Does the Probability Density \( P(r) \) Depend on Time for a Given Wavefunction?

[JamesJames]

Will P(r) depend on time? Explain your reasoning.

The wavefunction is
$$\frac{1}{\sqrt{2}} (\psi_{2,1,-1}+\psi_{2,1,1})$$

$$\frac{1}{16}\,{\frac {r{e^{-1/2\,ra}}\sin \left( \theta \right) \left( {e^{-i \phi}}-{e^{i\phi}} \right) \sqrt {2}\sqrt {{\pi }^{-1}}}{\sqrt {{a}^{3 }}a}}$$

Guys, this is really urgent and I am genuinely lost here...any help would be really appreciated. I can show why P(r) does not depend on time quite easily but how do I show or explain that P(r) depends/does not depend on time?

Please guys, I really need the help.

James

 

[dextercioby]

I can show why P(r) does not depend on time quite easily but how do I show or explain that P(r) depends/does not depend on time?

I'll let u make up your mind.Either u can or not show that baby is time independent??Astronauts from ISS could see that from space...Why can't u?

Daniel...

 

[JamesJames]

I would say time was used in deriving the solutions to this, the Hydrogen atom, problem. The Shrodinger equation was the time dependent Shordinger equation.

But how is this related to P(r)?

 

[JamesJames]

When we take the probaility density, the exponent containing time has an i in front of it so I would say that P(r) does NOT contain the time factor. But so far, all I have seen is P(r) so I am not sure how to EVEN WRITE P(r) let alone tell if it depends on time.

James

 

[dextercioby]

I would say time was used in deriving the solutions to this, the Hydrogen atom, problem. The Shrodinger equation was the time dependent Shordinger equation.

But how is this related to P(r)?

1.Why the hell didn't u sau that in the first place??There u have a superposition of 2 possible states,both sharing the same energry and enterng the expression of the time dependent wave function with the same exp(i"blablabla") factor.Once u take the square modulus,that time dependence would cancel out,sine one is with plus,and the other with minus.So there won't be any time dependence of the probability function.
I sincertely hope u know the definiton of P(\vec{r}),else i would be talking in vain... :grumpy:

 

[JamesJames]

Now I get it...that helps. Now, what would a sketch of P($$\phi$$) for this state look like? I simplified the exponentials that contain $$\phi$$ and get $$-2i sin \phi$$. Am I supposed to do this when sketching P($$\phi$$)? I am unable to see what P($$\phi$$) would like for this state.

 

[JamesJames]

Really, anything would help me for constructing $$P(\phi)$$.

 

[dextercioby]

Now I get it...that helps. Now, what would a sketch of P($$\phi$$) for this state look like? I simplified the exponentials that contain $$\phi$$ and get $$-2i sin \phi$$. Am I supposed to do this when sketching P($$\phi$$)? I am unable to see what P($$\phi$$) would like for this state.

If u have the addition of the 2 wave functions (both solutions of the Schroedinger equation for the H atom),and the \phi dependence in the wave functions is proportional to exp(im\phi),therefore,by substituting m=1 and m=-1 and adding them,u'll be left with the function cosine \phi,instead of "i times sine\phi".I believe your first expression for the time independent wave function is wrong,since adding exp (i\phi) and exp(-i\phi) would require a plus between them (it's adding,right??)
So sketching P(r,\theta,\phi) only as afunction of \phi is something like sketching the graph of constant times (cosine squared) of "x" which is not difficult at all.
Your problem is a classical one in applicative QM,as i remember doing it last year at one seminar of QM.But we were asked to sketch the entire orbitals.

Good luck!

 

[JamesJames]

According to a table that I am looking at (in the book by Brehm and Mullin), there is a minus sign between the two exponentials. I am adding but one of the $$Y_{lm}$$ s contains a minus sign so that converts the plus to minus which is why I have the sin$$\phi$$ rather than the cosine function.

 

[dextercioby]

According to a table that I am looking at (in the book by Brehm and Mullin), there is a minus sign between the two exponentials. I am adding but one of the $$Y_{lm}$$ s contains a minus sign so that converts the plus to minus which is why I have the sin$$\phi$$ rather than the cosine function.

We're both right here.We've touched a sensitive point af mathematics,the phase conventions for the spherical harmonics.I used the conventions from Abramowitz,Segun and apparently the autors of your book chose a different convention.In that case,your functions are valid.
Anyhow,mathematicians and physicists didn't decide which convention to adopt,so that's why mathematical expressions for h atom wavefunctions differ from on to another.I would reccomend for the applicative part of QM:Sieg.Fluegge:"Practical QM".2 volumes.The best in the field.

Daniel.

 

[JamesJames]

But how would I prove that what I get is the same as what you get? i.e. how could I show that the function I use gives $$[cos (\phi)]^{2}$$ which is what you are getting? Should I just ignore the minus sign and say VOILA ? Cause then I can plot the $$[cos(\phi)]^{2}$$ function easily...it is the 2i in front of the resulting sine function that was causing me discomfort !

 

[dextercioby]

But how would I prove that what I get is the same as what you get? i.e. how could I show that the function I use gives $$[cos (\phi)]^{2}$$ which is what you are getting? Should I just ignore the minus sign and say VOILA ? Cause then I can plot the $$[cos(\phi)]^{2}$$ function easily...it is the 2i in front of the resulting sine function that was causing me discomfort !

These different phase conventions cannot change the final outcome (they're usually chosen of the form that,when squared,should yield "1");the final is not cosine squared of phi but sine squared of phi.
That "i" dissappears when considering the probability function which is the sqared modulus of the wave function.So that "-2i sin\phi" would go into 4 sine squared of phi.

That does it,i guess.

Daniel.

 

[JamesJames]

what about the negative sign ..? [-2isin($$phi) ]^{2}$$ gives -4[sin($$phi)]^{2}$$...how can there be a negative probability density?

 

[dextercioby]

what about the negative sign ..? [-2isin($$phi) ]^{2}$$ gives -4[sin($$phi)]^{2}$$...how can there be a negative probability density?

It can't and it isn't.And that's a rule.
When considering the square modulus of a complex nummer which happens to have the an imaginary part (which is the case here,it actually misses the "real" part),try first to calculate the complex conjugate of the number.Then apply the famous formula $|z|^2=zz^*$.
It will never work by simply taking the square of the original complex number in the case of "i" including numbers.
I hope u know how to build "z*" for any complex number.If u can't/don't,u'll no chance of ever making head or tail of any QM calculation.

I hope u put the book reccomandation on a piece of paper and remember to consult it whenever "devius/awkward looking" QM problems might stay in your way to an absolute happiness.

Daniel.

 

[JamesJames]

That was a stupid mistake on my part..I apologize for the inconvenience. I get it now. I have another question...hope you aren't annoyed beyond control by now.

Spin and symmetry transformations: Consider a spin one half particle
a) Find the two properly normalized eigenstates of $$S_{z}= \frac{\hbar}{2}\sigma_{z}$$

b) Find the two properly normalized eigenstates of $$S_{x}= \frac{\hbar}{2}\sigma_{x}$$

c) Construct the operator that rotates spin vectors by $$2\pi$$ about the $$\hat z$$ axis. hint...you might think this is the identity operator

d) Construct the operator that rotates spin vectors by $$\pi /2$$ about the $$\hat y$$ axis. hint...you might think this is the identity operator

I managed to get a and b but am lost as to where to begin for c) or d).

Any suggestions?

James

 

[JamesJames]

Are these somehow going to contain a factor of $$exp(i \phi)$$ or something? I am just guessing here.

Any hepl would be really useful to me.

James

 

[JamesJames]

Guys, any suggestions would help me..please, I am genuinely confused about this.

James