Requesting help with the uncertainty principle equation

[dyn]

Hi
For 2 Hermitian operators A and B using the Cauchy-Schwarz inequality and assuming the expectation values of A and B are zero I get
(ΔA)²(ΔB)² ≥ (1/4)|<(AB+BA)>|² + (1/4)|<(AB-BA)>|²

Now both terms on the RHS are positive so why is this inequality usually just written with only the commutator term , dropping the anti-commutator term ? Surely it could be written with only the 1st term instead ? And why is it not normally written with both terms ?
Thanks

 

[BvU]

I get
(ΔA)2(ΔB)2 ≥ (1/4)|<(AB+BA)>|2 + (1/4)|<(AB-BA)>|2

Could you show us how ?

 

[mfb]

Often <AB+BA> = 0 but the inequality stays true if you drop a term that cannot be negative.

You derived the most general form, the Schrödinger uncertainty relation.

 

[dyn]

Often <AB+BA> = 0 but the inequality stays true if you drop a term that cannot be negative.

You derived the most general form, the Schrödinger uncertainty relation.

But the commutator can sometimes be zero and both terms cannot be negative so why is the anti-commutator term usually dropped from the uncertainty relation ?

 

[mfb]

The commutator is rarely zero in cases where the uncertainty relation is used.
If it is zero and the anticommutator is also zero you don't need the uncertainty relation. $\geq 0$ is trivial.

 

[dyn]

The commutator is rarely zero in cases where the uncertainty relation is used.
If it is zero and the anticommutator is also zero you don't need the uncertainty relation. $\geq 0$ is trivial.

Is it not the case that the anti-commutator is also rarely zero ? In which case why is that term general dropped ?

 

[PeroK]

Is it not the case that the anti-commutator is also rarely zero ? In which case why is that term general dropped ?

You could have the UP for a particular state, but in that case you have an equality. I.e. $(\Delta A)^2 (\Delta B)^2 =$ something definite.

Where the UP comes in is usually for any state of a system. And, you may be able to find a state where $\langle AB + BA \rangle = 0$. In that case, in general, the inequality will involve the commutator.

Note that its an inequality. Most of the time the uncertainty will be greater. But, if you need a general inequality for a system, then it reduces to the commutator term only.

(You might like to try to find a system where the expected value of the anti-commutator is never zero, in which case there is actually a stronger UP.)

Finally, the position-momentum UP, for example, is valid for all systems and all states. In which case, you can't do better than the HUP. I.e. there are systems and states for which equality applies.

 

[dyn]

Note that its an inequality. Most of the time the uncertainty will be greater. But, if you need a general inequality for a system, then it reduces to the commutator term only.
.

Why does a general inequality reduce to the commutator term only ? What happens to the anti-commutator term ?

 

[PeroK]

Why does a general inequality reduce to the commutator term only ? What happens to the anti-commutator term ?

Because it's an inequality. You can always drop a term. E.g.

If $a \ge b + c$ Where $c \ge 0$, then $a \ge b$.

 

[dyn]

But both terms are non-negative so why always drop the anti-commutator ? Is it just convention ?

 

[PeroK]

But both terms are non-negative so why always drop the anti-commutator ? Is it just convention ?

No. Look at the full inequality in the above link.

For example, with only the anti-commutator term the HUP would become:

$\sigma_x \sigma_p \ge 0$

Which doesn't say anything.

 

[kith]

But both terms are non-negative so why always drop the anti-commutator ? Is it just convention ?

For canonical conjugates, the (expectation value of the) commutator is $i \hbar$ for all states. It doesn't make sense to drop this term and keep only the anti-commutator term because you can't find a state where the resulting inequality becomes an equality.

For the Heisenberg uncertainty principle (where only the commutator term is kept), there are states where the inequality becomes an equality (the coherent states of the harmonic oscillator, for example).

Still, it might be interesting to restrict oneself to states where the anti-commutator doesn't vanish and ask what kind of states minimize the Schrödinger uncertainty principle . The so-called squeezed states have this property (Trifonov 2001). Trifonov also talks about such states for noncanonical observables but I have never heard of them.

In the end, the significance of all these states is whether they can be prepared in the lab and have interesting properties.

 

[dyn]

The UP I quoted in #1 contains a commutator term and an anti-commutator term. Both terms are non-negative so I do not understand why the UP is generally shown with only the commutator term.
Also with the usual form involving only the commutator term if [ A , B ] = 0 then this says (ΔA)²(ΔB)² ≥ 0 but it cannot be zero because if [A, B] = 0 then AB+BA does not equal zero

 

[PeroK]

The UP I quoted in #1 contains a commutator term and an anti-commutator term. Both terms are non-negative so I do not understand why the UP is generally shown with only the commutator term.
Also with the usual form involving only the commutator term if [ A , B ] = 0 then this says (ΔA)²(ΔB)² ≥ 0 but it cannot be zero because if [A, B] = 0 then AB+BA does not equal zero

You've dropped the $\langle A \rangle \langle B \rangle$ term from the expression involving the anti-commutator. If $A$ and $B$ commute, you can find a shared eigenbasis and the whole anti-commutator term is 0:

$\frac12 \langle AB + BA \rangle - \langle A \rangle \langle B \rangle = 0$

 

[dyn]

If A and B don't commute as with x and p_x then why is the anti-commuter term generally dropped ?

 

[kith]

If A and B don't commute as with x and p_x then why is the anti-commuter term generally dropped ?

It isn't always dropped, see the paper I cited above. But it's a question of usefulness.

The way the HUP is often used is "if the spread in position is such and such, the spread in momentum needs to be at least such and such". This enables quick estimates and you don't even need to know the exact state. An example would be "I know that the particle is located in the lab, so the spread in momentum needs to be at least of the order hbar / length of the lab".

You can't reason like this with the Schrödinger uncertainty relation because the anti-commutator depends on the state. For the same reason, uncertainty relations for noncanonical observables aren't used much in practise.