# Can the Uncertainty Principle equation be derived from the Schrodinger equation? Kindly show all steps?

Feb 7, 2018

$\Delta A \Delta B \ge \frac{1}{2} | i \left\langle\psi | \left[\hat{A} \text{, } \hat{B}\right] | \psi\right\rangle |$

In principle ANY normalized wave function can be used...

The uncertainty principle has nothing to do with Schrodinger. It is purely statistical... the variance product of two arbitrary observables is given by:

${\left(\Delta A\right)}^{2} {\left(\Delta B\right)}^{2} = \left\langle{\left(\hat{A} - \left\langleA\right\rangle\right)}^{2}\right\rangle \left\langle{\left(\hat{B} - \left\langleB\right\rangle\right)}^{2}\right\rangle$ $\text{ } \boldsymbol{\left(1\right)}$

where $\left\langleA\right\rangle = \left\langle\psi | \hat{A} | \psi\right\rangle = {\int}_{\text{allspace" psi^"*}} \hat{A} \psi d \tau$ is the expectation value of $A$.

Since observables must be Hermitian (i.e. their adjoint is equal to themselves, and the operator can move through $\psi$),

$\left\langle{\left(\hat{A} - \left\langleA\right\rangle\right)}^{2}\right\rangle = \left\langle\psi | {\left(\hat{A} - \left\langleA\right\rangle\right)}^{2} | \psi\right\rangle$

$= \left\langle\left(\hat{A} - \left\langleA\right\rangle\right) \psi | \left(\hat{A} - \left\langleA\right\rangle\right) \psi\right\rangle$

If we let

<< (hatA - << A >>) psi | = << f | $\text{ } \boldsymbol{\left(2\right)}$

be the $f$ bra vector and

| (hatB - << B >>) psi >> = | g >>  $\text{ } \boldsymbol{\left(3\right)}$

be the corresponding ket vector for $B$, then by the Cauchy-Schwartz inequality,

$| \left\langlef | g\right\rangle {|}^{2} \le \left\langlef | f\right\rangle \left\langleg | g\right\rangle$, $\text{ } \boldsymbol{\left(4\right)}$

the squared inner product between two different functions is less than or equal to the multiplication of the inner product amongst themselves.

That means that from $\left(2\right)$, $\left(3\right)$, and $\left(4\right)$,

$\left[{\left(\Delta A\right)}^{2} {\left(\Delta B\right)}^{2} = \left\langlef | f\right\rangle \left\langleg | g\right\rangle\right] \ge | \left\langlef | g\right\rangle {|}^{2}$. $\text{ } \boldsymbol{\left(5\right)}$

We have then by distributing the inner product terms and using $\left(2\right)$ and $\left(3\right)$:

$\left\langlef | g\right\rangle = \left\langle\psi | \left(\hat{A} - \left\langleA\right\rangle\right) \left(\hat{B} - \left\langleB\right\rangle\right) | \psi\right\rangle$

$= \left\langle\psi | \hat{A} \hat{B} - \hat{A} \left\langleB\right\rangle - \left\langleA\right\rangle \hat{B} + \left\langleA\right\rangle \left\langleB\right\rangle | \psi\right\rangle$

$= \left\langle\psi | \hat{A} \hat{B} | \psi\right\rangle - \left\langle\psi | \left\langleB\right\rangle \hat{A} | \psi\right\rangle - \left\langle\psi | \left\langleA\right\rangle \hat{B} | \psi\right\rangle + \left\langle\psi | \left\langleA\right\rangle \left\langleB\right\rangle | \psi\right\rangle$

Expectation values are constants, which move out, and $\psi$ is normalized by assumption. Here we also assume that $\psi$ is an eigenstate of the two observables.

$= \left\langle\psi | \hat{A} \hat{B} | \psi\right\rangle - \cancel{\left\langleB\right\rangle \left\langle\psi | \hat{A} | \psi\right\rangle - \left\langleA\right\rangle \left\langle\psi | \hat{B} | \psi\right\rangle} + \left\langleA\right\rangle \left\langleB\right\rangle {\cancel{\left\langle\psi | \psi\right\rangle}}^{1}$

$\implies \left\langlef | g\right\rangle = \left\langle\psi | \hat{A} \hat{B} | \psi\right\rangle - \left\langleA\right\rangle \left\langleB\right\rangle$ $\text{ } \boldsymbol{\left(6\right)}$

Also, the adjoint of $\left(6\right)$ gives

$\left\langleg | f\right\rangle = {\left\langlef | g\right\rangle}^{\text{†}}$

$= {\left(\left\langle\psi | \hat{A} \hat{B} | \psi\right\rangle - \left\langleA\right\rangle \left\langleB\right\rangle\right)}^{\text{†}}$

$= \left\langle\psi | \hat{B} \hat{A} | \psi\right\rangle - \left\langleA\right\rangle \left\langleB\right\rangle$ $\text{ } \boldsymbol{\left(7\right)}$

Now, we note that $f$ and $g$ may very well be complex. We have

$| z {|}^{2} = {x}^{2} + {y}^{2} \ge {y}^{2}$ $\text{ } \boldsymbol{\left(8\right)}$

and that

${z}^{\text{*}} = x - i y$. $\text{ } \boldsymbol{\left(9\right)}$

So, $z - {z}^{\text{*}} = x + i y - x + i y = 2 i y$ from $\left(9\right)$, and

${y}^{2} = {\left[\frac{z - {z}^{\text{*}}}{2 i}\right]}^{2}$

Therefore, from $\left(8\right)$,

$| z {|}^{2} \ge {\left[\frac{z - {z}^{\text{*}}}{2 i}\right]}^{2}$ $\text{ } \boldsymbol{\left(10\right)}$

and if $z = \left\langlef | g\right\rangle$, then consider $\left(6\right)$ and $\left(7\right)$ so that

$\left\langlef | g\right\rangle - \left\langleg | f\right\rangle = \left\langle\psi | \hat{A} \hat{B} | \psi\right\rangle - \left\langleA\right\rangle \left\langleB\right\rangle - \left\langle\psi | \hat{B} \hat{A} | \psi\right\rangle + \left\langleA\right\rangle \left\langleB\right\rangle$

$= \left\langle\psi | \hat{A} \hat{B} - \hat{B} \hat{A} | \psi\right\rangle$

$= \left\langle\psi | \left[\hat{A} \text{, } \hat{B}\right] | \psi\right\rangle$ $\text{ } \boldsymbol{\left(11\right)}$

where $\left[\hat{A} \text{, } \hat{B}\right] = \hat{A} \hat{B} - \hat{B} \hat{A}$ is the quantum commutator of $\hat{A}$ and $\hat{B}$.

Combine $\left(5\right)$, $\left(7\right)$, and $\left(10\right)$ to see that:

${\left(\Delta A\right)}^{2} {\left(\Delta B\right)}^{2} \ge | \left\langlef | g\right\rangle {|}^{2} \ge {\left(\frac{\left\langlef | g\right\rangle - \left\langleg | f\right\rangle}{2 i}\right)}^{2}$

We would get as a result, by inserting $\left(6\right)$, $\left(7\right)$, and $\left(11\right)$:

${\left(\Delta A\right)}^{2} {\left(\Delta B\right)}^{2} \ge | \frac{1}{2 i} \left\langle\psi | \left[\hat{A} \text{, } \hat{B}\right] | \psi\right\rangle {|}^{2}$,

or

$\Delta A \Delta B \ge | \frac{1}{2 i} \left\langle\psi | \left[\hat{A} \text{, } \hat{B}\right] | \psi\right\rangle |$.

Or, since we are in absolute value bars, we could rewrite this as

$\textcolor{b l u e}{\Delta A \Delta B \ge \frac{1}{2} | i \left\langle\psi | \left[\hat{A} \text{, } \hat{B}\right] | \psi\right\rangle |}$

and that is the general uncertainty relation for two observables $A$ and $B$.

CHALLENGE: Show that if we inserted $A = \hat{x}$ and B = hatp_x = -iℏ(del)/(delx), we would get ℏ"/"2 like expected.