How Do Complex Conjugates Affect Quantum Harmonic Oscillator Calculations?

In summary, the conversation discusses some difficulties in solving two problems involving the harmonic oscillator. The first problem involves using the wave function for when n=1, which leads to confusion about the complex conjugate and the cancellation of exponential functions in the calculation of <x>. The second problem involves calculating <p>, where there is confusion about using the operator function or <p>=m*d<x>/dt. The conversation also mentions conflicting advice from a grad student and the teacher.
  • #1
Tuneman
41
0
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 don't 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 canceled 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 canceled 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
bump thanks in advance
 
  • #3
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 don't 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 canceled 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
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, disappear 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 don't 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
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, disappear 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 don't 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
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 that's 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 wouldn't 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 don't 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 I am a little confused :(.
 
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  • #7
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 that's 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 wouldn't 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 don't 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 I am a little confused :(.

You'll get the same answer no matter which method you use.
 
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  • #8
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 don't see how that's odd or that becuase its odd, it would equal zero. but that's 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
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 don't see how that's odd or that becuase its odd, it would equal zero. but that's 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 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
ok, so I undestand now why <x>=0, and why <p>=0

So let's 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 won't 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 got to get me one of those equation editors :)
 
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  • #12
Tuneman said:
= integral {a[1]((mw)/(hbar))*x^4*e^(-(mw)/(hbar))x^2


and that won't 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 got to 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.
 

FAQ: How Do Complex Conjugates Affect Quantum Harmonic Oscillator Calculations?

What is quantum mechanics?

Quantum mechanics is a branch of physics that deals with the behavior of matter and energy at a very small scale, such as atoms and subatomic particles. It describes how particles behave and interact with each other through the use of mathematical equations.

Why is quantum mechanics important?

Quantum mechanics is important because it helps us understand the fundamental nature of matter and energy. It has also led to the development of many technological advancements, such as transistors, lasers, and computers.

What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This is because the act of measuring one property affects the other, making it impossible to have complete knowledge of both simultaneously.

What are quantum states?

Quantum states are the possible states that a particle can exist in, such as position, momentum, and spin. These states are described by mathematical equations and can change when a particle interacts with other particles or is observed/measured.

How does quantum mechanics differ from classical mechanics?

Classical mechanics is a branch of physics that deals with the behavior of macroscopic objects, while quantum mechanics deals with the behavior of microscopic particles. In classical mechanics, objects have definite positions and velocities, while in quantum mechanics, particles exist in multiple states simultaneously and have unpredictable behaviors.

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