Proof of Kaku (8.18): Completing the Square and Using Spiegel's Result

[rocdoc]

In the following there is a proof, for positive values of $a$ only, of (8.18) of Kaku, reference 1, I quote'
$$\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~~~~~~(8.18)$$
'. Kaku says this result can be proved by completing the square.
$$iap^2+ibp = ia(p^2+\frac{b}{a}p )$$
$$~~~~~~~~~~~~~~~~~~~~~~~~=ia[~(p+\frac{b}{2a})^2-\frac{b^2}{4a^2}]$$
Let
$$\frac{b}{2a}=\alpha~~~~~~~~~(1)$$
$$\frac{-b^2}{4a}=\beta~~~~~~~~~~(2)$$
$$iap^2+ibp = i[~a(p+\alpha)^2~+~\beta]$$
Let
$$I=\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2~+~\beta] }\mathrm{d}p$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=e^{i\beta}~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p~~~~~~~(3)$$
Concentrating upon the integral part of (3)
$$\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p=\int_{-\infty}^\infty \cos[a(p+\alpha)^2]\mathrm{d}p +i\int_{-\infty}^\infty \sin[a(p+\alpha)^2]\mathrm{d}p~~~~~~~~~~(4)$$
Substituting $q=p+\alpha$ gives
$$\int_{-\infty}^\infty~\cos[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\cos(aq^2)\mathrm{d}q$$
$$\int_{-\infty}^\infty~\sin[a(p+\alpha)^2]\mathrm{d}p=\int_{-\infty}^\infty~\sin(aq^2)\mathrm{d}q$$
Use Spiegel, see Reference 2 result 15.50, which I quote '
$$15.50~~~~~~~~\int_0^\infty~\cos(ax^2)\mathrm{d}x = \int_0^\infty~\sin(ax^2)\mathrm{d}x= \frac{ 1} {2} \sqrt \frac{ \pi} {2a} $$
' ,for positive $a$ hence as the integrands in 15.50 are even functions of $x$
$$\int_{-\infty}^\infty~\cos(ax^2)\mathrm{d}x = \int_{-\infty}^\infty~\sin(ax^2)\mathrm{d}x= \sqrt \frac{ \pi} {2a}~~~~~~(5)$$
Using (5) in (4)
$$~~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {2a} +i \sqrt \frac{ \pi} {2a}$$
In polar form
$$~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ \pi} {a} e^{i\pi/4}$$
Also, as
$$ ~~~~~~~~~~~ e^{i\pi/4}=\sqrt i$$
$$~~~\int_{-\infty}^\infty~e^{ i[~a(p+\alpha)^2] }\mathrm{d}p= \sqrt \frac{ i\pi } {a} ~~~~(6)$$
Hence, using (6) and (2) in (3)
$$I=e^{i\beta} \sqrt \frac{ i\pi } {a}=e^{-ib^2/4a} \sqrt \frac{i \pi} {a}$$
I.E.
$$~~~~~~~~~~~~~~~\int_{-\infty}^\infty~\mathrm{d}p~e^{iap^2+ibp}=\sqrt \frac{i\pi}{a}e^{-ib^2/4a}~~~~~~~~Q.E.D.$$References
1) M.Kaku , QuantumField Theory, A Modern Introduction , Oxford University Press, Inc. , 1993.
2) M. R. Spiegel, Ph.D. , Mathematical Handbook , McGraw-Hill , Inc. , 1968.

 

[Greg Bernhardt]

Why do you feel this is interesting to share?

 

[rocdoc]

Why do you feel this is interesting to share?

Readers of the following thread might be interested

https://www.physicsforums.com/threads/decomposing-a-certain-exponential-integral.944617/

Also, those who are interested in path integrals in quantum theory and (8.18) in particular, might be interested in the material.

 

[samalkhaiat]

Readers of the following thread might be interested

https://www.physicsforums.com/threads/decomposing-a-certain-exponential-integral.944617/

Also, those who are interested in path integrals in quantum theory and (8.18) in particular, might be interested in the material.

You put a lot of efforts into something almost everybody knows how to do. Since you are talking about path integral, the following integral also appears in path integrals

\int_{0}^{\infty} \ e^{-(a/x^{2}) - bx^{2}} \ dx = ?

And since not everybody knows how to do it, maybe you could show us how to do it.

 

[Demystifier]

You put a lot of efforts into something almost everybody knows how to do. Since you are talking about path integral, the following integral also appears in path integrals

\int_{0}^{\infty} \ e^{-(a/x^{2}) - bx^{2}} \ dx = ?

And since not everybody knows how to do it, maybe you could show us how to do it.

May I ask where does this integral appear in QM? What physical situation does it describe?

 

[samalkhaiat]

May I ask where does this integral appear in QM? What physical situation does it describe?

It shows up in the simplest physical system we know of. Just take the Fourier transform of the free particle propagator

\hat{G}(E ,x) = \int_{0}^{\infty} \ dt \ e^{\frac{i}{\hbar}Et} \ G(x,t ; 0) ,

and you will see it.

 

[Demystifier]

It shows up in the simplest physical system we know of. Just take the Fourier transform of the free particle propagator

\hat{G}(E ,x) = \int_{0}^{\infty} \ dt \ e^{\frac{i}{\hbar}Et} \ G(x,t ; 0) ,

and you will see it.

I understand that it can appear in the calculation of the propagator, e.g. in the Padmabhan's QFT textbook, Eqs. (1.81)-(1.82). But in this context it does not appear in the path integral, or at least I don't see that it does. Of course, by "path integral" I mean the functional integral over all paths, not an ordinary integral over one path.

 

[Demystifier]

A related very useful page: https://en.wikipedia.org/wiki/Common_integrals_in_quantum_field_theory

 

[rocdoc]

Might a full proof of (8.18) involve using the following primitive, which is given as result 7.4.32 of Abramowitz and Stegun, see pg.303 of Reference 2? I quote
'$$\int e^{-(ax^2+2bx+c)}\:\mathrm{d}x=\frac{1} {2}\sqrt{\frac{\pi} {a} }e^{ \frac{b^2-ac}{a}} erf(\sqrt ax+\frac{b} {\sqrt a})+const. ~~~~~~~~ (a\neq0) $$'here erf(z) denotes the error function. This primitive is true for complex valued $a,b~\text {and}~ c$.

Reference

2) Handbook of Mathematical Functions , Eds M.Abramowitz and I.A. Stegun , Dover Publications, Inc., New York , Ninth Printing , Nov. 1970.

 

[samalkhaiat]

I understand that it can appear in the calculation of the propagator, e.g. in the Padmabhan's QFT textbook, Eqs. (1.81)-(1.82). But in this context it does not appear in the path integral, or at least I don't see that it does. Of course, by "path integral" I mean the functional integral over all paths, not an ordinary integral over one path.

What exactly are you claiming in here? Are you saying that the expression

G(x,t ; y) = \lim_{N \to \infty} \left( \frac{m}{2\pi i \hbar \epsilon}\right)^{\frac{N+1}{2}} \int dx_{1}\cdots dx_{N} \ \exp \left( \frac{im}{2 \hbar \epsilon}\sum_{j = 0}^{N}(x_{j+1}-x_{j})^{2}\right) ,

(where x_{N+1}= x, \ \ x_{0}= y, \ \epsilon = t / N+1) is not the free particle path integral? In these days even undergraduate students can do the x_{i}’s integrations to obtain G(x,t;y) = \sqrt{\frac{m}{2 \pi i \hbar t}} \ \exp \left( \frac{im}{2 \hbar t}(x-y)^{2}\right) .
Of course, the reason for taking the Fourier transform of G(x,t;0) is way above the level of this thread and has to do with studying the analytic properties of the Feynman-Kac formula which Feynman obtained from his path integral. So, the integral in #4 does show up in the Path Integral formalism, hence the reason for asking the OP to "help us" calculating it . In other words I wanted to give him a hint that he has been doing very trivial, if not useless, stuff in this thread.

And please don’t point out a "textbook" for me because I only post things that I fully understand if not expert in.

 

[Demystifier]

What exactly are you claiming in here? Are you saying that the expression

G(x,t ; y) = \lim_{N \to \infty} \left( \frac{m}{2\pi i \hbar \epsilon}\right)^{\frac{N+1}{2}} \int dx_{1}\cdots dx_{N} \ \exp \left( \frac{im}{2 \hbar \epsilon}\sum_{j = 0}^{N}(x_{j+1}-x_{j})^{2}\right) ,

(where x_{N+1}= x, \ \ x_{0}= y, \ \epsilon = t / N+1) is not the free particle path integral? In these days even undergraduate students can do the x_{i}’s integrations to obtain G(x,t;y) = \sqrt{\frac{m}{2 \pi i \hbar t}} \ \exp \left( \frac{im}{2 \hbar t}(x-y)^{2}\right) .
Of course, the reason for taking the Fourier transform of G(x,t;0) is way above the level of this thread and has to do with studying the analytic properties of the Feynman-Kac formula which Feynman obtained from his path integral. So, the integral in #4 does show up in the Path Integral formalism, hence the reason for asking the OP to "help us" calculating it . In other words I wanted to give him a hint that he has been doing very trivial, if not useless, stuff in this thread.

I am claiming that the integral of the form in your post #4 may appear after one evaluates the path integral, not in the path integral. Unlike you, I always allow the possibility that I am wrong, but so far you didn't show that in this particular case. (In other discussions with you, which look to me like discussions with Dr. House, sometimes you do show that I am wrong, in which case I acknowledge it.)

And please don’t point out a "textbook" for me because I only post things that I fully understand if not expert in.

I have no idea why do you put "textbook" in quotes, perhaps because you think that this particular textbook does not deserve to be called so? Anyway, if I feel that I cannot explain something better than already explained in some textbook or published paper, which indeed is often the case, I will always refer to such a reference. I don't know about you, but my mission here is not to show that I am smarter than anybody else.

By the way, since your name is so difficult to write, may I call you Dr. House?
(If you don't know who Dr. House is, which would be hard to believe, I would recommend you to learn about him because I'm sure you will like him.)

 

[samalkhaiat]

I am claiming that the integral of the form in your post #4 may appear after one evaluates the path integral, not in the path integral. Unlike you, I always allow the possibility that I am wrong, but so far you didn't show that in this particular case. (In other discussions with you, which look to me like discussions with Dr. House, sometimes you do show that I am wrong, in which case I acknowledge it.)I have no idea why do you put "textbook" in quotes, perhaps because you think that this particular textbook does not deserve to be called so? Anyway, if I feel that I cannot explain something better than already explained in some textbook or published paper, which indeed is often the case, I will always refer to such a reference. I don't know about you, but my mission here is not to show that I am smarter than anybody else.

By the way, since your name is so difficult to write, may I call you Dr. House?
(If you don't know who Dr. House is, which would be hard to believe, I would recommend you to learn about him because I'm sure you will like him.)

I really don’t have the time for this. In #5 you asked me:

May I ask where does this integral appear in QM? What physical situation does it describe?

In #6 I answered your first question but not the second because I assumed you know that we do some times need to take the Fourier transform and that Fourier transform defines an isomorphism.

Then comes this in #7

… I mean the functional integral over all paths, not an ordinary integral over one path.

This made no sense to me, for I have never asked you to integrate over “one path”. So, I had to tell you, in #10, that the propagator is obtained by doing all the x_{i}’s integrations and taking the limit N \to \infty, i.e., summing over all possible paths (histories) connecting y = x(0) and x = x(t). I also mentioned one possible the reason for taking the Fourier transform.Then, in #11, you transformed everything into a Drama about you and me. Sir, give me a break for I only speak about physics and mathematics on these forums. So, I believe doing the integral in #4 is a lot cuter than spending time making dramas.

 

[Greg Bernhardt]

The OP appears not in this discussion any longer and discussion has turned uncivil. Closing this down.