easier if the expression were [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] i.e. to diffentiatie the function?! I'll be grateful for your help! no you cannot simply multiply sin(theta) by sin(phi) to simplify, they are separate unknown variables. the expression could not be [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] your sin(theta) should be sin squared(theta). Link to comment Share on other sites More sharing options...

ChemSiddiqui Posted October 18, 2009 Author Share Posted October 18, 2009 well this is why I am slighty confused. The question goes to operate wavefunction by an operator to check if the wavefunction is an eigenfunction of the operator. The operator is [math] U = \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] and the function is [math] f(\theta, \phi) = sin(\theta)sin(\phi) [/math] Now I know that I will have to use the product rule , but I dont know how, becuase the operator is very strange. Its like its saying that diffentiate the function f by [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] Maybe i havent understood the question properly? Link to comment Share on other sites More sharing options...

toastywombel Posted October 19, 2009 Share Posted October 19, 2009 I just gave my two bits on basis of what I know from algebra/trig. I want to help you answer you question but my math skills are limited lol. Someone more qualified that me please help this man! Link to comment Share on other sites More sharing options...

timo Posted October 19, 2009 Share Posted October 19, 2009 For a start: I think U is not an operator but a terrible mess . Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters? E.g. if U was [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] then f would obviously be an eigenfuntion. Link to comment Share on other sites More sharing options...

ChemSiddiqui Posted October 22, 2009 Author Share Posted October 22, 2009 For a start: I think U is not an operator but a terrible mess . Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters?E.g. if U was [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] then f would obviously be an eigenfuntion. You know what atheist, you got it right. The way the question was written down I didnt understand it . U is exactly as you suggested; [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] but I am not sure yet if it is an eigenfunction, so I will do the calculations but I know how to do that. Thanks a lot. Merged post follows: Consecutive posts mergedjust one more question. Its the double derivative question; [math] \frac {d^2}{dx^2} e^\frac{-x^2}{2}[/math] and this is what I did, frst taking the first derivate: [math] \frac {d}{dx} e^\frac{-x^2}{2} = \frac{-1}{2}e^\frac{-x^2}{2} [/math] Now taking the second derivative; [math] \frac {d^2}{dx^2} (\frac{-1}{2}e^\frac{-x^2}{2})=\frac{-x}{4} e^\frac{-x^2}{2} [/math] could anybody confirm id this is right or not? any help most appreciated! Link to comment Share on other sites More sharing options...

Bignose Posted October 22, 2009 Share Posted October 22, 2009 and this is what I did, frst taking the first derivate: [math] \frac {d}{dx} e^\frac{-x^2}{2} = \frac{-1}{2}e^\frac{-x^2}{2} [/math] chain rule [math] \frac{df(g(x))}{dx}= \frac{df}{dg}\frac{dg}{dx} [/math] letting [math]f(y)=e^y[/math] and [math]g(x)=-x^2[/math], then [math]\frac{dg}{dx}[/math] is clearly going to be be proportional to x to the 1st power. This is definately missing from your expression above. Link to comment Share on other sites More sharing options...

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