# Why don't we use the Schrodinger Uncertainty Principle?

• I
I have read that the Schrodinger Uncertainty Principle is an extension of Heisenberg's. So, why don't we use the Schrodinger Uncertainty Principle instead of Heisenberg's?
Thanks!

blue_leaf77
Homework Helper
Schrodinger Uncertainty Principle
Which one is that?

dextercioby
Homework Helper
The only reasonable answer to the posed question is that:
2. Robertson's formulation is simpler, thus better suitable for a textbook written for a greater audience.

The only reasonable answer to the posed question is that:
2. Robertson's formulation is simpler, thus better suitable for a textbook written for a greater audience.
Which one do researchers use?

dextercioby
Homework Helper
I wouln't know, I ain't one. But this is well-established physics since 1929 anyway.

I wouln't know, I ain't one. But this is well-established physics since 1929 anyway.
oh, ok. So, this also affects [x,p]=ihbar?

mfb
Mentor
Which one do researchers use?
Most of the time: none. The uncertainty principle alone is rarely sufficient to determine what happens, and if you use the more powerful tools of quantum mechanics the uncertainty principle is satisfied automatically.

blue_leaf77
Homework Helper
this also affects [x,p]=ihbar?
That's a fundamental commutation relation, so it will not change. In fact this relation affects the particular form of the uncertainty relation involving position and momentum.

blue_leaf77
Homework Helper
I have read that the Schrodinger Uncertainty Principle is an extension of Heisenberg's. So, why don't we use the Schrodinger Uncertainty Principle instead of Heisenberg's?
Thanks!
I think it's just a matter of convenience. The general form of the uncertainty principle is like ##(\Delta A \Delta B)^2 \geq C_1 + C_2## where ##C_1## and ##C_2## are real, positive numbers. Now, these two unequalities are true ##(\Delta A \Delta B)^2 \geq C_1## and ##(\Delta A \Delta B)^2 \geq C_2##. Most people, however, choose to use the commutator as the minimum value because I think in most cases, it has special simple form for a given pair of noncommuting observables.

Joker93
George Jones
Staff Emeritus
Gold Member
I like what Griffiths wrote in his elementary particles book,"When you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."

vanhees71, Joker93 and Demystifier
That's a fundamental commutation relation, so it will not change. In fact this relation affects the particular form of the uncertainty relation involving position and momentum.
So, does it affect Δx*Δp>=hbar/2?

blue_leaf77
Homework Helper
So, does it affect Δx*Δp>=hbar/2?
In your question, what affects what?

In your question, what affects what?
We get that inequality for the product of the uncertainty of the momentum and position from the standard inequality(Heisenberg's inequality). Is it more correct to use the other inequality?

blue_leaf77
Homework Helper
Is it more correct to use the other inequality?
The other inequality, which is more general, was derived without omitting anything (or at least it omits less frequent than the simpler Heisenberg inequality). This means, the general uncertainty principle is always automatically satisfied for any state. Since it is always satisfied, getting rid of one the terms in the lower bound of the inequality will not affect its validity.

The other inequality, which is more general, was derived without omitting anything (or at least it omits less frequent than the simpler Heisenberg inequality). This means, the general uncertainty principle is always automatically satisfied for any state. Since it is always satisfied, getting rid of one the terms in the lower bound of the inequality will not affect its validity.
So, why even consider using the more general inequality if the standard inequality gives a lower bound?

blue_leaf77
Homework Helper
general inequality
Which one are you terming as "general", the Schroedinger's or Heisenberg's uncertainty?

atyy
So, why even consider using the more general inequality if the standard inequality gives a lower bound?
Don't worry about the inequality, it is not (that) important. As blue_leaf77 said in post #9, the key point is the commutation relation which is fundamental, and from which all the inequalities can be derived. All the inequalities are more or less variations of each other.

Joker93
Which one are you terming as "general", the Schroedinger's or Heisenberg's uncertainty?
I think the more general is Schroedinger's. I am referring to Heisenberg's as the standard one! :D

blue_leaf77
$$(\sigma_A \sigma_b)^2 \geq \left( \frac{1}{2} \langle \{A,B\}\rangle - \langle A \rangle \langle B \rangle \right)^2$$
$$(\sigma_A \sigma_b)^2 \geq \left( \frac{1}{2i}\langle[A,B]\rangle \right)^2$$