Oscillation of a bound particle in a superposition of states

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  • #1
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Homework Statement


A bound particle is in a superposition state:
[tex] \psi(x)=a[\varphi_1(x)e^{-i\omega_1t}+\varphi_2(x)e^{-i\omega_2t}] [/tex]
Calculate [itex] <x> [/itex] and show that the position oscillates.

Homework Equations


[tex] <x>=\int_{-\infty}^{\infty} \psi(x) x \psi^*(x) \mathrm{d}x [/tex]

The Attempt at a Solution


[itex] <x>=\int_{-\infty}^{\infty} \psi(x) x \psi^*(x)\mathrm{d}x=\int_{-\infty}^{\infty} x[|\varphi_1(x)|^2 +x|\varphi_2(x)|^2+2\varphi_1(x)\varphi_2(x)\cos(\tilde{\omega}t) ]\mathrm{d}x [/itex]
where [itex] \tilde{\omega}=\omega_1 - \omega_2 [/itex],
I have assumed that [itex] \varphi_1(x) [/itex] and [itex] \varphi_2(x) [/itex] are real functions (is this a valid assumption? I think it is, because if there were some imaginary component it could just go into the phase).

Here I got stuck.
what I think i need to do is:
1)say that [itex]\int_{-\infty}^{\infty} x|\varphi_1(x)|^2 \mathrm{d}x [/itex] is zero (same for the second state). I think it's true because of a mathematical trick. maybe odd function over a symmetric interval type of thing? but im not sure if i can say [itex] \varphi(x) [/itex] has a defined parity,

2) define the oscillation amplitude [itex]A=\int_{-\infty}^{\infty} 2x\varphi_1(x)\varphi_2(x)\mathrm{d}x [/itex].
If this is true, what guarantees that the integral is finite?

Is this correct? is it too generalized and there was anything more specific i can do?
I feel like i'm being stumped by mathematical properties and not the physics :/
Thanks
R.
 

Answers and Replies

  • #2
kuruman
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The important thing to focus on is that the wavefunctions describe a bound particle. What does this say to you about their behavior at infinity? The assumptions that the wavefunctions are real or even are not needed to answer the question, so assume that they are not and see what you get.
 
  • #3
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The important thing to focus on is that the wavefunctions describe a bound particle. What does this say to you about their behavior at infinity? The assumptions that the wavefunctions are real or even are not needed to answer the question, so assume that they are not and see what you get.

Thanks for the response!
So, I understand that the functions are localized and drop to zero at infinity, which means [itex] \int_{-\infty}^{\infty} x|\varphi_1(x)|^2 \mathrm{d}x [/itex] is the average location of the site. so if we call [itex] \int_{-\infty}^{\infty} x|\varphi_1(x)|^2 \mathrm{d}x =-d [/itex] and [itex] \int_{-\infty}^{\infty} x|\varphi_2(x)|^2 \mathrm{d}x=+d [/itex], the amplitude should be [itex]d[/itex] as well, to show oscillations between the sites. how do i see this?

Also, if the function are not real, I cannot form a Cosine term, so why isn't this assumption needed?
 
  • #4
kuruman
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... so if we call ##\int_{-\infty}^{\infty} x|\varphi_1(x)|^2 \mathrm{d}x =-d##...
It's OK to call one integral ##-d## if you wish, but what is your justification for calling the other ##+d##? What would happen if you called it ##Bob## or just ##b##? In fact it is best that you give it a different name and then interpret what you get in the end.
Also, if the function are not real, I cannot form a Cosine term ...
Oscillations are not exclusively described by cosines. Making the functions real simplifies the math, but you still get a sinusoidal with complex functions. Just multiply ##\psi(x)## by its complex conjugate and see what you get.
 
  • #5
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It's OK to call one integral ##-d## if you wish, but what is your justification for calling the other ##+d##? What would happen if you called it ##Bob## or just ##b##? In fact it is best that you give it a different name and then interpret what you get in the end.
Can I not assume without loss of generality that the states are symmetrical about some zero point?

Oscillations are not exclusively described by cosines. Making the functions real simplifies the math, but you still get a sinusoidal with complex functions. Just multiply ##\psi(x)## by its complex conjugate and see what you get.

Here is my attempt:
## \int_{-\infty}^{\infty} x[\varphi_1(x)\varphi_2(x)^* e^{-i(\omega_1-\omega_2)t} + \varphi_1(x)^*\varphi_2(x)e^{+i(\omega_1-\omega_2)t}]\mathrm{d}x=##
let ## g(x)=\varphi_1(x)\varphi_2(x)^*=a(x)+ib(x)##, then i get
##\int_{-\infty}^{\infty} x*2Re[\varphi_1(x)\varphi_2(x)^* e^{-i(\omega_1-\omega_2)t}]\mathrm{d}x##
This is some function of the form ##a(x)\cos(\tilde{\omega}t)+b(x)\sin(\tilde{\omega}t)##
Is this correct?
It seems very general, is there anything I can assume to simplify this?
If i go back to my first assumption, that I can take the phase out of ##\varphi_i## and be left with a real function, i can allow a phase in the argument, and just write this whole thing as ##\cos(\omega t+\phi)## where ##\phi## is the phase difference between the two states.
 
  • #6
PeroK
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Can I not assume without loss of generality that the states are symmetrical about some zero point

Here is my attempt:
## \int_{-\infty}^{\infty} x[\varphi_1(x)\varphi_2(x)^* e^{-i(\omega_1-\omega_2)t} + \varphi_1(x)^*\varphi_2(x)e^{+i(\omega_1-\omega_2)t}]\mathrm{d}x=##
let ## g(x)=\varphi_1(x)\varphi_2(x)^*=a(x)+ib(x)##, then i get
##\int_{-\infty}^{\infty} x*2Re[\varphi_1(x)\varphi_2(x)^* e^{-i(\omega_1-\omega_2)t}]\mathrm{d}x##
This is some function of the form ##a(x)\cos(\tilde{\omega}t)+b(x)\sin(\tilde{\omega}t)##
Is this correct?
It seems very general, is there anything I can assume to simplify this?
If i go back to my first assumption, that I can take the phase out of ##\varphi_i## and be left with a real function, i can allow a phase in the argument, and just write this whole thing as ##\cos(\omega t+\phi)## where ##\phi## is the phase difference between the two states.

I think you are on the right track and nearly there. You just need to tidy up what you have. I feel it is easier to assume the functions are real in the first instance.

You could also assume in the first instance that your last integral converges - and prove this separately if necessary.

Remember these are definite integrals - i.e numbers, not functions of ##x##.
 
  • #7
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I think you are on the right track and nearly there. You just need to tidy up what you have. I feel it is easier to assume the functions are real in the first instance.

You could also assume in the first instance that your last integral converges - and prove this separately if necessary.

Remember these are definite integrals - i.e numbers, not functions of ##x##.
Thanks,

If i do make the assumption these function are real (just wanted to verify that I am allowed to do that), my question boils down to, what do i make of ##
A=\int_{-\infty}^{\infty} 2x\varphi_1(x)\varphi_2(x)\mathrm{d}x##.
Should the amplitude indeed be half the distance between the two sites (so that at some time, the particle is exactly in a site) or is it bigger? smaller? what do i make of this?
 
  • #8
PeroK
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Thanks,

If i do make the assumption these function are real (just wanted to verify that I am allowed to do that), my question boils down to, what do i make of ##
A=\int_{-\infty}^{\infty} 2x\varphi_1(x)\varphi_2(x)\mathrm{d}x##.
Should the amplitude indeed be half the distance between the two sites (so that at some time, the particle is exactly in a site) or is it bigger? smaller? what do i make of this?
You don't need to make anything of it, except that the integral is finite.
 
  • #9
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You don't need to make anything of it, except that the integral is finite.
Ok, so in general it can be bigger/smaller/equal to the half the distance between the sites?
and only given specific functions I can see what it actually is?

also, could you please tell me what you think about my assumption that without loss of generality, I can call the average location of the sites -d and +d, symmetrical about an arbitraty zero point?
 
  • #10
PeroK
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Ok, so in general it can be bigger/smaller/equal to the half the distance between the sites?
and only given specific functions I can see what it actually is?

also, could you please tell me what you think about my assumption that without loss of generality, I can call the average location of the sites -d and +d, symmetrical about an arbitraty zero point?

You can only assume that the expected value of ##x## in the two eigenstates is finite: ##x_1, x_2## or ##\langle x \rangle_1, \langle x \rangle_2## would be the notation I'd use.

The last integral you could denote by ##x_{12}## or ##\langle x \rangle_{12}##

In general, you can't say anything more about these values.
 
  • #11
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You can only assume that the expected value of ##x## in the two eigenstates is finite: ##x_1, x_2## or ##\langle x \rangle_1, \langle x \rangle_2## would be the notation I'd use.

why? If I call ##\langle x \rangle_1=-d##, then ##\langle x \rangle_2## is just some distance away, which I can call 2d.
do i have a mistake here? or is it just convention to not do this?
 
  • #12
PeroK
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why? If I call ##\langle x \rangle_1=-d##, then ##\langle x \rangle_2## is just some distance away, which I can call 2d.
do i have a mistake here? or is it just convention to not do this?

You can always change your origin, but I don't think that is necessary or particularly logical in this case.
 
  • #13
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You can always change your origin, but I don't think that is necessary or particularly logical in this case.
Ok, thank you.
I thought it to be nice, just for the sake of symmetry.
 
  • #14
kuruman
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If the expectation value can be brought into the form ##x(t) = A + B\cos(\omega t+\delta)##, the origin about which it is symmetric is implicit in the expression and can be cast in terms of the spatial integrals.
 
  • #15
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If the expectation value can be brought into the form ##x(t) = A + B\cos(\omega t+\delta)##, the origin about which it is symmetric is implicit in the expression and can be cast in terms of the spatial integrals.

So i see that ##A=\langle x_1 \rangle + \langle x_2 \rangle## and ##B=\int_{-\infty}^{\infty} 2x\varphi_1(x)\varphi_2(x)\mathrm{d}x##.
is the point about the particle oscillates A? but isn't that outside of the range between the two sites? shouldn't that point be somewhere in between the sites? I feel like i'm missing something silly.
 
  • #16
PeroK
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So i see that ##A=\langle x_1 \rangle + \langle x_2 \rangle## and ##B=\int_{-\infty}^{\infty} 2x\varphi_1(x)\varphi_2(x)\mathrm{d}x##.
is the point about the particle oscillates A? but isn't that outside of the range between the two sites? shouldn't that point be somewhere in between the sites? I feel like i'm missing something silly.

What if ##x_1 = x_2 = 0##, which would be the case in the Simple Harmonic Oscillator?
 
  • #17
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What if ##x_1 = x_2 = 0##, which would be the case in the Simple Harmonic Oscillator?
right. that is why i wanted to have them be -d and d, for symmetry purposes.
but if they are not zero, or symmetric about an origin?

if ##x_1 = a>0, x_2 =b>0 ## than the oscillations would be about a+b. how does that make sense?
 
  • #18
PeroK
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right. that is why i wanted to have them be -d and d, for symmetry purposes.
but if they are not zero, or symmetric about an origin?

if ##x_1 = a>0, x_2 =b>0 ## than the oscillations would be about a+b. how does that make sense?

You're forgetting the factor ##a## in the original superposition!
 
  • #19
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You're forgetting the factor ##a## in the original superposition!
how does that change anything? it's just a normalization constant which is applied to everything.
(the a in the superposition is not the same "a" for ##x_1=a##).

I guess it is always possible to redefine the origin to be a point halfway between a and b.
I guess that change the amplitude somehow, implicitly.
But i'm not sure if I'm still comfortable with this... I'm pretty sure there is something wrong my reasoning.

if ##\langle x \rangle = x_1 +x_2 + A\cos(\omega t+\delta)##, doesn't that mean the oscillations are about ##x_1 +x_2 ##, which definitely NOT the half way point ##\frac{x_1 +x_2}{2}##?
sk-b2ac380f2c5456d81fe4e582b5dac757.jpg

what am i missing?
 

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  • #20
PeroK
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how does that change anything? it's just a normalization constant which is applied to everything.
(the a in the superposition is not the same "a" for ##x_1=a##).

I guess it is always possible to redefine the origin to be a point halfway between a and b.
I guess that change the amplitude somehow, implicitly.
But i'm not sure if I'm still comfortable with this... I'm pretty sure there is something wrong my reasoning.

if ##\langle x \rangle = x_1 +x_2 + A\cos(\omega t+\delta)##, doesn't that mean the oscillations are about ##x_1 +x_2 ##, which definitely NOT the half way point ##\frac{x_1 +x_2}{2}##?
View attachment 221606
what am i missing?

##|a|^2## is missing from your equations. For normalised eigenfunctions, ##|a|^2 = 1/2##.
 
  • #21
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##|a|^2## is missing from your equations. For normalised eigenfunctions, ##|a|^2 = 1/2##.
But that is just a normalization factor. how does that change anything?
 
  • #22
PeroK
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The Attempt at a Solution


[itex] <x>=\int_{-\infty}^{\infty} \psi(x) x \psi^*(x)\mathrm{d}x=\int_{-\infty}^{\infty} x[|\varphi_1(x)|^2 +x|\varphi_2(x)|^2+2\varphi_1(x)\varphi_2(x)\cos(\tilde{\omega}t) ]\mathrm{d}x [/itex]

This should be:

[itex] <x>=|a|^2 \int_{-\infty}^{\infty} \psi(x) x \psi^*(x)\mathrm{d}x= |a|^2 \int_{-\infty}^{\infty} x[|\varphi_1(x)|^2 +x|\varphi_2(x)|^2+2\varphi_1(x)\varphi_2(x)\cos(\tilde{\omega}t) ]\mathrm{d}x [/itex]
 
  • #23
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This should be:

[itex] <x>=|a|^2 \int_{-\infty}^{\infty} \psi(x) x \psi^*(x)\mathrm{d}x= |a|^2 \int_{-\infty}^{\infty} x[|\varphi_1(x)|^2 +x|\varphi_2(x)|^2+2\varphi_1(x)\varphi_2(x)\cos(\tilde{\omega}t) ]\mathrm{d}x [/itex]
Ohhh....... It took me some time, but the light bulb moment finally came.
Thank you so much!
 

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