To derive the Heisenberg uncertainty principle analytically for specific operators such as *x* and *p* _{x}, the following argument can be used. The squared standard deviations of position and momentum observables are

$$<\Delta {x}^{2}>={\displaystyle \int {\psi}^{\ast}{\left(x-x<x>\right)}^{2}\psi dV},$$

(B.1)
$$<\Delta {p}_{x}^{2}>={\displaystyle \int {\psi}^{\ast}{\left(-i\u0127\frac{\partial}{\partial x}-<{p}_{x}>\right)}^{2}\psi dV}.$$

(B.2)
Taking the average values to be 0, (<*x*> = <*p* _{x}> = 0), we find

$$<\Delta {p}_{x}^{2}><\Delta {x}^{2}>={\displaystyle \int {\psi}^{\ast}}\left(-{\u0127}^{2}\frac{{\partial}^{2}}{\partial {x}^{2}}\right)\psi dr{\displaystyle \int {\psi}^{\ast}}{x}^{2}\psi dV.$$

(B.3)
Integration by parts permits us to write

$$\int {\psi}^{\ast}}\frac{{\partial}^{2}\psi}{\partial {x}^{2}}dV=-{\displaystyle \int \frac{\partial {\psi}^{\ast}}{\partial x}\frac{\partial \psi}{\partial x}}dV,$$

(B.4)
$$<\Delta {p}_{x}^{2}><\Delta {x}^{2}>={\u0127}^{2}{\displaystyle \int \frac{\partial {\psi}^{\ast}}{\partial x}}\frac{\partial \psi}{\partial x}dV{\displaystyle \int {\psi}^{\ast}}{x}^{2}\psi dV.$$

(B.5)
Now, using the Schwartz inequality, we obtain

$$\int f{f}^{\ast}dV{\displaystyle \int {g}^{\ast}}}gdV\ge {\left[\frac{1}{2}\left({\displaystyle \int f{g}^{\ast}dV+{\displaystyle \int g{f}^{\ast}dV}}\right)\right]}^{2}.$$

(B.6)
Setting *f* ≡ ∂ψ/∂*x* and *g* ≡ *x*ψ in this expression, one finds

$$\begin{array}{c}<\Delta {p}_{x}^{2}><\Delta {x}^{2}>\ge \frac{{\u0127}^{2}}{4}{\left({\displaystyle \int \frac{\partial \psi}{\partial x}x{\psi}^{\ast}dV}+{\displaystyle \int x\psi \frac{\partial {\psi}^{\ast}}{\partial x}dV}\right)}^{2}\\ \ge \frac{{\u0127}^{2}}{4}{\left({\displaystyle \int x\frac{\partial}{\partial x}\left(\psi {\psi}^{\ast}\right)dV}\right)}^{2}=\frac{{\u0127}^{2}}{4}\\ \ge \frac{{\u0127}^{2}}{4}.\end{array}$$

(B.7)
(p.265)
The final integral above is again evaluated by parts and is equal to -1. Defining the uncertainties in momentum and position as $\Delta {p}_{x}\equiv {<\Delta {p}_{x}^{2}>}^{1/2}$ and Δ*x* ≡ <Δ *x* ^{2}>^{1/2}, we obtain

$$\Delta {p}_{x}\Delta x\ge \frac{\u0127}{2}.$$

(B.8)
The operators *x* and *p* _{x} provide one example of two Hermitian operators *Â* and *B̂* with a commutator that is Hermitian and nonzero. Note that this results in an uncertainty principle between *Â* and *B̂*. Next, this result is extended to the general case.

Consider two Hermitian operators *Â* and *B̂* with a Hermitian commutator

$$\left[\widehat{A},\widehat{B}\right]=i\widehat{C}.$$

(B.9)
It is shown below that the standard deviations Δ*A* and Δ*B* have the product

$$\Delta A\Delta B\ge \frac{1}{2}\left|<C>\right|.$$

(B.10)
The standard deviations (uncertainties) in *Â* and *B̂* are defined as

$$\Delta A\equiv \widehat{A}-<A>,$$

(B.11)
$$\Delta B\equiv \widehat{B}-<B>,$$

(B.12)
hence observed values of their squares are

$$<{(\Delta A)}^{2}>=<\psi \Delta A|\Delta A\psi >={\Vert \Delta A\psi \Vert}^{2}$$

(B.13)
and

$$<{(\Delta B)}^{2}>=<\psi \Delta B|\Delta B\psi >={\Vert \Delta B\psi \Vert}^{2}.$$

(B.14)
The Schwartz inequality is equivalent to the inequality expressing the fact that the inner product of two vector operators, namely <ψΔ*A*ǀΔ*B*ψ>, is less than the product of their lengths, given by ǀΔ*A*ψǁǀΔ*B*ψǁ. That is,

$$||\Delta A\psi |{|}^{2}||\Delta B\psi |{|}^{2}\ge |<\psi \Delta A|\Delta B\psi >{|}^{2},$$

(B.15)
or

$${(\Delta A)}^{2}{(\Delta B)}^{2}\ge {\left|<\psi \Delta A|\Delta B\psi >\right|}^{2}={\left|<\psi |\Delta A\Delta B|\psi >\right|}^{2}.$$

(B.16)
The last step above is due to the Hermiticity of Δ*A*. The product Δ*A*Δ*B* on the right hand side of this result can be reexpressed in terms of the commutator of *Â* and *B̂*, because any operator can be written as a linear combination of two Hermitian
(p.266)
operators:

$$\begin{array}{c}\Delta A\Delta B=\frac{1}{2}(\Delta A\Delta B+\Delta B\Delta A)+\frac{1}{2}[\Delta A,\Delta B]\\ =\frac{1}{2}(\Delta A\Delta B+\Delta B\Delta A)+\frac{1}{2}[A,B]\\ =\widehat{D}+\frac{i}{2}\widehat{C}.\end{array}$$

(B.17)
Hence,

$${(\Delta A)}^{2}{(\Delta B)}^{2}\ge {\left|<\psi \left|\widehat{D}+\frac{i}{2}\widehat{C}\right|\psi >\right|}^{2}={\left|<D>+\frac{i}{2}<C>\right|}^{2}.$$

(B.18)
Because *D̂* and *Ĉ* are Hermitian, their expectation values are real. Consequently, the squared expression on the right above is simply

$${(\Delta A)}^{2}{(\Delta B)}^{2}\ge {\left|<D>\right|}^{2}+\frac{1}{4}{\left|<C>\right|}^{2}\ge \frac{1}{4}{\left|<C>\right|}^{2}.$$

(B.19)
The uncertainties in *A* and *B* therefore satisfy the general relation

$$(\Delta A)(\Delta B)\ge \frac{1}{2}\left|<C>\right|.$$

(B.20)