Why Does the Expectation Value Depend on the Space Used?

[cjellison]

I had thought that the expectation value would be the same...whether you did it in momentum space or position space. Could someone explain what is going on in this particular problem?

$$\psi (x) = \sqrt{b} e^{-b |x| + i p_0 x / \hbar }$$

Taking the Fourier transform, I can get this function in momentum space.

$$\phi (p) = \sqrt{\frac{2b}{\pi \hbar}} \thickspace \frac{\hbar^2 b}{\hbar^2b^2+(p-p_0)^2}$$

Now, if I compute $<p^2>$, I should get the same value no matter which space I choose to do it in. However, this does not seem to be the case. I computed these by hand...and by Mathematica...so I am fairly confident that they are correct (just confusing).

[tex] \begin{align*}<br /> <p^2> &= \int \phi^* p^2 \phi \, dp = b^2 \hbar^2 + p_0^2\\<br /> <p^2> &= \int \psi^* \left(\frac{\hbar}{i}\right)^2 \frac{\partial^2 \psi}{\partial x^2} \, dx = -b^2 \hbar^2 + p_0^2<br /> \end{align*}<br /> [/itex]<br /> <br /> As you can see...and please carry out the computations if you doubt them...the expectation value seems to depend on the space in which I computed it. I thought that this was NOT supposed to happen. Any ideas?[/tex]

 

[Galileo]

$$<p^2> &= \int \phi^* p^2 \phi \, dp = b^2 \hbar^2 + p_0^2$$

I get:

$$\langle p^2 \rangle = b\sqrt b(\hbar^2 b+p_0^2)$$
I'm having trouble getting Maple to evaluate the other one...

 

[dextercioby]

Are u both sure with those integrals??I mean,the initial wave function is different for the intervals minus infinity to zero and o to plus infinity.This should be taken into consideration for all three integrals (the Fourier transform and the 2 EV-s).
I would advise you to do those integrals by hand,or maybe with the help of an integrals book.
One other thing:

$$\langle p^2 \rangle = b\sqrt b(\hbar^2 b+p_0^2)$$

That's definitely wrong.Check out the units.The constant "b" has the dimension length to the minus one.

 

[lalbatros]

Hi

The wave function you consider has no derivative for x=0.
Its second-derivative should be a Dirac function for x=0.

I think your strange results are related to this.
Probably you dropped the terms related to the singular behaviour at x=0.

Also, be careful with Mathematica because such problems can easily be missed.
By the way, I would be interrested to know how you did the integral in Mathematica.
How did you let Mathematica know that the integral converges for x->Infinity?

Michel

 

[cjellison]

Here we go with the details:

$$\psi (x) = \sqrt{b} e^{-b |x| + i p_0 x / \hbar }$$

Then,

$$\begin{align*}<br /> \phi (p) &= \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} \psi(x) e^{-i p x /<br /> \hbar} \, dx \\<br /> &= \frac{1}{\sqrt{2\pi \hbar}} \left( \int_{-\infty}^{0} \sqrt{b} e^{bx + ip_0 x/\hbar}<br /> e^{-i p x / \hbar} \, dx + \int_{0}^{\infty} \sqrt{b} e^{-bx + ip_0 x/\hbar}<br /> e^{-i p x / \hbar} \, dx \right) \\ &= \sqrt{\frac{2 b}{\pi \hbar}} \thickspace<br /> \frac{\hbar^2 b}{\hbar^2b^2 + (p - p_0)^2}<br /> \end{align*}$$

Now,

$$\begin{align*}<br /> <p^2> &= \int_{-\infty}^{\infty} \phi^* p^2 \phi \, dp\\&\\<br /> &= \frac{2b^3\hbar^3}{\pi} \int_{-\infty}^{\infty}<br /> \frac{p^2}{[b^2\hbar^2+(p-p_0)^2]^2} \, dp\\&\\<br /> &= \frac{2b^3\hbar^3}{\pi} \, \frac{\pi (b^2\hbar^2+p_0^2)}{2b^3\hbar^3}\\&\\<br /> &= b^2 \hbar^2 + p_0^2<br /> \end{align*}$$

If you want to do this in Mathematica, you don't have to do any special tricks.
Just set up the problem and integrate. However, you must use the Assuming[]
command. For example:

Assuming[ b > 0 && pz > 0 && h > 0, Integrate[p^2 /(b^2 h^2 + (p -pz)^2)^2, {p, -Infinity, Infinity}]]

In the code above, I am letting $h$ be $\hbar$ and
$pz$ be $p_0$.

The wave function you consider has no derivative for x=0.
Its second-derivative should be a Dirac function for x=0.

I think your strange results are related to this.

I suspect you are correct, but I would like some help figuring out that exact
location of the mistake. I would like to think that the x=0 point wouldn't
matter since it is set of measure zero...but perhaps not.

Now, we do it in position space. Sorry, I don't have the in-between
steps...basically you have to break the integral up into two separate integrals.
I will provide Mathematica code to get the answer.

[tex] \begin{align*}<br /> <p^2> &= \int_{-\infty}^{\infty} \psi^* \left(\frac{h}{i}\right)^2<br /> \frac{\partial^2 \psi}{\partial x^2} \, dx\\<br /> &= \int_{-\infty}^{0} \sqrt{b} e^{bx - ip_0 x/\hbar}\left(\frac{h}{i}\right)^2<br /> \frac{\partial^2}{\partial x^2}(\sqrt{b} e^{bx + ip_0 x/\hbar}) \, dx+<br /> \int_{0}^{\infty} \sqrt{b} e^{-bx - ip_0 x/\hbar}\left(\frac{h}{i}\right)^2<br /> \frac{\partial^2}{\partial x^2}(\sqrt{b} e^{-bx + ip_0 x/\hbar}) \, dx\\ <br /> &=-\frac{1}{2}(b \hbar + i p_0)^2 + -\frac{1}{2}(b\hbar-i p_0)^2\\<br /> &=-b^2\hbar^2 + p_0^2<br /> \end{align*}<br /> [/itex]<br /> <br /> Sample mathematica code...I could have defined functions, but I took the lazy way out. This code should work in Mathematica 5.<br /> <br /> ---begin code----<br /> <br /> psineg = Sqrt<b> Exp[b x + I pz x /h];<br /> psinegcon = Sqrt<b> Exp[b x - I pz x /h];<br /> psipos = Sqrt<b> Exp[-b x + I pz x /h];<br /> psiposcon = Sqrt<b> Exp[-b x - I pz x /h];<br /> <br /> Assuming[h > 0 && pz > 0 && b > 0, Integrate[psinegcon (h/I)^2 D[psineg, {x, 2}], {x, -Infinity, 0}]]<br /> <br /> Assuming[h > 0 && pz > 0 && b > 0, Integrate[psiposcon (h/I)^2 D[psipos, {x, 2}], {x, 0, Infinity}]]<br /> <br /> ---end code---<br /> <br /> Then just add the two results...or have Mathematica do it.<br /> <br /> Anyway, what we see is that the expectation value is different depending on the space in which I computed it. I also believe that this could be related to the fact that $\psi$ is not differentiable at $x=0$. However, I also feel that the math is correct...so where did I go wrong? As far as $\psi$ not being differentiable at zero, it seems like this might not matter since I break the integral into two separate integrals...when I take the derivative over these secondary ranges, $\psi$ is differentiable---and the point x=0 is no longer an issue.</b></b></b></b>[/tex]

 

[lalbatros]

Hello

I think you can go through the same calculations, but you should not forget the contribution at x=0.
You simply need to take the following rules into account (I think they are ok, but check it):

$$\frac{d|x|}{dx} = 2H(x)-1$$​

$$\frac{dH(x)}{dx} = \delta(x)$$​

Where H(x) is the Heaviside function, (H(x)=0 for x<0 and 1 for x>0), and $$\delta(x)$$ is the Dirac distribution.

Thanks for your indications on Mathematica, I will try it at home this evening.
My version of Mathematica is more than 10 years old.
It does not know the "Assuming" function.
There is maybe a replacement, but I am not sure.

I hope you will get the correct results.
However, I am wondering now what is the physical meaning of all that?
Clearly the singularity at x=0 is associated with high frequency components in the momentum spectrum.

Your question triggered a vacuum in my recollections about quantum mechanics.
I can imagine that operators in different representations (x or p) are equal if their averages are always equal too, of course.
But why is it that p² in the different representations are obtained just by squaring in these two representations.
I think this is true for any function of operator.

 

[masudr]

It does matter if $$\psi$$ is not differentiable at 0. The only way it wouldn't matter is if as the integrand was evaluated tending to 0 from the right and tending to 0 from the left, the limit was the same. If this isn't the case, then you must find another way to evaluate the improper integral.

masud.

 

[jujio77]

It doesn't matter what you are calculating the expectation value of, you should always get the same result in any basis. Your choice of space is just a choice of basis which you integrate out.

 

[seratend]

I had thought that the expectation value would be the same...whether you did it in momentum space or position space. Could someone explain what is going on in this particular problem?

$$\psi (x) = \sqrt{b} e^{-b |x| + i p_0 x / \hbar }$$

Taking the Fourier transform, I can get this function in momentum space.

$$\phi (p) = \sqrt{\frac{2b}{\pi \hbar}} \thickspace \frac{\hbar^2 b}{\hbar^2b^2+(p-p_0)^2}$$

Now, if I compute $<p^2>$, I should get the same value no matter which space I choose to do it in. However, this does not seem to be the case. I computed these by hand...and by Mathematica...so I am fairly confident that they are correct (just confusing).

[tex] \begin{align*}<br /> <p^2> &= \int \phi^* p^2 \phi \, dp = b^2 \hbar^2 + p_0^2\\<br /> <p^2> &= \int \psi^* \left(\frac{\hbar}{i}\right)^2 \frac{\partial^2 \psi}{\partial x^2} \, dx = -b^2 \hbar^2 + p_0^2<br /> \end{align*}<br /> [/itex]<br /> <br /> As you can see...and please carry out the computations if you doubt them...the expectation value seems to depend on the space in which I computed it. I thought that this was NOT supposed to happen. Any ideas?[/tex]

$$<br /> A very good example to understand better the use of the unbounded operators and the basis change (or the Fourier transforms if you prefer).<br /> <br /> You have not the same results because lim_p p^2.phi(p) is not zero thus p^2.phi(p) is not square integrable. p^2|phi> does not belong to the hilbert space.<br /> Thus you cannot apply simply the basis transformation to p^2.phi(p) (you need to use distribution theory,i.e. enlarge the hilbert space, to recover the result).<br /> <br /> Seratend.$$