Graduate Found a new formula of Dirac calculus

[Demystifier]

I have found a new formula in Dirac calculus. The formula is elementary, so probably I'm not the first who found it. Yet, I have never seen it before. As many other formulas in Dirac calculus, it is not rigorous in the sense of functional analysis. Rather, it is a formal equality, which is only well defined under integrals, while otherwise it should be used with a grain of salt. But the formula is very elegant and, I believe, can be very useful.

Let $\hat{x}$ be an operator with a continuous spectrum, such as the position operator. Its eigenstates $|x\rangle$ obey $\hat{x}|x\rangle=x|x\rangle$, where $x$ are the eigenvalues of $\hat{x}$. The operator has a formal spectral decomposition
$$\hat{x}=\int dx\, x |x\rangle\langle x|$$
On the other hand, we can also write a formal identity
$$\int dx\, x\delta(\hat{x}-x)=\hat{x}$$
Comparing the two equation above, one finds the formula
$$|x\rangle\langle x|=\delta(\hat{x}-x)$$
which seems to be something new, or at least new to me.

Have you seen this formula before? For the mathematically inclined, have you got any idea how to make it more rigorous?

 

[gentzen]

On the other hand, we can also write a formal identity
$$\int dx\, x\delta(\hat{x}-x)=\hat{x}$$
Comparing the two equation above,

Well, since $\hat{x}$ is supposed to be an operator, $\delta(\hat{x}-x)$ is "not yet" defined. So your "formal identity" could instead be used as a definition of (the meaning of) $\delta(\hat{x}-x)$.

one finds the formula
$|x\rangle\langle x|=\delta(\hat{x}-x)$
which seems to be something new, or at least new to me.

Have you seen this formula before? For the mathematically inclined, have you got any idea how to make it more rigorous?

Are we sure that comparing the two equations really gives anything? I tried to make it more rigorous, started by rewritting the "definition" of $\delta(\hat{x}-x)$ to a definition for $\delta(\hat{y}-x)$
$$\int dx\, x\delta(\hat{y}-x)=\hat{y}$$

Then I wanted to introduce test function to be integrated over, but then I no longer understood why we should be able to conclude from two integrals being equal that their integrants were equal too.

 

[Demystifier]

Then I wanted to introduce test function to be integrated over, but then I no longer understood why we should be able to conclude from two integrals being equal that their integrants were equal too.

For an arbitrary test function $f(x)$ we have
$$\int dx\, f(x) \delta(\hat{x}-x)=f(\hat{x})$$
that's how the functional $\delta(\hat{x}-x)$ is defined. Using my formula
$$\delta(\hat{x}-x)=|x\rangle\langle x|$$
this becomes
$$\int dx\, f(x) |x\rangle\langle x|=f(\hat{x})$$
which is nothing but the standard spectral decomposition of the operator $f(\hat{x})$. The formula is therefore valid for any test function, so in a sense it is valid even without the integral.

 

[Demystifier]

By the way, I have found this in an attempt to find a counterexample to the statement than any operator (in the Hilbert space of one particle in one dimension) can be written as $F(\hat{x}, \hat{p})$. It occurred to me that $|x\rangle\langle x|$ could be a counterexample, so I found that it's not a counterexample, if $F$ can be a generalized function (i.e. functional).

 

[gentzen]

The formula is therefore valid for any test function, so in a sense it is valid even without the integral.

So you succeded yourself to make it more rigorous. My mistake was to introduce $y$ in the wrong place. I should have kept $\hat{x}$ and renamed the free/integration variable to $y$. Then I can "rewrite" your answer, such that it reads:

For an arbitrary test function $f(x)$ we have
$$\int dy\, f(y) \delta(\hat{x}-y)=f(\hat{x})$$
that's how the functional $\delta(\hat{x}-y)$ is defined. Using "the proposed" formula
$$\delta(\hat{x}-y)=|y\rangle\langle y|$$
this becomes
$$\int dy\, f(y) |y\rangle\langle y|=f(\hat{x})$$
which is nothing but the standard spectral decomposition of the operator $f(\hat{x})$.

Of course, one can discuss whether $\delta(\hat{x}-x)$ or $|y\rangle\langle y|$ is more confusing. Perhaps a compromise would be to use $x'$ instead of $y$, and then claim that neither $\delta(\hat{x}-x')$ nor $|x'\rangle\langle x'|$ should be confusing.

 

[vanhees71]

For a spinless particle in non-relativistic QM the Heisenberg algebra of $\hat{\vec{x}}$ and $\hat{\vec{p}}$ is sufficient to describe all one-particle operators by definition. You can ensure this by analyzing the ray represenstations of the Galilei group.

 

[Demystifier]

My next project is to find a similar formula for $|x\rangle\langle x'|$.

 

[martinbn]

https://physics.stackexchange.com/questions/199757/can-operators-be-argument-of-dirac-delta-function