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

Tags:
1. May 4, 2016

### Joker93

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!

2. May 4, 2016

### blue_leaf77

Which one is that?

3. May 4, 2016

4. May 4, 2016

### dextercioby

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.

5. May 4, 2016

### Joker93

Which one do researchers use?

6. May 4, 2016

### dextercioby

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

7. May 4, 2016

### Joker93

oh, ok. So, this also affects [x,p]=ihbar?

8. May 4, 2016

### Staff: Mentor

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.

9. May 4, 2016

### blue_leaf77

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.

10. May 4, 2016

### blue_leaf77

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.

11. May 4, 2016

### George Jones

Staff Emeritus
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."

12. May 5, 2016

### Joker93

So, does it affect Δx*Δp>=hbar/2?

13. May 5, 2016

### blue_leaf77

In your question, what affects what?

14. May 5, 2016

### Joker93

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?

15. May 5, 2016

### blue_leaf77

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.

16. May 5, 2016

### Joker93

So, why even consider using the more general inequality if the standard inequality gives a lower bound?

17. May 5, 2016

### blue_leaf77

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

18. May 5, 2016

### atyy

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.

19. May 5, 2016

### Joker93

I think the more general is Schroedinger's. I am referring to Heisenberg's as the standard one! :D

20. May 5, 2016

### blue_leaf77

The general/Schroedinger inequality is what one ended up when deriving the lower bound for $(\Delta A \Delta B)^2$, i.e. it's the true lower bound. As I have tried to illustrate in post #10, the general uncertainty principle has the form $(\Delta A \Delta B)^2 \geq C_1 + C_2$ with $C_1$ and $C_2$ being positive semi-definite. Knowing this, one can say $(\Delta A \Delta B)^2$ is always bigger than either $C_1$ or $C_2$ separately. In other words, both
$$(\sigma_A \sigma_b)^2 \geq \left( \frac{1}{2} \langle \{A,B\}\rangle - \langle A \rangle \langle B \rangle \right)^2$$
and
$$(\sigma_A \sigma_b)^2 \geq \left( \frac{1}{2i}\langle[A,B]\rangle \right)^2$$
are true. But IMO, people tend to prefer using the second one because a commutator usually has simpler form than an anti-commutator Take for example, the momentum and position operators. Their commutator is a constant while their anti-commutator ... I don't know myself, may be it doesn't have a state-independent form.