Fourier Integral involving a position operator

[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.

 

[meopemuk]

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.

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.

 

[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 https://www.amazon.com/gp/sitbv3/reader/103-8485930-5330225?ie=UTF8&p=S002&asin=0131118927&tag=pfamazon01-20 (though if you want to buy, I would suggest finding a cheaper provider )

 

[olgranpappy]

Hello.

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

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

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

 

[maverick280857]

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 https://www.amazon.com/gp/sitbv3/reader/103-8485930-5330225?ie=UTF8&p=S002&asin=0131118927&tag=pfamazon01-20 (though if you want to buy, I would suggest finding a cheaper provider )

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.

 

[maverick280857]

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

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

 

[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 $q^{-1/2}$?

 

[meopemuk]

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 $q^{-1/2}$?

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.

 

[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).

 

[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 $$t$$.

 

[meopemuk]

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).

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.

 

[Hurkyl]

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

 

[olgranpappy]

what substitution?

 

[Hurkyl]

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)

 

[olgranpappy]

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

 

[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?)

 

[CompuChip]

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

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
$$I^2 = \frac{\pi}{-i q} \left. e^{-i q r^2} \right|_0^\infty = \frac{\pi}{-i q}$$.

 

[olgranpappy]

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

nope, I'm not... you are...

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 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 $$-\alpha=iq$$.

 

[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?

 

[CompuChip]

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)

Did you try that already? Did it work out?

 

[olgranpappy]

Yeah, it works formally.

But, as I was saying before, I think that it makes more sense to choose, say, the branch cut of the integrand as running up the imaginary axis and then deform the integration contour to run up one side and down the other of the cut.

In that case you actually do have an honest-to-god real Gaussian integral that really converges instead of oscillating.

 

[maverick280857]

olgranpappy, could you please elaborate on what you said? How do you reconcile with the definition of Fourier transforms of functions that are singular. I can see how you could do it for delta functions and therefore possibly for generalized functions, but how do you do it for something like

$$\frac{1}{\sqrt{t}}$$

When t goes through negative values, this becomes complex. But at t = 0, the integrand is singular.

 

[olgranpappy]

olgranpappy, could you please elaborate on what you said? How do you reconcile with the definition of Fourier transforms of functions that are singular. I can see how you could do it for delta functions and therefore possibly for generalized functions, but how do you do it for something like

$$\frac{1}{\sqrt{t}}$$

When t goes through negative values, this becomes complex. But at t = 0, the integrand is singular.

Yes it is singular, but it is not so singular that it is not integrable. It diverges more slowly than 1/t and so you can, actually, integrate it. For example,
$$\int_0^{a}\frac{1}{\sqrt{t}}=2\sqrt{a}$$
is perfectly finite.

 

[olgranpappy]

the real problem is that the singularity at zero is a branch point. for example, if I choose my branch cut to run from zero to $$i\infty$$ along the imaginary axis, and I choose to define this Fourier transform by integrating along a path that runs along the real axis almost everywhere, but goes *underneath* (a little semcircle) the point at zero, then I can deform my contour as I said in my previous post and perform the integration.

 

[jostpuur]

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
$$I^2 = \frac{\pi}{-i q} \left. e^{-i q r^2} \right|_0^\infty = \frac{\pi}{-i q}$$.

This derivation is flawed (not only the minus sign). The expression
$$e^{-iqr^2}$$
does not approach zero in the limit $r\to\infty$. I tried to start talking about the derivation of this integral formula here https://www.physicsforums.com/showpost.php?p=1381952&postcount=53

 

[Hurkyl]

$1 / \sqrt{|t|}$ doesn't have a branch point. That's the function appearing in Wikipedia's table; the presumption, I think, is that that's what you meant to write.

 

[olgranpappy]

but we are talking about

$$1/\sqrt{t}$$

not

$$1/\sqrt{|t|}$$

 

[olgranpappy]

This derivation is flawed (not only the minus sign). The expression
$$e^{-iqr^2}$$
does not approach zero in the limit $r\to\infty$. I tried to start talking about the derivation of this integral formula here https://www.physicsforums.com/showpost.php?p=1381952&postcount=53

Well, I don't have anything too clever to add, but I guess that I would start off by considering the function f(z)
$$f(z)=\int_{-\infty}^{\infty}d\tau e^{-z{\tau}^2}$$
which obviously exists for all values of Im(z) and for Re(z)>0 (I.e., in the
right half-plane) and is simply equal to (in the RHP)
$$f(z)=\sqrt{\frac{\pi}{z}}$$
But, the above is an expression which is valid in the entire z-plane and just has a branch point at zero and a branch cut wherever we like--it's the analytic continuation of our function which was defined only in the RHP to the entire plane.