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

##### 1 Answer

#DeltaADeltaB >= 1/2|i << psi | [hatA", "hatB] | psi >>|#

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:

#(DeltaA)^2(DeltaB)^2 = << (hatA - << A >>)^2 >><< (hatB - << B >>)^2 >># #" "bb((1))# where

#<< A >> = << psi | hatA | psi >> = int_"allspace" psi^"*" hatA 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

#<< (hatA - << A >>)^2 >> = << psi | (hatA - << A >>)^2 | psi >>#

#= << (hatA - << A >>) psi | (hatA - << A >>)psi >>#

If we let

#<< (hatA - << A >>) psi | = << f |# #" "bb((2))#

be the

#| (hatB - << B >>) psi >> = | g >> # #" "bb((3))#

be the corresponding ket vector for **Cauchy-Schwartz inequality**,

#|<< f | g >>|^2 <= << f | f >> << g | g >># ,#" "bb((4))# 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

#[(DeltaA)^2(DeltaB)^2 = << f | f >> << g | g >>] >= |<< f | g >>|^2# .#" "bb((5))#

We have then by distributing the inner product terms and using

#<< f | g >> = << psi |(hatA - << A >>)(hatB - << B >>) | psi >>#

#= << psi | hatAhatB - hatA<< B >> - << A >> hatB + << A >> << B >> | psi >>#

#= << psi | hatAhatB | psi >> - << psi | << B >>hatA | psi >> - << psi | << A >> hatB | psi >> + << psi | << A >> << B >> | psi >>#

**Expectation values are constants**, which move out, and *normalized* by assumption. Here we also assume that

#= << psi | hatAhatB | psi >> - cancel(<< B >> << psi | hatA | psi >> - << A >> << psi | hatB | psi >>) + << A >> << B >> cancel(<< psi | psi >>)^(1)#

#=> << f | g >> = << psi | hatAhatB | psi >> - << A >><< B >># #" "bb((6))#

Also, the **adjoint** of

#<< g | f >> = << f | g >>^"†"#

#= (<< psi | hatAhatB | psi >> - << A >><< B >>)^"†"#

#= << psi | hatBhatA | psi >> - << A >><< B >># #" "bb((7))#

Now, we note that

#|z|^2 = x^2 + y^2 >= y^2# #" "bb((8))#

and that

#z^"*" = x - iy# .#" "bb((9))#

So,

#y^2 = [(z-z^"*")/(2i)]^2#

Therefore, from

#|z|^2 >= [(z-z^"*")/(2i)]^2# #" "bb((10))#

and if

#<< f | g >> - << g | f >> = << psi | hatAhatB | psi >> - << A >><< B >> - << psi | hatBhatA | psi >> + << A >><< B >>#

#= << psi | hatAhatB - hatBhatA | psi >>#

#= << psi | [hatA", "hatB] | psi >># #" "bb((11))# where

#[hatA", "hatB] = hatAhatB - hatBhatA# is thequantum commutatorof#hatA# and#hatB# .

Combine

#(DeltaA)^2(DeltaB)^2 >= |<< f | g >>|^2 >= ((<< f | g >> - << g | f >>)/(2i))^2#

We would get as a result, by inserting

#(DeltaA)^2(DeltaB)^2 >= | 1/(2i) << psi | [hatA", "hatB] | psi >>|^2# ,

or

#DeltaADeltaB >= | 1/(2i) << psi | [hatA", "hatB] | psi >>|# .

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

#color(blue)(DeltaADeltaB >= 1/2|i << psi | [hatA", "hatB] | psi >>|)#

and that is the **general uncertainty relation** for two observables

*CHALLENGE: Show that if we inserted* *and* *we would get #ℏ"/"2# like expected.*