# Fourier Integral involving a position operator

1. Jul 22, 2007

### maverick280857

Hello.

I'm teaching myself quantum mechanics. I want to understand the meaning of the following integral representation:

$$q^{-\frac{1}{2}} = \kappa \int_{-\infty}^{\infty}\frac{dt}{\sqrt{t}}exp(itq)$$

where $q$ is the quantum mechanical position operator. I know that this is a Fourier Integral, but I want to understand it in a deeper way. I don't know much about operators in such integrals, so I also want to know how to evaluate the right hand side...I'm using the book by Atkinson and Hounkonnou and this is from a problem which asks to prove that

$$\left[q^{-\frac{1}{2}},p\right] = -\frac{1}{2}i\hbar q^{-\frac{3}{2}}$$

What is the significance of the integral and of the negative powers?

Thanks.

Last edited: Jul 22, 2007
2. Jul 22, 2007

### meopemuk

Operators $q$ and $p$ do not commute

$$\left[q,p\right] = i\hbar$$

Using this commutator you can easily prove

$$\left[q^2,p\right] = 2i\hbar q$$

$$\left[q^n,p\right] = ni\hbar q^{n-1}$$

and, in general, for any function $f(q)$ that can be expressed as a Taylor series around $q=0$

$$\left[f(q),p\right] = i\hbar \frac{d}{dq} f(q)$$ (1)

The problem is that function $f(q) = q^{-1/2}$ is singular at $q=0$, so this method formally doesn't work. My guess is that authors are trying to prove that eq. (1) still can be applied to $f(q) = q^{-1/2}$ by expressing this function as a Fourier integral. Then eq. (1) can be applied to the function $exp(itq)$ (which has a regular behavior at $q=0$) in the integrand, and performing the inverse Fourier transform you should get your commutator equation.

Usual rules of algebra and calculus can be applied to operators and their functions as well. The only trouble arises when you have functions of two or more operators, which do not commute with each other. Then you need to worry about the order of factors. However, this difficulty is not present in your problem.

Eugene.

3. Jul 22, 2007

### CompuChip

If you want to self-learn quantum mechanics, I can recommend the book by David J. Griffiths. It's very accessible. The first part really focuses on solving the Schroedinger equation in different cases, to give you a feel of how QM works, while the second part formalizes everything from the first parts (introducing brakets and operators). The rest of the book covers the most important subjects: identical particles, perturbation theory, scattering, etc.

Index (and sample) can be found here (though if you want to buy, I would suggest finding a cheaper provider )

4. Jul 22, 2007

### olgranpappy

what is $$\kappa$$? is it $$\frac{1}{2\sqrt{\pi i}}$$?

5. Jul 22, 2007

### maverick280857

Thanks, I have Griffiths' QM book too and I read it as well. I went up to the third chapter, but I haven't completed it yet. I recently came across this book and decided to check it out.

6. Jul 22, 2007

### maverick280857

$\kappa$ is a constant that they evaluate by setting $s=t{\textbf q}$. I'll check the value in the book and reply back.

Last edited: Jul 22, 2007
7. Jul 22, 2007

### maverick280857

Thanks Eugene. I understand the first few parts (that p and q do not commute) and also the interpretation of positive integral powers of p and q (as repetitive applications of the operators). As for negative integral powers, I can think of them as repeated applications of the inverse operators. But what about the physical interpretation of fractional powers of operators?

For now, I think that if I apply $q^{-\frac{1}{2}}$ twice and write

$$q^{-1} = q^{-\frac{1}{2}}q^{-\frac{1}{2}}$$

then $q^{-\frac{1}{2}}$ is a kind of half inverse operator. I could extend this to other negative and positive fractional powers. Is this correct?

EDIT: How do I show that the right hand side indeed equals $q^{-1/2}$?

Last edited: Jul 22, 2007
8. Jul 22, 2007

### meopemuk

You are absolutely right. But there is an easier way to think about functions of operators. You can imagine operator as a matrix. This is an infinite matrix, of course. Hermitian operators (like q) have Hermitian matrices. They have the property of diagonalizability. This means that there is a basis in the Hilbert space in which the matrix is diagonal. In this representation, it is easy to define *any* function of your operator. Just apply this function to all diagional matrix elements and you are done. So, you defined arbitrary functions of the operator in one specific basis. By doing usual basis transformations, you can extend this definition to all bases.

The above method may be not simple to implement in practice, but it does show you that you don't need to worry about the existence of functions of operators. You can apply to these functions all usual rules of algebra and calculus.

Eugene.

9. Jul 22, 2007

### maverick280857

Thanks meopemuk, now how do I convince myself that the right hand side (the integral) indeed equals $q^{-1/2}$.

EDIT: I want to understand how does one define the fourier transform of a singular function. I'm guessing I have to read more about distributions. Can you please walk me through the evaluation of this integral and also suggest some text(s) where I could learn more about all this (to get the mathematical background).

Last edited: Jul 22, 2007
10. Jul 22, 2007

### olgranpappy

the Fourier transform of a function like $$\frac{1}{\sqrt{t}}$$ doesn't exactly make sense (to me, at least) if you just write down
$$\int_{-\infty}^{\infty}\frac{dt}{\sqrt{t}}e^{i\omega t}$$
I think you need to specify what exactly is going on with the branch cut when $$\frac{1}{\sqrt{t}}$$ is considered as a function of complex $$t$$.

11. Jul 22, 2007

### meopemuk

The Fourier transform that you wrote seems to be correct (see table entry #210 in http://en.wikipedia.org/wiki/Fourier_transform#Square-integrable_functions ). However, I can't tell you off the top of my head how to prove it. You'll probably need to do some contour integration tricks. Check references on the Wikipedia site. They will teach you how to do such integrals.

Eugene.

12. Jul 22, 2007

### Hurkyl

Staff Emeritus
Well, this particular integral appears to be straightforward via substitution.

13. Jul 22, 2007

### olgranpappy

what substitution?

14. Jul 22, 2007

### Hurkyl

Staff Emeritus
Well, I started with |t|=s^2. The form of the integrand suggests this as a possible thing to try; one reason is because |t| appears under the square root. Another is because that's the substitution that makes the integrand entirely exponential.

(Of course, I had to break the integral into two pieces)

15. Jul 22, 2007

### olgranpappy

...oh, good call... and then one just has to use the fact that
$$\int_{-\infty}^{\infty}dse^{iqs^2}=\sqrt{\frac{\pi}{-iq}}$$

16. Jul 23, 2007

### maverick280857

Okay, so we have to treat t as a complex number. I can see how to do this using the substitution $|t| = s^2$. We could write

$$t = s^{2}e^{i\phi}$$

but what about the integration limits? Anyway, proceeding as such

$$\int_{-\infty}^{\infty}\frac{1}{s}e^{i\frac{\phi}{2}}e^{itq}2sds$$

which is a (complex) Gaussian integral, except for a phase factor. But I don't understand how to fix the integration limits here. How do you convert this to a contour integral? (I know how to do contour integrals). And I'm still not convinced how I can take a Fourier Transform of a function that isn't absolutely integrable. I guess its possible to construct the transform by treating it as a distribution...but I don't know much about distributions either.

(Or do you mean that it should be |t| under the square root sign rather than t?)

Last edited: Jul 23, 2007
17. Jul 23, 2007

### CompuChip

I think you're missing a minus sign there in the exponent.

I think you can prove it completely analogous to the real case. Let
$$I = \int_{-\infty}^\infty e^{-iqs^2} ds = \int_{-\infty}^\infty e^{-iqt^2} dt.$$
Then
$$I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-iqs^2} e^{iqt^2} ds dt = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-iq(s^2+t^2)} ds dt$$
and now substitute $r^2 = s^2 + t^2$ and go to polar coordinates:
$$I^2 = \int_0^{2\pi} d\phi \cdot \int_0^\infty r e^{-i q r^2} dr$$
and because the integrand is the derivative of $e^{-iqr^2}$ - apart from a constant factor - you get
$$I^2 = \frac{\pi}{-i q} \left. e^{-i q r^2} \right|_0^\infty = \frac{\pi}{-i q}$$.

Last edited: Jul 23, 2007
18. Jul 23, 2007

### olgranpappy

nope, I'm not... you are...
... you have
$$\pi \int_{0}^{\infty}d(r^2)e^{-iqr^2} =\pi (\frac{e^{-iqr^2}}{-iq})|_{0}^{\infty}$$
the evaluation at the *lower* limit equal to zero gives back the correct minus sign.

I.e., in general
$$\int e^{-\alpha x^2}=\sqrt{\frac{\pi}{\alpha}}$$
just substitute in $$-\alpha=iq$$.

Last edited: Jul 23, 2007
19. Jul 23, 2007

### maverick280857

Thanks CompuChip and olgranpappy, I know how to do the Gaussian Integral

$$\int_{-\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}$$

but what I want to know is how to evaluate the particular Fourier integral at hand. First of all, how does one deal with singular functions? How can we define Fourier transforms of such functions? Next, how does one prove the existence of such integrals and finally, how does one solve them?

Last edited: Jul 23, 2007
20. Jul 23, 2007

### CompuChip

Did you try that already? Did it work out?