Undergrad Conjugate variables: two descriptions (Link?)

[nomadreid]

TL;DR

To describe conjugate variable pairs in order to derive uncertainty principles (not counting the Energy-time one), there appear to be two descriptions: one by their relationship by a Fourier transform (modulo a constant), and another by the lack of commutativity of the corresponding operators. What is the link between these two descriptions?

If I understand correctly (a big caveat), one shows that if one can get from one function to the other via a Fourier transform and multiplication by a constant, then the width of the corresponding Gaussian wave of one gets larger as that of the other gets smaller, and vice-versa, and by a bit of manipulation you have a pair of variables to which an uncertainty principle applies (and which are non-commuting).

Alternatively, if the variables correspond to two non-commuting Hermitian operators, then by direct algebraic manipulation (e.g., the Cauchy-Schwarz inequality) and some clever substitutions, the uncertainty principle results in a straightforward manner.

I am sure that the connection between the two methods is straightforward, but it eludes me. I will be grateful for any indications to set me on the right path.

 

[vanhees71]

The only example of an uncertainty relation which is due to Fourier transformation is the well-known position-momentum uncertainty relation, where of course you have a Fourier transformation in the literal sense between position- and momentum-space wave functions.

For the general uncertainty relation for any two observables you don't need a concrete representation and transformations between different representations but just representation-free Hilbert-space theory:
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
where $\Delta A$ and $\Delta B$ are the standard deviations of the observables $A$ and $B$, represented by their self-adjoint operators $\hat{A}$ and $\hat{B}$ with respect to any (pure or mixed!) state.

 

[nomadreid]

Thanks, vanhees71. I thought that the variables of any two single-variable functions related by Fourier transforms would always obey an uncertainty principle (even though the conjugate pair for the Heisenberg uncertainty principle happens to be position and momentum, and the lower limit might be another number than ħ/2).

Be that as it may, the question remains: if you have two ways of deriving that particular instance, one way by treating the corresponding functions of the Heisenberg uncertainty principle conjugate pair (position , momentum via their relationship through the Fourier transform, and another way as treating the corresponding operators in Hilbert space, can one translate one treatment into the other treatment?

There are links; to take one at random, the expectation value of the momentum basis p is related to the corresponding momentum operator P by ∫^∞_-∞ p|φ(p)|²dp = <φ|P|φ> -- but I do not know how to use such relations to see a clear connection between the two methods, either formally or intuitively.

 

[vanhees71]

Of course, there's only one theory called QM. The relation between the representation-free formulation (Dirac 1926) and the formulation as wave-mechanics is by express all vectors in terms of their "components" wrt. the (generalized) position eigenvectors, i.e.,
$$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle,$$
where I assumed you use the Schrödinger picture of time evolution, i.e., the entire time evolution is with the state ket and the position operator is time independent.

From the Heisenberg algebra for position and momentum [edit: corrected in view of #6]
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \delta_{jk},$$
you can deduce easily that $\hat{p}_j$ is the generator for spatial translations in $j$-direction and thus the momentum operator in position representation is
$$\hat{\vec{p}} \psi(t,\vec{x}):=\langle \vec{x} |\hat{\vec{p}}|\psi(t) \rangle =-\mathrm{i} \vec{\nabla} \psi(t,\vec{x}).$$
From this you get the generalized momentum eigenfunctions to be
$$\vec{\hat{p}} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x})=\langle \vec{x} | \vec{p} \rangle =\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}),$$
where I "normalized" the momentum eigenstates "to a $\delta$ distribution",
$$\langle \vec{p}_1|\vec{p}_2 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}_1}^*(\vec{x}) u_{\vec{p}_2}(\vec{x})=\delta^{(3)}(\vec{p}_1-\vec{p}_2)$$
and the momentum-wave function
$$\tilde{\psi}(t,\vec{p}) = \langle \vec{p}|\psi(t) \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \vec{p}|\vec{x} \rangle \langle \vec{x}|\psi(t) \rangle = \int_{\mathbb{R}} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) \psi(t,\vec{x}),$$
i.e., the momentum wave function is a Fourier transform of the position wave function (and vice versa).

 

[nomadreid]

Thanks very much, vanhees71. I believe that answers my question.

 

[DrClaude]

From the Heisenberg algebra for position and momentum
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i},$$

The last one should read
$$[\hat{x}_j,\hat{p}_j]=\mathrm{i}$$

 

[nomadreid]

Ah, yes. I overlooked that typo.
Thanks, DrClaude.

 

[vanhees71]

The last one should read
$$[\hat{x}_j,\hat{p}_j]=\mathrm{i}$$

My formula should read
$$[\hat{x}_j,\hat{p}_k]=\mathrm{i} \delta_{jk}.$$
I'll correct it in the original posting too. Thanks for pointing out the error!