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Intro to Quantum Homework Help! Please!

  • Thread starter Tuneman
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I am having trouble with 2 problems about the harmonic oscilator.

First of all the question tells me to use the wave function for when n=1

So I have:
Wavefunction = A[1]*r^1/2 * x * e^-((r^2)/2)

where r= (mw/(hbar))^1/2
I am wondering when I am multiplying this by the complex conjugate, is the complex conjugate going to have a e^+((r^2)/2). I dont think I am but for some reason, in a solution finding <x> which equals ](integral) psi* (times) x (times) psi] those exponential functions did not appear. So I was wondering how they cancelled out when you calculate <x>

also it asks me to calculate <p>, which I know equals m*d<x>/dt. My question is, if my <x> does not depend on t, because the time dependent part of the equation was cancelled out when solving for <x>, how can I find d<x>/dt?

Any help would be appreciated, I'm sure my questions or equations aren't too clear, so if you have any questions, I will try to clarify. Thank you.
 
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  • #2
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bump thanks in advance
 
  • #3
Tom Mattson
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Tuneman said:
where r= (mw/(hbar))^1/2
Just a little nitpick: You forgot the 'x'.

I am wondering when I am multiplying this by the complex conjugate, is the complex conjugate going to have a e^+((r^2)/2).
No, [itex]e^{-r^2/2}[/itex] is a real-valued function. Real quantities are equal to their own complex conjugates.

I dont think I am but for some reason, in a solution finding <x> which equals ](integral) psi* (times) x (times) psi] those exponential functions did not appear. So I was wondering how they cancelled out when you calculate <x>
That's because those exponential functions (I assume you mean [itex]e^{-ikx}[/itex]) are complex-valued, and in general are not equal to their own complex conjugate.

also it asks me to calculate <p>, which I know equals m*d<x>/dt.
For a state function [itex]\psi(x)[/itex] and an operator [itex]\hat{O}[/itex] the expectation value [itex]\left<\hat{O}\right>[/itex] is given by:

[tex]\left<\hat{O}\right>=\int_{-\infty}^{\infty}\psi^*\hat{O}\psi dx[/tex]

Use your wavefunction and the fact that the momentum operator is [itex]\hat{p}=-i\hbar\frac{d}{dx}[/itex]. Try to use a symmetry argument when it comes time to integrate.
 
  • #4
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Ok, so this is what I tought,

but what I ment was that the real expoential value does not show up in the answer I was provided.


The problem is I was told by a grad student that

<x> = integral of a[1]^2*((hw)/(hbar))*x^3

so somehow that e^(-(r^2)/2) that should have come up in both the wavefunction and the complex conjugate, dissapear when evaluating. Also I was told I could use <p>=m*d<x>/dt, and use <T> (kenetic energy) = <p^2>/2m

My answer, and theirs, for <x> does not depend on t, so I am dumbfounded on how they got their solution and how they expect me to apply it.

I came here thinking the same thing you told me. I dont understand how the intstructor and company were able to come up with that solution for <x> and also, how I can use <x>=m*d<x>/dt. My teacher is a real stickler to keep wth his method, do you know of any way they could be correct? Thank you for you help.
 
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  • #5
Tom Mattson
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Tuneman said:
but what I ment was that the real expoential value does not show up in the answer I was provided.
What, the answer for <x>? Of course it doesn't appear there, because you're doing a definite integral. That doesn't mean that a real exponential does not equal its own complex conjugate.

The problem is I was told by a grad student that

<x> = integral of a[1]^2*((hw)/(hbar))*x^3

so somehow that e^(-(r^2)/2) that should have come up in both the wavefunction and the complex conjugate, dissapear when evaluating.
Have you tried to evaluate it? If so, then how far have you gotten?

Also I was told I could use <p>=m*d<x>/dt, and use <T> (kenetic energy) = <p^2>/2m

My answer, and theirs, for <x> does not depend on t, so I am dumbfounded on how they got their solution and how they expect me to apply it.
Why are you dumbfounded? Surely you know how to differentiate a constant function!

I came here thinking the same thing you told me. I dont understand how the intstructor and company were able to come up with that solution for <x> and also, how I can use <x>=m*d<x>/dt. My teacher is a real stickler to keep wth his method, do you know of any way they could be correct? Thank you for you help.
I can't say if your teacher is correct without knowing his answer. What did he get for <x> and for <p>?
 
  • #6
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Tom Mattson said:
What, the answer for <x>? Of course it doesn't appear there, because you're doing a definite integral. That doesn't mean that a real exponential does not equal its own complex conjugate.

ya, let me clarify, what he gave me was that
<x> = Integral {Sumation: a[n]*((mw)/(hbar)*x^2n+1 dx}

I got rid of the sumation becuase N=1 here, so I evaluated this to
<x> = integral { a[1]*(mw/(hbar)*x^3 }

so my answer was <x> = a[1]*((mw)/(hbar)*1/4*x^4

so thats why I have gotten confused tho exponent, from the time independent wavefunction (e^(-r^2)/2) dissapears before i integrate according to him. The only reason for that, that I could come up with was that if somehwere in psi* I had e^(r^2/2).






Tom Mattson said:
Why are you dumbfounded? Surely you know how to differentiate a constant function!
Well wouldnt that mean that then <p>=0? and thus kenetic energy=0 and I was told that KE and PE were going to both equal to 1/2 the total energy. So i dont think i can have a zero KE.


Tom Mattson said:
I can't say if your teacher is correct without knowing his answer. What did he get for <x> and for <p>?
He didn't really give me the the final answer, he just gave me that hint. And I asked him ok, well when I calculate <p>, can I use <p>=mdx/dt, and he said yes. But I got a hold of someone and said to use the operator function for <p>. So as you can see im a little confused :(.
 
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  • #7
Tom Mattson
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Tuneman said:
ya, let me clarify, what he gave me was that
<x> = Integral {Sumation: a[n]*((mw)/(hbar)*x^2n+1 dx}

I got rid of the sumation becuase N=1 here, so I evaluated this to
<x> = integral { a[1]*(mw/(hbar)*x^3 }

so my answer was <x> = a[1]*((mw)/(hbar)*1/4*x^4
That's wrong, <x> is not a function of x. As I said before you're doing a definite integral over x. x cannot be in the answer.

so thats why I have gotten confused tho exponent, from the time independent wavefunction (e^(-r^2)/2) dissapears before i integrate according to him. The only reason for that, that I could come up with was that if somehwere in psi* I had e^(r^2/2).
It sounds like whoever was helping you is extremely confused himself. Look, you've got an integral of the form:

[tex]
\int_{-\infty}^{\infty}x^3e^{-ax^2}dx
[/tex]

Since the integrand is odd, the integral--and hence <x>--is obviously zero.

Well wouldnt that mean that then <p>=0?
Yes, it would. You would get the exact same answer if you did it the way I recommended.

and thus kenetic energy=0 and I was told that KE and PE were going to both equal to 1/2 the total energy. So i dont think i can have a zero KE.
The KE isn't zero, because:

[tex]\left<\hat{T}\right>\neq\frac{\left<\hat{p}\right>^2}{2m}[/tex]

Rather, we have:

[tex]\left<\hat{T}\right>=\frac{\left<\hat{p}^2\right>}{2m}[/tex]

which is completely different.

He didn't really give me the the final answer, he just gave me that hint. And I asked him ok, well when I calculate <p>, can I use <p>=mdx/dt, and he said yes. But I got a hold of someone and said to use the operator function for <p>. So as you can see im a little confused :(.
You'll get the same answer no matter which method you use.
 
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  • #8
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Tom Mattson said:
That's wrong, <x> is not a function of x. As I said before you're doing a definite integral over x. x cannot be in the answer.



It sounds like whoever was helping you is extremely confused himself. Look, you've got an integral of the form:

[tex]
\int_{-\infty}^{\infty}xe^{-ax^2}dx
[/tex]

Since the integrand is odd, the integral--and hence <x>--is obviously zero.
So wait wouldn't I have: integral A[1]^2*(mw/(hbar)*x^3*e^-r^2. And you're saying that over -infinity to infinity that equals zero becuase the inegrand is odd? i dont see how thats odd or that becuase its odd, it would equal zero. but thats the least of my worries.



The KE isn't zero, because:

[tex]\left<\hat{T}\right>\neq\frac{\left<\hat{p}\right>^2}{2m}[/tex]

Rather, we have:

[tex]\left<\hat{T}\right>=\frac{\left<\hat{p}^2\right>}{2m}[/tex]

which is completely different.
ok so then does <p^2>=m d<x^2>/dt equal zero? becuase 0^2=0?

how do I compute T then?'



also I have that to find <v>

where <v>=1/2*m*w^2*integral{psi*(times)x^2(times)psi}

so now if I know that integral{psi*(times)x(times)psi}=0

wouldn't <v>=0? or am i missing something?
 
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  • #9
Tom Mattson
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Tuneman said:
So wait wouldn't I have: integral A[1]^2*(mw/(hbar)*x^3*e^-r^2.
You're right, that should be x3 in there. There was a typo in my integral, but it's fixed now.

And you're saying that over -infinity to infinity that equals zero becuase the inegrand is odd? i dont see how thats odd or that becuase its odd, it would equal zero. but thats the least of my worries.
The integrand is clearly odd, because if you substitute [itex]-x[/itex] in for [itex]x[/itex], you pick up an overall negative sign on the integrand. And it is a well-known fact from calculus that definite integrals over odd integrands vanish. You need to review your fundamentals.

ok so then does <p^2>=m d<x^2>/dt equal zero? becuase 0^2=0?
No, because [itex]\left<\hat{p}^2\right>\neq0[/itex].

how do I compute T then?'
Put the expression for [itex]\hat{T}[/itex] into the following equation:

[tex]\left<\hat{O}\right>=\int_{-\infty}^{\infty}\psi^*\hat{O}\psi dx[/tex]

You will find that it is not zero.

also I have that to find <v>

where <v>=1/2*m*w^2*integral{psi*(times)x^2(times)psi}
This looks very strange. What is v, and why are you putting x^2 in the integrand?

so now if I know that integral{psi*(times)x(times)psi}=0

wouldn't <v>=0? or am i missing something?
This doesn't make any sense at all.

It is true that:

[tex]\int_{-\infty}^{\infty}\psi^*x\psi dx=0[/tex]

But this does not in any way imply that:

[tex]\int_{-\infty}^{\infty}\psi^*x^2\psi dx=0[/tex]

and in fact the last integral is wrong.
 
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  • #10
Tom Mattson
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Tom Mattson said:
This looks very strange. What is v, and why are you putting x^2 in the integrand?
Never mind this bit, I get it now. I was thinking that v stood for velocity (which is why I thought it strange, this being QM and all). But clearly v stands for potential energy. I was thrown off because PE is usually represented by a capital V.

The rest of my comments are unchanged though.
 
  • #11
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ok, so I undestand now why <x>=0, and why <p>=0

So lets see if I have <x^2> straightened out.

<x^2>= integral {psi*(times)x^2(times)psi}

= integral {a[1]((mw)/(hbar))*x^4*e^(-(mw)/(hbar))x^2


and that wont equal zero becuase its a even function. I would probibaly solve this through integration by parts? It would take a few steps though...no?

so once I have <x^2> I could try <p^2>=md<x^2>/dt

OR

<p^2> = Integral {psi* (times) -i^2 (times) hbar^2 (times) d^2psi/dx^2}


after calculating that

To find <T> I use:

<T> = <p^2>/2m

and then for average potential I use

<V>= 1/2*m*w^2*[integral {psi* (times) x^2 (times) psi}


Sry for the strange way I write the equations of the interweb, I gotta get me one of those equation editors :)
 
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  • #12
Tom Mattson
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Tuneman said:
= integral {a[1]((mw)/(hbar))*x^4*e^(-(mw)/(hbar))x^2


and that wont equal zero becuase its a even function.
Yes.

I would probibaly solve this through integration by parts? It would take a few steps though...no?
No.

Go to an integral table and look up the following class of integrals:

[tex]\int_0^{\infty}x^ne^{-ax^2}dx[/tex]

You can use that when calculating both the PE and the KE.

Sry for the strange way I write the equations of the interweb, I gotta get me one of those equation editors :)
The equation editor is built right into this website, and you can use it. If you click on any of my equations you will see a pop-up that has the source code. Follow this link tutorials and more samples.
 

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