Shankar's treatment of Dirac's delta-function

[tymarats]

No matter how hard I tried, I could not find anything but the praise for Shankar's Principles of Quantum Mechanics, and not a single soul ever had any problems with this book... Hmm... Am I the only one who thinks that Shankar's explanations are at times a little... well, let's say, counter-intuitive? Don't get me wrong, I still think this is a great book (even though I'm still a QM newbie), but for me, it's has a number of points which are presented in somewhat unusual manner.

Maybe the most prominent (and the first, I think) example of this is the treatment of Dirac's delta function in the Mathematical introduction, pages 60-63 (Second Edition). Does anyone else find it extremely counter-intuitive to define delta as delta(x-x') integrated over x' (!) to give unit value? Of course, delta is even, so it really doesn't matter if it's defined this way, or delta(x'-x), but still... Every textbook I've _ever_ seen defines this as delta(x-T), where T is simple delay, and integration is always done over t, which is completely intuitive (at least to me). BTW, while we're at these pages, I was also a little uncomfortable with his asking for a "proof" in 1.10.2 (altogether with the Taylor's series expansion hint) - what _I_ have learned in school is pretty well summarized in http://en.wikipedia.org/wiki/Dirac's_delta_function#Composition_with_a_function

My question is: did anyone else have these subtle discomforts while reading this book, or is it just my inability to adapt to a different (Yale?) teaching style?

 

[G01]

Maybe I'm missing something, but you're complaint seems to be more about style than anything else.

x' ( or T) is a dummy variable. It doesn't matter what you call it.

Also, $\delta (x-x')$ is the standard why of writing the Dirac delta. So, Shankar is following convention.

 

[Chopin]

Your comment about T being a delay makes me think your background may be signal processing? FIR filters are just one of the ways in which delta functions are used--in QM they pop up in a lot of contexts, not all of them involving time. You'll often see them in commutation relationships, spatial eigenvalue equations, etc. So it's more general to just call the variables x and x', and let their meaning be determined by context.

 

[Ben Niehoff]

Maybe Shankar's (standard) convention will make more sense if you consider the fact

$$f(x) = \int_{-\infty}^{\infty} f(x') \delta(x - x') \; dx'$$

In physics, we typically use primes to denote dummy variables like this. Otherwise we might run out of letters...

 

[tymarats]

So, it is just me, eh? :)

Well, it's not about the style, and it's not about the SP background -- I believe I mislead you by using variable name $T$ -- should have been $x_0$. Also, I was using the term "delay" in a generalized sense, not having to do anything with time, but rather with shifting the function over x-axis.

Let me get back to the point. If (by Shankar) $x'$ is the domain of integration i.e. axis, and $x$ is any given point on that axis (I would also appreciate if he started his derivation so that these variable names come the other way round, but that's irrelevant now), it seems to me that the standard convention is (note the order of variables under delta):

$$f(x) = \int_{-\infty}^{\infty} f(x') \delta(x' - x) \; dx'$$

Why? Any given function $g(x')$, shifted by $x_0$ (to the right, or to the left, depending on sign of $x_0$) gives $\tilde{g}(x') \equiv g(x'-x_0)$. Now, there exists a function:

$$\delta(x')$$

which is called Dirac delta function, and it can be used to sample function $f(x')$ at point $x'=0$. If we want a function that will sample $f(x')$ in some given point $x$ on $x'$, we just shift Dirac delta function by $x$, to obtain:

$$\delta(x'-x)$$

Isn't this way more intuitive than $\delta(x-x')$ (given that $x'$ is the axis, and x is the point of delta's burst on that axis, as Shankar puts it)? I mean, who defines functions as $g(x_0-x)$ anyway? :)

I'm really interested in your replies on this matter - even if it seems like knit-picking, nobody ever mentioned such points in the book, so I have a feeling that my intuition is somewhat too narrow :)

EDIT: @G01: You are right, $\delta(x-x')$ really is the standard way of denoting delta, but it's usually the product $f(x)\delta(x-x')$ that is integrated over $dx$ to obtain $f(x')$.

 

[G01]

So, it is just me, eh? :)

Well, it's not about the style, and it's not about the SP background -- I believe I mislead you by using variable name $T$ -- should have been $x_0$. Also, I was using the term "delay" in a generalized sense, not having to do anything with time, but rather with shifting the function over x-axis.

Let me get back to the point. If (by Shankar) $x'$ is the domain of integration i.e. axis, and $x$ is any given point on that axis (I would also appreciate if he started his derivation so that these variable names come the other way round, but that's irrelevant now), it seems to me that the standard convention is (note the order of variables under delta):

$$f(x) = \int_{-\infty}^{\infty} f(x') \delta(x' - x) \; dx'$$

Why? Any given function $g(x')$, shifted by $x_0$ (to the right, or to the left, depending on sign of $x_0$) gives $\tilde{g}(x') \equiv g(x'-x_0)$. Now, there exists a function:

$$\delta(x')$$

which is called Dirac delta function, and it can be used to sample function $f(x')$ at point $x'=0$. If we want a function that will sample $f(x')$ in some given point $x$ on $x'$, we just shift Dirac delta function by $x$, to obtain:

$$\delta(x'-x)$$

Isn't this way more intuitive than $\delta(x-x')$ (given that $x'$ is the axis, and x is the point of delta's burst on that axis, as Shankar puts it)? I mean, who defines functions as $g(x_0-x)$ anyway? :)

I'm really interested in your replies on this matter - even if it seems like knit-picking, nobody ever mentioned such points in the book, so I have a feeling that my intuition is somewhat too narrow :)

EDIT: @G01: You are right, $\delta(x-x')$ really is the standard way of denoting delta, but it's usually the product $f(x)\delta(x-x')$ that is integrated over $dx$ to obtain $f(x')$.

OK. I'm confused. So you are suggesting that instead of writing:

$$f(x)=\int f(x')\delta(x-x')$$

We write:

$$f(x')=\int f(x)\delta(x'-x)$$

Is this what your suggesting?

If so, I don't think one is more intuitive than the other. I don't see the difference. It's just a naming convention.

 

[tymarats]

No :)

If you read my post carefully, I'm actually suggesting:

$$f(x) = \int_{-\infty}^{\infty} f(x') \delta(x' - x) \; dx'$$

which is qualitatively different than what is originally used. -- it is not the matter of naming. The two functions $\delta(x'-x)$ and $\delta(x-x')$ are equivalent, but only because of evenness of [tex]\delta[/itex]. The difference is, with integration over $dx'$ (as in Shankar's book), the first notion intuitively shows the shift $x$ in $\delta(x')$, while the second one does not.[/tex]

 

[G01]

No :)

If you read my post carefully, I'm actually suggesting:

$$f(x) = \int_{-\infty}^{\infty} f(x') \delta(x' - x) \; dx'$$

which is qualitatively different than what is originally used. -- it is not the matter of naming. The two functions $\delta(x'-x)$ and $\delta(x-x')$ are equivalent, but only because of evenness of [tex]\delta[/itex]. The difference is, with integration over $dx'$ (as in Shankar's book), the first notion intuitively shows the shift $x$ in $\delta(x')$, while the second one does not.[/tex]

$$<br /> Oh. OK. I see what you are saying. I do not know why Skankar chooses to write it one way vs. the other.<br /> <br /> However, I still think this is not really much of an issue. Anyone studying the Dirac delta has studied enough math to know what a shift in a function looks like, even it it isn't written in the most blatant way.$$

 

[C. H. Fleming]

Shankar's form is the canonical form for a convolution integral.

 

[tymarats]

Hm, it makes sense - seeing it that way, $\delta$ is defined exactly as it should be, though most of the time he is actually dealing with (and drawing) its shifted reflections. Seems fine, though I would still appreciate if he mentioned that fact. Thank you!