Meaning of Dirac Delta function in Quantum Mechanics

[Ibraheem]

If I have a general (not a plain wave) state $$|\psi\rangle$$, then in position space :

$$\langle \psi|\psi\rangle = \int^{\infty}_{-\infty}\psi^*(x)\psi(x)dx$$
is the total probability (total absolute, assuming the wave function is normalized)
So if the above is correct, does that mean
$$\langle x|x' \rangle= \delta(x-x')$$
is also the total probability ?
if that is the case, then the total probability is infinite! is that why plane waves are called unphysical? or is there more to it ?

 

[dextercioby]

First there is something you could learn: user \langle and \rangle instead of < and >, respectively.
Then I could say you have the right understanding at a beginner's level.

 

[Ibraheem]

Sorry what I meant was this
$$\langle x|x \rangle= \delta(x-x)=\delta(0)$$
is the total probability
But
$$\langle x|\psi\rangle$$ is a wave function $$\psi(x)$$
so is it correct to say

$$\langle x|x' \rangle= \delta(x-x')$$ is a wave function too ??

 

[vanhees71]

Of course not! Wave functions can provide only probabilities if their modulus squared is integrable, i.e., you can choose the normalization such that
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=1.$$
A "position eigenstate" is not a proper function but a distribution (in the sense of "generalized function"), i.e., you cannot even square it, and the expression $\delta(0)$ makes no sense at all. It's undefined.

It's purpose is to map a function in a certain dense subspace, which defines the part of the Hilbert space of square integrable function, where the position (and also the momentum) operators are defined, and for these functions you have
$$\int_{\mathbb{R}} \mathrm{d} x' \delta(x-x') \psi(x')=\psi(x).$$
It's also true that the generalized position eigenstates (where again I used the word "generalized" in the sense of distributions) obey
$$\langle x|x' \rangle=\delta(x-x').$$
The same holds for momentum eigenstates
$$\langle p|p' \rangle=\delta(p-p').$$
You also may know that the generalized momentum eigenstates in the position representation read
$$u_p(x)=\langle x|p \rangle=\frac{1}{(2 \pi \hbar)^{1/2}} \exp \left (\frac{\mathrm{i} x p}{\hbar} \right).$$
Again this is not a square-integrable function, i.e., you cannot normalize it since $|u_p(x)|^2=1/(2 \pi \hbar)$, which is not integrable over the entire real line, but still you can use the completeness relation in the form
$$\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x| = \int_{\mathbb{R}} \mathrm{d} p |p \rangle \langle p |=\hat{1}.$$
This implies that you can write any proper wave function in the domain, where posision and momentum operators are defined either in position and momentum representation, via
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{(2 \pi \hbar)^{1/2}} \exp \left (\frac{\mathrm{i} p x}{\hbar} \right) \tilde{\psi}(p),$$
which is just a Fourier transform of the momentum-space wave function. Also the inverse follows immediately from this formalism
$$\tilde{\psi}(p) = \langle p|\psi=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{(2 \pi \hbar)^{1/2}} \exp \left (-\frac{\mathrm{i} p x}{\hbar} \right) \psi(x),$$
which is also well known from the theory of Fourier transformations.

 

[mikeyork]

If I have a general (not a plain wave) state $$|\psi\rangle$$, then in position space :

$$\langle \psi|\psi\rangle = \int^{\infty}_{-\infty}\psi^*(x)\psi(x)dx$$
is the total probability (total absolute, assuming the wave function is normalized)
So if the above is correct, does that mean
$$\langle x|x' \rangle= \delta(x-x')$$
is also the total probability ?
if that is the case, then the total probability is infinite! is that why plane waves are called unphysical? or is there more to it ?

The Dirac delta-function is a density function. You must integrate over a range to get the probability of that range occurring.

 

[bhobba]

The Dirac delta-function is a density function. You must integrate over a range to get the probability of that range occurring.

Hmmm - its not even a function so I am not quite sure what you mean.

To the OP - this is an I level thread and the following should be understandable at that level:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

IMHO its something all applied mathematicians and physicists should know.

Thanks
Bill

 

[mikeyork]

Hmmm - its not even a function so I am not quite sure what you mean.

I guess that's why it's called the Dirac delta-FUNCTION. As I recall Dirac defined his FUNCTION by the requirements $\delta(x) = 0$ for $x \ne 0$ and $\int_{-\infty}^\infty \delta(x)dx = 1$. (The Principles Of Quantum Mchanics, section 15, p58.) If Dirac thinks it's a function that's good enough for me.

 

[PeterDonis]

If Dirac thinks it's a function that's good enough for me.

Dirac used the word "function" in a way that current math and physics convention does not, strictly speaking, support. That's why @bhobba said the delta "function" is not a function. According to current convention, a function must have a well-defined value at every point in its domain; $\delta(x)$ does not have a well-defined value at $x = 0$. ("Infinity" or whatever you want to call it is not a well-defined value.)

Since you don't actually need to know the value of $\delta(x)$ at $x = 0$ in order to do physics (at least not in the context of this thread), the fact that this value is not well-defined does not matter for (most) physics purposes. That, I think, is what you referred to when you said $\delta(x)$ is a "density function". The usual term for this by current convention is "distribution", as in the title of the book @bhobba linked to. So when speaking informally, many physicists will in fact use the word "function" to refer to $\delta(x)$. But @bhobba was perfectly correct to point out that this is not, strictly speaking, justified; $\delta(x)$ does not have all of the properties that a function is assumed, mathematically, to have, and this can become important in certain contexts (which is why there is a whole theory of distributions to make all this rigorous and formalize what these properties are and what impact they have).

 

[mikeyork]

Dirac acknowledges that when he calls it an improper function. And the reason for its existence, without having to talk about its value at x=0, is in the integral part of its two-part definition since that enables its use in QM as what in probability theory is called a probability density function or p.d.f. (the derivative of a probability distribution). Strictly speaking of course, it is as Dirac explains, the infinitessimally narrow limit of a probability density function.

 

[weirdoguy]

Dirac acknowledges that when he calls it an improper function.

Doesn't matter, it's still not a function and anyone who learn QM should have that in the back of their head. It's 2017, things moved on since Dirac.
Actually during my studies (Warsaw University, Poland) we had a course on functional analysis and distribution theory before we had Quantum Mechanics 1 lectures. It made things much easier.

 

[Ibraheem]

I was not being precise enough in the second comment.
$\langle x|x \rangle= \delta(x-x)=\delta(0)$
I should not have said total probability. What I meant was relative probability since it can never be normalized to 1. So it is physically meaningless, I guess. I am asking this just to clarify the use of the Bra-Ket Notation and the idea of wave function in my head, since Shankar refers to $\langle x|x' \rangle= \delta(x-x')$ as a wave function of a particle at x' see page 154 Principals of Quantum Mechanics.

 

[HAYAO]

It is never out of place to make more precise the meaning of the language we use. The OP is not the only one who might possibly read this thread or benefit from it.

I'm just stopping by to say this. Ignore it if you don't like it.

As much as I can agree that knowing the precise definition of something is important and should be understood, that does not mean we cannot make any discussion about it. Like Heaviside said, "Am I to refuse to eat because I do not fully understand the mechanism of digestion?"

Dirac's delta "function" were considered nonsense in the field of mathematics until L. Schwartz provided a theory in distribution that made its existence valid. Yes, it is not a "function" precisely, but it's still valid and we can still perform mathematical operations to it or with it and treat it similarly to a function. The way bhobba puts it makes it sound like you can't because "it is not a function". Of course, I know that's not what he means. But the fact that Dirac's Delta Function is not a "function" is merely an annotation to the discussion here and practically should be treated as so IMO.

 

[bhobba]

I guess that's why it's called the Dirac delta-FUNCTION. As I recall Dirac defined his FUNCTION by the requirements $\delta(x) = 0$ for $x \ne 0$ and $\int_{-\infty}^\infty \delta(x)dx = 1$. (The Principles Of Quantum Mchanics, section 15, p58.) If Dirac thinks it's a function that's good enough for me.

You will benefit from reading the first few pages of Von-Neumann - Mathematical Foundations of QM - its very enlightening about the so called Dirac Delta Function.

Von-Neumann never could, or possibly even could not bother, in sorting it out. It took the combined efforts of some of the greatest mathematicians of the 20th century to do that eg Grothendieck, Schwartz and Gelfland culminating in the Generalized Eigenvalue Theorem. This stuff is NOT trivial, and glossed over in beginner and even some intermediate texts. Advanced texts like Ballentine however do not.

It now has many many applied applications including probability theory eg it's how discreet and continuous PDF's are reconciled.

Thanks
Bill

 

[bhobba]

we had a course on functional analysis and distribution theory before we had Quantum Mechanics 1 lectures. It made things much easier.

It sure does.

I studied functional analysis but not distribution theory before Von-Neumann and Dirac. Bad bad move. Von-Neumann was, correctly for his time, scathing of Dirac's use of that function - yet Dirac was elegant, beautiful and worked - it was generally used by physicists - correctly IMHO - Von-Neumann was a breeze after studying as I had done, Hilbert Spaces - but used a different approach and was more used by mathematicians. I took a sojourn for quite an extended period in sorting it all out reading some pretty heavy tomes such as Gelflad's on Generalized functions. I came out the other end knowing the answer - but I would have been better off learning more actual QM and sorting it out a bit later. Nowadays we have texts that explain it all from the start eg:
https://www.amazon.com/dp/0821846302/?tag=pfamazon01-20

You should study that AFTER a book like Ballentine, but before that, as I said, all physicists/applied mathematicians should know distribution theory. It has many advantages eg it makes Fourier transforms a snap - otherwise it gets bogged down in issues of convergence.

Thanks
Bill

 

[bhobba]

But the fact that Dirac's Delta Function is not a "function" is merely an annotation to the discussion here and practically should be treated as so IMO.

Applied mathematicians/physicists do that all the time, correctly IMHO. But it does confuse thinking students. It confused me when I first came across it in a course on differential equations. I of course asked my professor. He knew me and sort of smiled - I will leave you to sort that one out. I didn't get around to it at the time but later, as explained above, it reared its ugly head again in QM where I took the time to sort it out.

That's why I have formed the view it should be part of all applied mathematicians/physicists education. After that - yes one speaks quite informally about it as a function, and correctly so IMHO. However the sense I got from the OP is he is a thinking student like I was and needed a careful answer. I was not, and apologize if anyone took it that way, trying to suggest others obviously knowledgeable in QM that posted didn't know this. I was simply fleshing out what's really going on by pointing out density function isn't quite the correct usage of language for this.

Thanks
Bill

 

[bhobba]

Hmmm - its not even a function so I am not quite sure what you mean.

Rereading this I think I didn't express myself well.

I should have said - even though it is often called a function when speaking informally, it really isn't, so the use of density function is a bit ambiguous for a beginning/intermediate student

Thanks
Bill

 

[PeterDonis]

Moderator's note: A few off topic posts have been deleted. Please refrain from arguing about whether mentioning properties of the Dirac delta is appropriate.

 

[vanhees71]

I guess that's why it's called the Dirac delta-FUNCTION. As I recall Dirac defined his FUNCTION by the requirements $\delta(x) = 0$ for $x \ne 0$ and $\int_{-\infty}^\infty \delta(x)dx = 1$. (The Principles Of Quantum Mchanics, section 15, p58.) If Dirac thinks it's a function that's good enough for me.

Well, but it's a contradiction. There can't be any function with the properties you state. In fact the $\delta$ distribution (it's a distribution in the sense of a generalized function) has been invented much earlier by Sommerfeld around 1910, but Dirac has made use of it in QT so much that the name stuck. The mathematicians immediately jumped up and told everybody that such a "function" doesn't exist, but since Dirac's method was so convenient and lead to sensible results, the mathematicians right away brought order into the mess and constructed an entire new subfield of mathematics, called "functional analysis". It's a perfect example for a fruitful inspiration of a huge field in mathematics from applications of mathematics (in this case in theoretical physics).

It is good to use the modern correct names of objects also in the non-rigorous way to calculate sometimes in physics, because it keeps you aware that there is something special to be handled with some care. Although the hand-waving robust way physicists treat the separable Hilbert space as if it were a finite-dimensional unitary vector space works astonishingly well, there are pitfalls, you have to be aware, not to get confused.

E.g., it's quite easy to see that the Heisenberg algebra $[\hat{x},\hat{p}]=\mathrm{i} \hbar$ cannot be realized in a finite-dimensional unitary vector space and that neither the "eigenvectors" of $\hat{x}$ nor of $\hat{p}$ are true Hilbert-space vectors, but they live in the dual of the dense subspace, where these operators are commonly defined. It's a good exercise to think about this by trying to evaluate the trace of the commutator on both sides of the Heisenberg relation!

 

[bhobba]

the mathematicians right away brought order into the mess and constructed an entire new subfield of mathematics, called "functional analysis". It's a perfect example for a fruitful inspiration of a huge field in mathematics from applications of mathematics (in this case in theoretical physics).

They sure did.

Being an applied mathematician and interested in physics, as a student sometimes professors and students used to joke a bit and call pure math types 'puerile mathematicians' and other such things in good humor. However math and physics actually interact with each providing stimulus to the other - sometimes physicists provide problems/issues mathematicians need to solve and conversely. Both arts benefit greatly from each other. For example Witten is a great physicist yet won the Fields medal. I am sure he was greatly honored, but might have felt like Rutherford when he won the Nobel in Chemistry rather than physics - honored, of course, but he was a physicist not a chemist.

Occasionally we have people who are both great mathematicians and physicists like Poincare and Von-Neumann but of recent times they don't seem as common. Maybe its just increasing specialization we see in all areas - I don't really know. The best I can think of currently at their peak (Witten is starting to get a bit old these days at 66) is Terry Tao - however his applied stuff isn't physics but signal processing.

Thanks
Bill