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Thanks!

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Thanks!

- #2

blue_leaf77

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Which one is that?Schrodinger Uncertainty Principle

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https://en.wikipedia.org/wiki/Uncer...2.80.93Schr.C3.B6dinger_uncertainty_relationsWhich one is that?

2Robertson–Schrödinger uncertainty relations

The second point of the wikipedia page(check contents)

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1. Textbook writers are traditionalist.

2. Robertson's formulation is simpler, thus better suitable for a textbook written for a greater audience.

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Which one do researchers use?

1. Textbook writers are traditionalist.

2. Robertson's formulation is simpler, thus better suitable for a textbook written for a greater audience.

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

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

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mfb

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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.Which one do researchers use?

- #9

blue_leaf77

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

- #10

blue_leaf77

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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.

Thanks!

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George Jones

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- #12

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So, does it affect Δx*Δp>=hbar/2?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.

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blue_leaf77

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

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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?In your question, what affects what?

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blue_leaf77

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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 isIs it more correct to use the other inequality?

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

- #17

blue_leaf77

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

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atyy

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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.

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

- #20

blue_leaf77

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$$

(\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.

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