General Uncertainty Relation -- Why drop anticommutator?

[Aziza]

In the derivation of the generalized uncertainty principle (as pgs 1-2 of here), there is an anticommutator term that is dropped at the end, leaving just the commutator part...this is said to "strengthen" the relation, as both terms are positive.

I don't understand this. So we basically have C>A+B, and we are dropping B. Then we are lowering the lower limit of the uncertainty, which I interpret as making the relation 'looser' rather than 'stronger'. Can anyone try convincing me otherwise?

Thanks!

 

[atyy]

Well, wikipedia seems to agree with you because it says "The Robertson uncertainty relation immediately follows from a slightly stronger inequality, the Schrödinger uncertainty relation" http://en.wikipedia.org/wiki/Uncertainty_principle

Also, http://damtp.cam.ac.uk/user/eal40/teach/QM2012/7comm.pdf "the last term can only make the inequality stronger".

 

[Aziza]

Thanks! haha wow first time I am 'right' here!

Sakurai also says dropping the term makes the relation stronger. That is where I first saw this, and didn't think it possible for this quantum god to be wrong lol.

 

[atyy]

Sakurai also says dropping the term makes the relation stronger. That is where I first saw this, and didn't think it possible for this quantum god to be wrong lol.

Me too, now I'm doubting wikipedia lol.

 

[vanhees71]

Hm, I don't see, how you can make the Robertson uncertainty relation stronger, because the equal sign usually can be reached. For the position-momentum uncertainty relation these are the Gaussian wave packets. The most simple derivation of the uncertainty relation just uses the positive definiteness of the scalar product. We assume that the system is prepared in a pure state represented $|\psi \rangle$ (normalized to 1).

Assume that $\langle A \rangle=\langle B \rangle$ (expectation values wrt. $|\psi \rangle$). Otherwise consider the operators $\hat{A}-\langle A \rangle \hat{1}$ and $\hat{B}-\langle B \rangle \hat{1}$.

Then we have to evaluate the standard deviations by
$$\langle \Delta A^2 \rangle=\langle \psi|\hat{A}^2|\psi \rangle, \quad \langle \Delta B^2 \rangle=\langle \psi|\hat{B}^2|\psi \rangle.$$
To that end we define the 2nd-order polynomial
$$P(\lambda)=\langle \psi|(\hat{A}-\mathrm{i} \lambda \hat{B})(\hat{A}+\mathrm{i} \lambda \hat{B})|\psi \rangle,$$
which is real, i.e., $P(\lambda) \in \mathbb{R}$ for $\lambda \in \mathbb{R}$. To prove this, note that for $\lambda \in \mathbb{R}$ the operator in the brackets is self-adjoint and thus
$$P(\lambda)=\langle (\hat{A}+\mathrm{i} \lambda \hat{B}) \psi|(\hat{A}+\mathrm{i} \lambda \hat{B}) \psi \rangle, \quad \lambda \in \mathbb{R} \qquad (*).$$
Since the scalar product is positive definite this implies $P(\lambda) \geq 0$ for $\lambda \in \mathbb{R}$. Multiplying out the operator square, you get
$$P(\lambda)=\lambda^2 \Delta B^2 + \Delta A^2 + \lambda \langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi \geq 0, \quad \lambda \in \mathbb{R}.$$
Note that $\mathrm{i} [\hat{A},\hat{B}]$ is self-adjoint and thus all coefficients are real, which again proves that $P(\lambda)$ is a real polynomial.

The discrimant of this quadratic polynomial must thus be negative or 0 (otherwise you'd have 2 real zeros, and the polynomial could not be positive semidefinite). This implies
$$\Delta A^2 \Delta B^2 \geq \frac{1}{4} \langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi\rangle^2$$
or
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \psi|\mathrm{i} [\hat{A},\hat{B}]|\psi\rangle|.$$
This is the general Robertson-uncertainty relation.

It also shows, when the equality sign holds. That's only the case if the discriminant of the quadratic equation is 0, and then the polynomial has a double zero at some real $\lambda$. Thus, due to (*) and the positive definiteness of the scalar product, the equality sign can hold if and only if there is a state $|\psi \rangle$, for which
$$(\hat{A}+\mathrm{i} \lambda \hat{B}) |\psi \rangle=0$$
for some $\lambda \in \mathbb{R}$.

[EDITED on 12/12/14 from here on:]

Let's take position and momentum as an example. This is most easily worked out in the position representation, where $\hat{x}=x$ and $\hat{p}=-\mathrm{i} \partial_x$, acting on the Hilbert space $L^2(\mathbb{R})$. To get the most general case, we look for "minimal-uncertainty states", with general averages for both $x$ and $p$, which we call $\bar{x}$ and $\bar{p}$. Then from (*) and the discussion thereafter we know that the equality sign in the Robertson uncertainty relation is valid, if there exists a real $\lambda$ and normalizable ket $|\psi \rangle$, which fulfills
$$[\lambda (\hat{p}- \mathrm{i} \bar{p} \hat{1})+(\hat{x}-\bar{x} \hat{1})]|\psi \rangle=0.$$
In position representation this reads
$$\lambda(\partial_x - \mathrm{i} \bar{p})+(x-\bar{x}) \psi(x)=0$$
or
$$\partial_x \psi(x)=\left [-\frac{x-\bar{x}}{\lambda} + \mathrm{i} \bar{p} \right ] \psi(x)$$
Separation of variables leads immediately to the solution
$$\psi(x)=A \exp \left [-\frac{(x-\bar{x})^2}{2 \lambda}+\mathrm{i} \bar{p} x \right ].$$
This Gaussian wave packet is a proper state for all $\lambda>0$. It's normalized for
$$A=\left (\frac{1}{\pi \lambda} \right )^{1/4}.$$

 

[atyy]

Hm, I don't see, how you can make the Robertson uncertainty relation stronger, because the equal sign usually can be reached. For the position-momentum uncertainty relation these are the Gaussian wave packets.

I googled and found this answer by Emilio Pisanty on Stack Exchange http://physics.stackexchange.com/questions/30248/what-is-expectation-values-of-this-anti-commutator. It seems the anti-commutator is zero for the usual Gaussian, but there are other minimum uncertainty states in which the anti-commutator is not zero.

 

[Matterwave]

Thanks! haha wow first time I am 'right' here!

Sakurai also says dropping the term makes the relation stronger. That is where I first saw this, and didn't think it possible for this quantum god to be wrong lol.

Can you maybe quote where you read this? It might just be a linguistic issue. The anti-commutator term is sometimes 0 and sometimes not 0, never negative, so in any case, the uncertainty is AT LEAST greater than the commutator term.

 

[Aziza]

Here you go. I uploaded the relevant part of the derivation from Sakurai's Modern QM book, and the eqn 1.4.53 he refers to (which is just the general uncertainty principle). I don't think it is a linguistic issue, though I may be wrong.

You will notice there is a footnote referred to, although this footnote just talks about notation conventions.

 

[thegreenlaser]

Here you go. I uploaded the relevant part of the derivation from Sakurai's Modern QM book, and the eqn 1.4.53 he refers to (which is just the general uncertainty principle). I don't think it is a linguistic issue, though I may be wrong.

You will notice there is a footnote referred to, although this footnote just talks about notation conventions.

From reading the passage, this just seems like a language issue. Given the context, "can only make the inequality stronger" would appear to mean "can only make the inequality more true." As in, if A>B+C, then A is definitely greater than C (or "even more greater than" C), so A>C is a "stronger" inequality in that sense. It's definitely not the wording I would choose, but I think I see where he's coming from. The problem, of course, is that most people would say that A>B+C is stronger than A>C because the first one gives more information about A.

In general, if I see an ambiguous phrase that can be interpreted in more than one way, I tend to go with the one that makes sense to me in terms of math/physics...

 

[Matterwave]

I agree with thegreenlaser. Here by saying "can only make the inequality stronger", Sakurai means "A is at least greater than B because the inclusion of C in this inequality can only make A even larger still".

By "stronger" he doesn't mean "more stringent", he means "can only make the inequality more true".