Filter Theory in Physics Notation

Sep 7, 2018   #Quantum Mechanics  #Signal Processing  #Filters 

I happened to have picked up quantum mechanics from Feynman’s lectures and Dirac’s book before I got around to my main stream theory of linear filters during my B.Tech. So I went through a somewhat weird path of understanding the math around filter theory based on the quant-mech math I’d picked up. I do think it is clearer than the way linear filter theory books deal with the topic, especially due to an explicit notation for a “signal” in the abstract that is independent of “a representation of a signal”.

Here I capture a gist of how it went into my head. Note that I’m not aiming for mathematical rigour, but more as an illustration of how good notations are technology for the mind - simplifying our ability to reason in a domain.

A “signal” is a vector, which we represent using the “braket” notation as \(\ket{\mathbf s}\).

Since a signal is entirely defined by its values at various times, we use \(\ket{t}\) to denote the basis vector for an impulse signal peaking only at time \(t\), where \(\braket{t'|t} = \delta(t-t')\). So we can decompose \(\ket{\mathbf s}\) as

$$\ket{\mathbf s} = \int{\ket{t}\braket{t|{\mathbf s}}dt} = \int{s(t)\ket{t}dt}$$

where \(\braket{t|{\mathbf s}} = s(t)\).

A linear filter \(\hat{H}\) is expressed as a linear operator on \({\ket{\mathbf s}}\) as \(\hat{H}{\ket{\mathbf s}}\).

If you denote by \(\hat{D}_t\) the operator that delays a signal by time \(t\), then we have -

$$\hat{D}_{t'}{\ket{t}} = {\ket{t+t'}}$$

which when applied to \(\ket{\mathbf s}\) gives -

$$\hat{D}_{t'}\ket{\mathbf s}$$

$$= \int{s(t)\hat{D}_{t'}\ket{t}dt}$$ $$= \int{s(t){\ket{t+t'}}dt}$$ $$= \int{s(t-t'){\ket{t}}dt}$$

Now, how do we characterize \(\hat{D}_t\)?

$$\braket{t'|\hat{D}_{t'}|t}$$ $$= {\langle t''}{\mid t+t'\rangle}$$ $$= \delta(t' + (t - t''))$$ $$= {\langle t''-t'}{\mid t\rangle}$$


$$\bra{t''}\hat{D}_{t'} = \bra{t''-t'}$$

Then a “time invariant” filter is (by definition) an operator \(\hat{H}\) that commutes with \(\hat{D}_t\).

$$[\hat{H},\hat{D}_t] = \hat{H}\hat{D}_t - \hat{D}_t\hat{H} = 0$$

To construct a frequency representation of a signal, we choose a basis \({\mid \omega \rangle}\) such that \(\langle \omega {\mid t \rangle} = e^{i\omega{t}}\).

So the Fourier representation of \(\mid {\mathbf s} \rangle\) is given by

$${\mid {\mathbf s} \rangle} = \int{{\mid\omega\rangle}{\langle\omega\mid}{{\mathbf s}\rangle}d\omega} = \int{S(\omega){\mid \omega\rangle}d\omega}$$


$$ S(\omega) = {\langle\omega{\mid {\mathbf s}\rangle}} = \int{{\langle\omega}{\mid{t}\rangle}{\langle t}{\mid {\mathbf s}\rangle}dt} = \int{s(t)\langle\omega{\mid t\rangle}dt} = \int{s(t)e^{i\omega t}dt} $$

If we apply a linear time-invariant operator \(\hat{H}\) to a signal \({\mid {\mathbf s} \rangle}\), we have

$$\hat{H}{\mid {\mathbf s} \rangle} = \int{\hat{H}{\mid t\rangle}{\langle{t}\mid}{{\mathbf s}\rangle}dt} = \int{s(t)\hat{H}{\mid t\rangle}dt}$$

So we can characterize \(\hat{H}\) using -

$${{\langle t\mid}\hat{H}{\mid {\mathbf s}\rangle}} = \int{s(t'){\langle t\mid}\hat{H}{\mid t'\rangle}dt'}$$

So we need to know -

$${\langle t\mid}\hat{H}{\mid t'\rangle}$$

$$= {\langle t\mid}\hat{H}\hat{D}_{t'}\ket{0}$$

$$= {\langle t\mid}\hat{D}_{t'}\hat{H}\ket{0}$$

$$= {\langle t\mid}\hat{D}_{t'}\hat{H}\ket{0}$$

$$= {\langle t-t'\mid}\hat{H}\ket{0}$$

The last expression tells us that we can completely characterize such a \(\hat{H}\) if we know \({\langle t\mid}\hat{H}{\mid 0\rangle} = h(t)\) which we call the “impulse response” of \(\hat{H}\).

For two such filters, \(\hat{H}\) and \(\hat{G}\), we can characterize their sequential application \(\hat{H}\hat{G}\) (which is also time invariant) if we know their individual responses, since -

$${\langle t\mid}\hat{H}\hat{G}{\mid 0\rangle}$$

$$= \int{{\langle t\mid}\hat{H}{\mid t'\rangle}{\langle t'\mid}\hat{G}{\mid 0\rangle}dt'}$$

$$= \int{h(t-t')g(t')dt'}$$

That gives the convolution expression.

Now what is the Fourier representation of \(\hat{H}\)? i.e. we want to find the components in the \({\mid\omega\rangle}\) basis.

Since \(\hat{H}{\mid{\mathbf s}\rangle} = \int{\hat{H}{\mid\omega\rangle}{\langle\omega\mid}{{\mathbf s}\rangle}d\omega} = \int{\hat{H}{\mid\omega\rangle}S(\omega)d\omega}\), we need to just find \(\hat{H}{\mid\omega\rangle}\) in the \({\mid\omega\rangle}\) basis.

$$\hat{H}{\mid\omega\rangle} = \int{{\mid{t}\rangle}{\langle{t}\mid}\hat{H}{\mid\omega\rangle}dt}$$

So -

$$\braket{t|\hat{H}|\omega}$$ $$ = \int{\braket{t|\hat{H}|t'}\braket{t'|\omega}dt'}$$ $$ = \int{\braket{t|\hat{H}|t'}e^{-i\omega t'}dt'}$$ $$ = \int{h(t-t'){e^{-i\omega t'}}dt'}$$ $$ = e^{-i\omega t}\int{h(t')e^{i\omega t'}dt'}$$ $$ = e^{-i\omega t}H(\omega)$$ $$ = {\langle{t}{\mid}}{\omega\rangle}{\langle\omega\mid}\hat{H}{\mid\omega\rangle}$$

where \(H(\omega) = \braket{\omega|\har{H}|\omega}\), which implies \(\hat{H}\ket{\omega} = H(\omega)\ket{\omega}\) - i.e. \(H(\omega)\) are the eigenvalues of \(\hat{H}\) with the eigenvectors being the basis vectors \(\ket{\omega}\). In other words, linear time invariant filters are “diagonal” in the \(\ket{\omega}\) basis.

So we have -

$$\braket{\omega|\hat{H}|{\mathbf s}}$$ $$ = \braket{\omega|\hat{H}|\omega}\braket{\omega|\mathbf s}$$ $$ = H(\omega)S(\omega)$$

… which means that filtering using linear time-invariant filters is akin to multiplication in the Fourier domain. Therefore we can characterize the product \(\hat{H}\hat{G}\) as -

$$\hat{H}\hat{G}\ket{\omega}$$ $$ = \hat{H}\ket{\omega}\braket{\omega|\hat{G}|\omega}$$ $$ = \ket{\omega}\braket{\omega|\hat{H}|\omega}\braket{\omega|\hat{G}|\omega}$$ $$ = H(\omega)G(\omega)\ket{\omega}$$ $$ = \hat{G}\hat{H}\ket{\omega}$$

which is the Fourier domain multiplication of filter characteristic functions. This makes it more apparent that linear time-invariant filters are commutative - i.e. \([\hat{H},\hat{G}] = \hat{H}\hat{G} - \hat{G}\hat{H} = 0\).