Appendix I: The Uncertainty Principle

${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }x^{2}\cdot |\psi (x)|^{2}\,dx-\left(\int _{-\infty }^{\infty }x\cdot |\psi (x)|^{2}\,dx\right)^{2}}$

${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }p^{2}\cdot |\varphi (p)|^{2}\,dp-\left(\int _{-\infty }^{\infty }p\cdot |\varphi (p)|^{2}\,dp\right)^{2}~.}$

Without loss of generality, we will assume that the means vanish, which just amounts to a shift of the origin of our coordinates. (A more general proof that does not make this assumption is given below.) This gives us the simpler form

${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }x^{2}\cdot |\psi (x)|^{2}\,dx}$

${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }p^{2}\cdot |\varphi (p)|^{2}\,dp~.}$

The function ${\displaystyle f(x)=x\cdot \psi (x)}$ can be interpreted as a vector in a function space. We can define an inner product for a pair of functions u(x) and v(x) in this vector space:

${\displaystyle \langle u\mid v\rangle =\int _{-\infty }^{\infty }u^{*}(x)\cdot v(x)\,dx,}$

where the asterisk denotes the complex conjugate.

With this inner product defined, we note that the variance for position can be written as

${\displaystyle \sigma _{x}^{2}=\int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\langle f\mid f\rangle ~.}$

We can repeat this for momentum by interpreting the function ${\displaystyle {\tilde {g}}(p)=p\cdot \varphi (p)}$ as a vector, but we can also take advantage of the fact that ${\displaystyle \psi (x)}$ and ${\displaystyle \varphi (p)}$ are Fourier transforms of each other. We evaluate the inverse Fourier transform through integration by parts:

${\displaystyle {\begin{aligned}g(x)&={\frac {1}{\sqrt {2\pi \hbar }}}\cdot \int _{-\infty }^{\infty }{\tilde {g}}(p)\cdot e^{ipx/\hbar }\,dp\\&={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }p\cdot \varphi (p)\cdot e^{ipx/\hbar }\,dp\\&={\frac {1}{2\pi \hbar }}\int _{-\infty }^{\infty }\left[p\cdot \int _{-\infty }^{\infty }\psi (\chi )e^{-ip\chi /\hbar }\,d\chi \right]\cdot e^{ipx/\hbar }\,dp\\&={\frac {i}{2\pi }}\int _{-\infty }^{\infty }\left[{\cancel {\left.\psi (\chi )e^{-ip\chi /\hbar }\right|_{-\infty }^{\infty }}}-\int _{-\infty }^{\infty }{\frac {d\psi (\chi )}{d\chi }}e^{-ip\chi /\hbar }\,d\chi \right]\cdot e^{ipx/\hbar }\,dp\\&={\frac {-i}{2\pi }}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {d\psi (\chi )}{d\chi }}e^{-ip\chi /\hbar }\,d\chi \,e^{ipx/\hbar }\,dp\\&=\left(-i\hbar {\frac {d}{dx}}\right)\cdot \psi (x),\end{aligned}}}$

where the canceled term vanishes because the wave function vanishes at infinity. Often the term ${\displaystyle -i\hbar {\frac {d}{dx}}}$ is called the momentum operator in position space. Applying Parseval's theorem, we see that the variance for momentum can be written as

${\displaystyle \sigma _{p}^{2}=\int _{-\infty }^{\infty }|{\tilde {g}}(p)|^{2}\,dp=\int _{-\infty }^{\infty }|g(x)|^{2}\,dx=\langle g\mid g\rangle .}$

The Cauchy–Schwarz inequality asserts that

${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}=\langle f\mid f\rangle \cdot \langle g\mid g\rangle \geq |\langle f\mid g\rangle |^{2}~.}$

The modulus squared of any complex number z can be expressed as

${\displaystyle |z|^{2}={\Big (}{\text{Re}}(z){\Big )}^{2}+{\Big (}{\text{Im}}(z){\Big )}^{2}\geq {\Big (}{\text{Im}}(z){\Big )}^{2}={\Big (}{\frac {z-z^{\ast }}{2i}}{\Big )}^{2}.}$

we let ${\displaystyle z=\langle f|g\rangle }$ and ${\displaystyle z^{*}=\langle g\mid f\rangle }$ and substitute these into the equation above to get

${\displaystyle |\langle f\mid g\rangle |^{2}\geq {\bigg (}{\frac {\langle f\mid g\rangle -\langle g\mid f\rangle }{2i}}{\bigg )}^{2}~.}$

All that remains is to evaluate these inner products.

${\displaystyle {\begin{aligned}\langle f\mid g\rangle -\langle g\mid f\rangle ={}&\int _{-\infty }^{\infty }\psi ^{*}(x)\,x\cdot \left(-i\hbar {\frac {d}{dx}}\right)\,\psi (x)\,dx\\&{}-\int _{-\infty }^{\infty }\psi ^{*}(x)\,\left(-i\hbar {\frac {d}{dx}}\right)\cdot x\,\psi (x)\,dx\\={}&i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\left[\left(-x\cdot {\frac {d\psi (x)}{dx}}\right)+{\frac {d(x\psi (x))}{dx}}\right]\,dx\\={}&i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\left[\left(-x\cdot {\frac {d\psi (x)}{dx}}\right)+\psi (x)+\left(x\cdot {\frac {d\psi (x)}{dx}}\right)\right]\,dx\\={}&i\hbar \cdot \int _{-\infty }^{\infty }\psi ^{*}(x)\psi (x)\,dx\\={}&i\hbar \cdot \int _{-\infty }^{\infty }|\psi (x)|^{2}\,dx\\={}&i\hbar \end{aligned}}}$

Plugging this into the above inequalities, we get

${\displaystyle \sigma _{x}^{2}\sigma _{p}^{2}\geq |\langle f\mid g\rangle |^{2}\geq \left({\frac {\langle f\mid g\rangle -\langle g\mid f\rangle }{2i}}\right)^{2}=\left({\frac {i\hbar }{2i}}\right)^{2}={\frac {\hbar ^{2}}{4}}}$

or taking the square root

${\displaystyle \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}~.}$

Note that the only physics involved in this proof was that ${\displaystyle \psi (x)}$ and ${\displaystyle \varphi (p)}$ are wave functions for position and momentum, which are Fourier transforms of each other. A similar result would hold for any pair of conjugate variables.

${\displaystyle {\hat {A}}}$

${\displaystyle \sigma _{A}^{2}=\langle ({\hat {A}}-\langle {\hat {A}}\rangle )\Psi |({\hat {A}}-\langle {\hat {A}}\rangle )\Psi \rangle .}$

we let ${\displaystyle |f\rangle =|({\hat {A}}-\langle {\hat {A}}\rangle )\Psi \rangle }$ and thus

${\displaystyle \sigma _{A}^{2}=\langle f\mid f\rangle \,.}$

Similarly, for any other Hermitian operator ${\displaystyle {\hat {B}}}$ in the same state

${\displaystyle \sigma _{B}^{2}=\langle ({\hat {B}}-\langle {\hat {B}}\rangle )\Psi |({\hat {B}}-\langle {\hat {B}}\rangle )\Psi \rangle =\langle g\mid g\rangle }$

for ${\displaystyle |g\rangle =|({\hat {B}}-\langle {\hat {B}}\rangle )\Psi \rangle .}$

The product of the two deviations can thus be expressed as

${\displaystyle \sigma _{A}^{2}\sigma _{B}^{2}=\langle f\mid f\rangle \langle g\mid g\rangle .}$

(1)

In order to relate the two vectors ${\displaystyle |f\rangle }$ and ${\displaystyle |g\rangle }$, we use the Cauchy–Schwarz inequality^[22] which is defined as

${\displaystyle \langle f\mid f\rangle \langle g\mid g\rangle \geq |\langle f\mid g\rangle |^{2},}$

and thus Eq. (1) can be written as

${\displaystyle \sigma _{A}^{2}\sigma _{B}^{2}\geq |\langle f\mid g\rangle |^{2}.}$

(2)

Since ${\displaystyle \langle f\mid g\rangle }$ is in general a complex number, we use the fact that the modulus squared of any complex number ${\displaystyle z}$ is defined as ${\displaystyle |z|^{2}=zz^{*}}$, where ${\displaystyle z^{*}}$ is the complex conjugate of ${\displaystyle z}$. The modulus squared can also be expressed as

${\displaystyle |z|^{2}={\Big (}\operatorname {Re} (z){\Big )}^{2}+{\Big (}\operatorname {Im} (z){\Big )}^{2}={\Big (}{\frac {z+z^{\ast }}{2}}{\Big )}^{2}+{\Big (}{\frac {z-z^{\ast }}{2i}}{\Big )}^{2}.}$

(3)

we let ${\displaystyle z=\langle f\mid g\rangle }$ and ${\displaystyle z^{*}=\langle g\mid f\rangle }$ and substitute these into the equation above to get

${\displaystyle |\langle f\mid g\rangle |^{2}={\bigg (}{\frac {\langle f\mid g\rangle +\langle g\mid f\rangle }{2}}{\bigg )}^{2}+{\bigg (}{\frac {\langle f\mid g\rangle -\langle g\mid f\rangle }{2i}}{\bigg )}^{2}}$

(4)

The inner product ${\displaystyle \langle f\mid g\rangle }$ is written out explicitly as

${\displaystyle \langle f\mid g\rangle =\langle ({\hat {A}}-\langle {\hat {A}}\rangle )\Psi |({\hat {B}}-\langle {\hat {B}}\rangle )\Psi \rangle ,}$

and using the fact that ${\displaystyle {\hat {A}}}$ and ${\displaystyle {\hat {B}}}$ are Hermitian operators, we find

${\displaystyle {\begin{aligned}\langle f\mid g\rangle &=\langle \Psi |({\hat {A}}-\langle {\hat {A}}\rangle )({\hat {B}}-\langle {\hat {B}}\rangle )\Psi \rangle \\[4pt]&=\langle \Psi \mid ({\hat {A}}{\hat {B}}-{\hat {A}}\langle {\hat {B}}\rangle -{\hat {B}}\langle {\hat {A}}\rangle +\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle )\Psi \rangle \\[4pt]&=\langle \Psi \mid {\hat {A}}{\hat {B}}\Psi \rangle -\langle \Psi \mid {\hat {A}}\langle {\hat {B}}\rangle \Psi \rangle -\langle \Psi \mid {\hat {B}}\langle {\hat {A}}\rangle \Psi \rangle +\langle \Psi \mid \langle {\hat {A}}\rangle \langle {\hat {B}}\rangle \Psi \rangle \\[4pt]&=\langle {\hat {A}}{\hat {B}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle +\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle \\[4pt]&=\langle {\hat {A}}{\hat {B}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle .\end{aligned}}}$

Similarly it can be shown that ${\displaystyle \langle g\mid f\rangle =\langle {\hat {B}}{\hat {A}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle .}$

Thus we have

${\displaystyle \langle f\mid g\rangle -\langle g\mid f\rangle =\langle {\hat {A}}{\hat {B}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle -\langle {\hat {B}}{\hat {A}}\rangle +\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle =\langle [{\hat {A}},{\hat {B}}]\rangle }$

and

${\displaystyle \langle f\mid g\rangle +\langle g\mid f\rangle =\langle {\hat {A}}{\hat {B}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle +\langle {\hat {B}}{\hat {A}}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle =\langle \{{\hat {A}},{\hat {B}}\}\rangle -2\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle .}$

We now substitute the above two equations above back into Eq. (4) and get

${\displaystyle |\langle f\mid g\rangle |^{2}={\Big (}{\frac {1}{2}}\langle \{{\hat {A}},{\hat {B}}\}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle {\Big )}^{2}+{\Big (}{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle {\Big )}^{2}\,.}$

Substituting the above into Eq. (2) we get the Schrödinger uncertainty relation

${\displaystyle \sigma _{A}\sigma _{B}\geq {\sqrt {{\Big (}{\frac {1}{2}}\langle \{{\hat {A}},{\hat {B}}\}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle {\Big )}^{2}+{\Big (}{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle {\Big )}^{2}}}.}$

This proof has an issue^[23] related to the domains of the operators involved. For the proof to make sense, the vector ${\displaystyle {\hat {B}}|\Psi \rangle }$ has to be in the domain of the unbounded operator ${\displaystyle {\hat {A}}}$, which is not always the case. In fact, the Robertson uncertainty relation is false if ${\displaystyle {\hat {A}}}$ is an angle variable and ${\displaystyle {\hat {B}}}$ is the derivative with respect to this variable. In this example, the commutator is a nonzero constant—just as in the Heisenberg uncertainty relation—and yet there are states where the product of the uncertainties is zero.^[24] (See the counterexample section below.) This issue can be overcome by using a variational method for the proof.,^[25][26] or by working with an exponentiated version of the canonical commutation relations.^[24]

Note that in the general form of the Robertson–Schrödinger uncertainty relation, there is no need to assume that the operators ${\displaystyle {\hat {A}}}$ and ${\displaystyle {\hat {B}}}$ are self-adjoint operators. It suffices to assume that they are merely symmetric operators. (The distinction between these two notions is generally glossed over in the physics literature, where the term Hermitian is used for either or both classes of operators. See Chapter 9 of Hall's book^[27] for a detailed discussion of this important but technical distinction.)