I Questions about the Dirac delta

1. Jul 25, 2017

Isaac0427

Hi,

Consider this definition of the Dirac delta:
$$\delta (x-q)=\lim_{a \rightarrow 0}\frac{1}{a\sqrt \pi}e^{-(x-q)^2/a^2}$$
First, this would make a normalized position eigenfunction
$$\psi (x)=\lim_{a \rightarrow 0}\frac{1}{\sqrt{a\sqrt \pi}}e^{-x^2/2a^2}$$
right?

If that is so, why do people say that the Dirac delta is a normalized position eigenfunction? The integral of the delta function squared clearly is not equal to one.

I have another question involving the Fourier transform, but it is dependent on the answer to this one, so I will ask it in this thread or a new thread after I receive an answer.

2. Jul 25, 2017

andrewkirk

Where did you get that definition? IIRC the definition as a limit has the pdf of the distribution $\mathcal N(q,a)$ inside the limit, in which case the definition would be
$$\delta(x-q) = \lim_{a\to 0} \frac1{a\sqrt{2\pi}} e^{-(x-q)^2/2a^2}$$

3. Jul 25, 2017

@andrewkirk You used $\sigma=a$ in the Gaussian distribution. @Isaac0427 simply used $\sigma=\frac{1}{\sqrt{2}} a$. (Both are appropriate delta functions).

4. Jul 25, 2017

Daniel Gallimore

This is a good definition of the Dirac delta; it has area $1$, always. However, I disagree with your normalized position eigenfunction. The Dirac delta is the normalized position eigenfunction.

To see why this is the case, we need to talk about orthonormality. If the spectrum of an operator is discrete (that is, there is a finite spacing between each eigenvalue) then its eigenvectors have true orthonormality; i.e., if $\left\lvert f_m\right>$ and $\left\lvert f_n\right>$ are eigenvectors of some operator with a discrete spectrum, then $$\left<f_m\big\vert f_n\right>=\delta_{mn}$$ However, if an operator's spectrum is continuous, then its eigenvectors can only achieve something that Griffiths calls "Dirac orthonormality": $$\left<f_{x'}\big\vert f_x\right>=\delta(x-x')$$ Considering the delta function here is zero everywhere except when $x=x'$, this is very close to true orthonormality.

What does this have to do with the formula you have provided for the delta function? It means that if the position eigenfunctions are really Dirac deltas, we should expect to get another Dirac delta when we take the inner product of two eigenfunctions.

The position eigenfunction should satisfy the equation $$xg_y(x)=yg_y(x)$$ where $y$ is the eigenvalue. We are looking for a function with the property that multiplying it by any $x$ is the same as multiplying it by $y$. Furthermore, we know that $g_y(x)=0$ wherever $x\not=y$. However, at $x=y$, $g_y(x)$ can equal any positive, nonzero value (if we chose a negative value, this would be somewhat redundant since the negative signs would just cancel in the equation above). This fits the bill for a Dirac delta function in $x$ centered at $y$. Since we don't yet know that the delta function is a normalized eigenfunction of the position operator, we'll (naively) slap a normalization constant in front of it. So $$g_y(x)=A\delta(x-y)$$

Now, the big revelation... eigenvectors of an operator with a continuous spectrum are not square integrable. (If you have a copy of Mathematica, you can see this for yourself. Take your definition of the Dirac delta, square it, and take the integral from $-\infty$ to $\infty$. If you then let $a\rightarrow0$, you'll see that it blows up.) However, we have another property that we need to verify: Dirac orthonormality. Our eigenvector is no good at all if it does not satisfy Dirac orthonormality. Note $$\left<g_{y'}\big\vert g_y\right>=|A|^2\int_{-\infty}^\infty\delta(x-y')\delta(x-y) \, dx=|A|^2\delta(y-y')$$ To satisfy Dirac orthonormality, we let $A=1$, so $$g_y(x)=\delta(x-y)$$ is the eigenfunction of the position operator. Unfortunately, normalization just doesn't makes sense in this context, so this is as good as we can get.

If we're to perform these same integrals with your definition of the Dirac delta (assuming $A=1$ for simplicity), $$\left<g_q'\big\vert g_q\right>=\lim_{a\rightarrow0}\lim_{a'\rightarrow0}\int_{-\infty}^\infty\frac{1}{a'\sqrt \pi}e^{-(x-q')^2/(a')^2}\frac{1}{a\sqrt \pi}e^{-(x-q)^2/a^2} \, dx=\quad\lim_{a\rightarrow0}\lim_{a'\rightarrow0}\frac{1}{aa'\sqrt{\frac{1}{a^2}+\frac{1}{(a')^2}}\sqrt{\pi}}e^{-\frac{(q-q')^2}{a^2+(a')^2}}$$ Though this may not look familiar at first, it behaves just we expect it too. Note that $\exp(-\frac{(q-q')^2}{a^2+(a')^2})$ becomes arbitrarily narrow as $a$ and $a'$ go to zero. Furthermore, $aa'\sqrt{\frac{1}{a^2}+\frac{1}{(a')^2}}\sqrt{\pi}$ goes to zero as $a$ and $a'$ go to zero, so the height of the gaussian blows up. This is precisely the behavior of a delta function. In fact, the behavior is identical to if we had gone ahead and written the resulting delta function according to your original definition: $$\delta(q-q')=\lim_{a \rightarrow 0}\frac{1}{a\sqrt \pi}e^{-(q-q')^2/a^2}$$

5. Jul 25, 2017

DeathbyGreen

The eigenvalue equation of the position operator is
$\hat{x}\psi(x') = x'\psi(x')$

In other words, the operator $\hat{x}$ acting on a wavefunction $\psi(x')$ located at position x' returns the wavefunction multiplied by the single value x'. A function that follows this form is the Dirac delta function.

$\hat{x}\delta(x-x') = x'\delta(x-x')$

In this case x is the continuous variable, and the delta function "picks out" the value of the variable $\hat{x}$ at the point where it is equal to the constant x', similar to how the operator $\hat{H}$ picks out the energy eigenvalue. In non-normalized language:

$|\psi_{x'}\rangle = A\delta(x-x')$

This function IS NOT square integrable, in other words its square diverges: $\int_{-\infty}^{\infty}|\delta(x-x')|^2dx \rightarrow \infty$.However, in quantum mechanics we work in a Hilbert space, and in this space we require our basis states to be orthonormal.

$\langle\psi_{x''}|\psi_{x'}\rangle = \delta_{x'',x'}$

So the overlap between a state at x'' and another at x' can only be non zero if $x''=x'$. For two position eigenfunctions:

$\langle\psi_{x''}|\psi_{x'}\rangle = |A|^2\int \delta(x-x'')\delta(x-x')dx=|A|^2\delta(x''-x')$

Choosing A=1 gives us this "orthonormality" that we are looking for. You can actually normalize the Dirac delta mathematically though, through some trickery. Because a free particle has no boundary conditions, one must take all space to be contained in a box $LxLxL$. After completing the calculation, you can take the limit as $L\rightarrow \infty$ and get a finite answer. Normalization is a property of a continuous function, which is why is breaks down with the Delta. An interesting exercise is to take the Fourier transform into momentum space; because a delta function says you know exactly where a particle is, when you Fourier transform you get the Heisenberg uncertainty principle showing an uncertainty in momentum space.

6. Jul 25, 2017

Isaac0427

I get what everyone is saying, but why have the position eigenfunction satisfy "Dirac orthonormality" and not be normalizable while you could define the position eigenfunction as $\psi_q(x)=\lim_{a\rightarrow 0}\frac{1}{\sqrt{a\sqrt\pi}}e^{-(x-q)^2/2a^2}$, which would follow the same "infinite when x=q and zero when $x\neq q$" rule as the Dirac delta does, still be an eigenfunction of the position operator, but unlike the Dirac delta, is normalized and $\left< \psi_p(x)|\psi_q(x)\right>=\delta_{pq}$ (i.e. the functions are actually orthonormal).

7. Jul 25, 2017

Staff: Mentor

No, it isn't. The fact that functions of the form you give with $a > 0$ are normalized does not mean that the limit of those functions as $a \rightarrow 0$ is normalized.

8. Jul 25, 2017

Isaac0427

Would it be correct to say that $$\lim_{a\rightarrow 0}\int_{-\infty}^{\infty}\left|\frac{1}{\sqrt{a\sqrt\pi}}e^{-(x-q)^2/2a^2}\right|^2dx=1$$

9. Jul 25, 2017

Isaac0427

Either way, I understand Dirac orthonormality. Would it still be correct to say that $$\psi(x)=\sum_j c_j \delta\left(x-x_j\right)$$ is normalized if $$\sum_j\left|c_j\right|^2=1$$

10. Jul 25, 2017

Staff: Mentor

No. In fact, this is equivalent to saying that the Dirac delta function is not square integrable. (Your limit function is just a redefined Dirac delta.)

11. Jul 26, 2017

Orodruin

Staff Emeritus
He is taking the limit after integrating.

12. Jul 26, 2017

David Olivier

We often view integration as a kind of continuous-case equivalent of summing, but when we integrate, there is the addition of a little piece of a thing that is absent in the summing: namely the $dx$.

In a scalar product, specifically, one sums over two sets of values, like in $\sum {f_i g_i}$; one could view the equivalent integral $\int dx f g$ as $\int (\sqrt {dx}\ f) (\sqrt {dx}\ g)$, or as $\sum (\sqrt {dx}\ f) (\sqrt {dx}\ g)$. If $f$ is a delta function, this makes sense of the idea that the “real” eigenvector is not $\delta$, but something like $\sqrt {dx}\ \delta$. The $\sqrt {dx}$ is the square root of an infinitesimal variation in $x$, that is somewhat akin to the $\sqrt a$ factor that the OP has trouble with, with $a$ an infinitely small width of the delta function.

All this is of course not at all rigorous math, but I think it makes heuristic sense. (And in any case, treating the delta as a function is not rigorous math to begin with :D )

13. Jul 26, 2017

Daniel Gallimore

I like what you've done here. While eigenfunctions of operators with continuous spectra are not normalizable, linear combinations of those eigenfunctions have the possibility of being normalizable. If a linear combination of eigenfunctions is able to construct an arbitrary (normalizable) wavefunction, then the eigenfunctions are complete. The eigenfunctions of the position operator have this convenient property. However, a linear combination of such eigenfunctions is not a discrete sum, but an integral: $$\psi(x)=\int_{-\infty}^\infty c(x_j)\delta(x-x_j) \, dx_j=c(x)$$ Notice that your coefficients must likewise vary continuously. However, since in this case the Dirac delta is centered at $x$, $c(x)$ is the only coefficient that matters.

As for your second statement, you should rephrase it so that an integral of $|c(x_j)|^2$ over $x_j$ equals one. Given this condition, your wavefunction should absolutely be normalized.

14. Jul 26, 2017

vanhees71

I think we have to still clarify a lot (mathematically and physically).

First of all as it stands in #1 the limit is misleading. It's of course 0 for $x \neq q$ but $\infty$ for $x=q$. The limit, however, has to be understood as a socalled weak limit, valid only "after the integral", i.e., it defines a functional, defined on test functions (you can choose very well-behaved beasts here like all $C^{\infty}$ functions with compact support), and as such a functional it has a very simple weak limit, namely the $\delta$ distribution.

For physics it's important to keep in mind that only square-integrable functions represent (pure) states but not a distribution like the Dirac $\delta$, which cannot even be squared (at least not in a naive way). For a very good introduction to all that issues, see the booklet

M. J. Lighthill, Introduction to Fourier Analysis and Generalized Functions, Cambridge University Press (1959)

15. Jul 26, 2017

Isaac0427

I do get this, but I am wondering about the discrete case.

16. Jul 26, 2017

Daniel Gallimore

There is no discrete case. $x$ (that is, position) is a continuous variable.

17. Jul 26, 2017

Isaac0427

But say that I want to construct a wavefunction for which the particle could only be at a or b, with an equal probablilty of being at each. Would that wavefunction be $$\psi (x)=\frac{1}{\sqrt 2}\left(\delta (x-a) + \delta (x-b)\right)$$

18. Jul 26, 2017

Isaac0427

Or, if I wanted to make the wavefunction be a discrete superposition of position eigenfunctions, would I have to make it a wavevector in vector space instead of a wavefunction in function space?

19. Jul 26, 2017

Daniel Gallimore

First of all, I'd like to draw attention to the fact that this wavefunction is still not normalizable, so it cannot represent a physical particle. In fact, you can add as many delta functions together as you like (even a countably infinite amount), but you will never get a normalizable wavefunction. If you defined some finite region around $a$ and $b$ where the particle could also be, then that wavefunction could be normalized; but as it stands, you are trying to assign probabilities to individual points on the axis. The fact of the matter is this: the probability of finding a particle at any specific point on the $x$ axis is proportional to $dx$. The proportionality constant is called the probability density, and it is intimately connected to the wavefunction. In fact, it is $|\psi|^2$. This is why the wavefunction $must$ be square integrable. Without that property, there is nothing you can say about probabilities.

20. Jul 26, 2017

Isaac0427

21. Jul 26, 2017

Orodruin

Staff Emeritus
Equations 0.4 and 0.6 do not apply, $\hat x$ is continuous, not discrete.

22. Jul 26, 2017

Isaac0427

Ah, I see what you are saying. Can you define position to be discrete, i.e. by using a vector space where the position operator is, say, a 3x3 matrix with only three eigenvalues?

23. Jul 26, 2017

Staff: Mentor

Hm, yes, I see that. In that case I think evaluating the equation as written depends on how we evaluate the limit. If we view the integral inside the limit as defining a function of $a$, then that function has a constant value of $1$ for all $a > 0$, but is undefined for $a = 0$. So taking the limit to be $1$ would be consistent with the values for $a > 0$, but I don't know if you can rigorously prove that the limit must be $1$.

24. Jul 26, 2017

Daniel Gallimore

A version of this is possible, but not in the way you are thinking. (And it will still definitely be the case that you can't just name off a finite or even countably infinite number of locations where the particle could be. You need to define finite regions where you can find the particle. They can be small regions, but they must be finite in order for the wavefunction to be normalizable.)

If we want to represent the position operator as a (row-major order) "matrix" within its own basis, we calculate $$x_{y'y}=\left< g_{y'}(x) \right\rvert x \left\lvert g_y(x) \right>=\int_{-\infty}^\infty\delta(x-y')x\delta(x-y) \, dx=y'\delta(y-y')$$ But since there are an uncountably infinite number of eigenvalues (i.e., choices of $y$), there are essentially an uncountably infinite number of "rows" and "columns." It just makes no sense to write it as a matrix. It is simply more useful to think of $y'$ as a value along the $y$ axis. For each value along the axis (which all happen to be eigenvalues, since the spectrum of $x$ is continuous), there is a delta function centered at that point multiplied by the value of that point. In this way, it behaves a lot like a diagonal matrix: move $y'$ over, go $y'$ up, make everything else zeros. This is great since we should expect "diagonal" behavior from an operator expressed as a "matrix" in its own basis. (The words "diagonal" and "matrix" have very precise and intuitive meanings if the spectrum of the operator is discrete, but for a continuous spectrum, the behavior is better interpreted as analogous.)

Suppose instead we wanted to know the matrix representation of the position operator in the energy eigenbasis of the harmonic oscillator $\{\left\lvert n\right>\}$. The elements of this basis are just the stationary states of the harmonic oscillator. The matrix elements are $$x_{nn'}=\left< n \right\rvert x \left\lvert n' \right>$$ which is a nightmare to compute. It won't be diagonal since you're expressing the operator in a basis other than its own eigenbasis, and it will have a countably infinite number of elements, but it will be a matrix.

25. Jul 26, 2017

George Jones

Staff Emeritus
Actually, this example is easy to compute, but other examples might not be so easy. To compute this example, write the position operator as a linear combination of raising and lowering operators.