# Is this statement of the Heisenberg uncertainty principle correct? [duplicate]

Is this correct?

The Heisenberg Uncertainty Principle is not a Principle or law of physics. It's one of many results you can work out from quantum theory with some math:

-- In quantum theory observables can be represented by Hermitian matrices.

-- If an observable of a system can be represented by a particular matrix at a particular instant, then all matrices of the same dimension represent observables of that system.

-- In a state specified by the vector |psi>, an observable X is sharp if and only if X|psi> = x|psi> for some real number x. In which case x is an eigenvalue of X and |psi> is an eigenvector of X.

Now let Y be any matrix that does not have |psi> among its eigenvectors. (For any vector, there exists an infinity of such matrices.)

If the actual state is |psi>, the observable Y cannot be sharp. (Because of the 'if and only if' above.)

EDIT: I'm trying to find out if this line of reasoning is correct. Those other questions don't address this.

• Possible duplicate of physics.stackexchange.com/q/35516, physics.stackexchange.com/q/100966, Jun 12, 2016 at 4:45
• Hi Curi, is this from Deutsch's book? I'm not familiar with the term sharp observable, though I guess it's reasonably clear from the context what is meant. Jun 12, 2016 at 4:51
• It is simpler and more accurately derived from the commutators of quantum mechanical operators . see for example sjsu.edu/faculty/watkins/Uncertainty.htm . It is a consequence of the quantum mechanical underlying framework, an envelope. Jun 12, 2016 at 4:57
• No this isn't in Deutsch's books. This is my understanding of the matter which I'd like to check. I understand sharp observables to be ones with the same value in all universes. I don't think the word is important to the point. Another explanation of sharp: philsci-archive.pitt.edu/4109/1/ObservableBuschLahti.pdf "Observables represented by positive operator measures which are not projection valued are referred to as generalized observables, or unsharp observables, while spectral measures and generally all projection valued measures are called standard, or sharp observables."
– curi
Jun 12, 2016 at 4:58
• well, are you getting the inequality quantitatively with your arguments? It is not that they are wrong, but do not give the exact relationships Jun 12, 2016 at 5:00

Fix a state $\left|\psi\right\rangle$. In this state, any linear operator $\hat{\mathcal{O}}$ on the Hilbert space (i.e. square matrix) has mean $\left\langle\hat{\mathcal{O}}\right\rangle:=\left\langle\psi\left|\hat{\mathcal{O}}\right|\psi\right\rangle$. Let $\hat{A},\,\hat{B}$ denote two operators for which $\hat{A},\,\hat{B},\,\hat{A}^2,\,\hat{B}^2$ have finite means in our chosen state. Then $\hat{A}$ has variance $\sigma_A^2=\left\langle\hat{A}^2\right\rangle-\left\langle\hat{A}\right\rangle^2$, and similarly with $B$.
The Cauchy-Schwarz inequality on the Hilbert space implies the Schrödinger uncertainty relation $$\sigma_A^2\sigma_B^2\geq\left|\frac{1}{2}\left\langle\left[\hat{A},\,\hat{B}\right]_+\right\rangle-\left\langle\hat{A}\right\rangle\left\langle\hat{B}\right\rangle\right|^2+\left|\frac{1}{2}\left\langle\left[\hat{A},\,\hat{B}\right]_-\right\rangle\right|^2$$(proven here), where $\left[\hat{A},\,\hat{B}\right]_\pm:=\hat{A}\hat{B}\pm\hat{B}\hat{A}$. From this we obtain the Robertson uncertainty relation $$\sigma_A\sigma_B\geq\left|\frac{1}{2}\left\langle\left[\hat{A},\,\hat{B}\right]_-\right\rangle\right|.$$
(Note: we call $\left[\hat{A},\,\hat{B}\right]_-$ a commutator, and may denote it $\left[\hat{A},\,\hat{B}\right]$. Similarly, we call $\left[\hat{A},\,\hat{B}\right]_+$ an anticommutator, and may denote it $\left\{\hat{A},\,\hat{B}\right\}$.)
If $\left[\hat{A},\,\hat{B}\right]_-$ is a nonzero multiple of the identity operator, we call $A,\,B$ conjugate variables, as happens for example with $A=x,\,B=p_x$. (I've not placed hats on $A,\,B$ here, because "they are conjugate variables" is actually a statement about the classical observables. You may wish to read how it's defined in terms of Lagrangian mechanics, Poisson brackets etc.) Conjugate variables have a positive lower bound on $\sigma_A\sigma_B$, so a state for which $\hat{A}$ has low variance is one for which $\hat{B}$ has high variance.