Notation Translations

Translating notation requires preserving the object and its action, not replacing one symbol by another. The same bracket can use opposite linearity conventions, the same Fourier transform can use momentum or wave number, and the same “generator” can differ by a sign and a factor of $\hbar$. A short test on a vector or a probability is often more reliable than visual resemblance between formulas.

Helpful background. Operator Theory Minimum supplies the distinction between an operator, its domain, and its adjoint. This page develops a translation procedure; the formalism crosswalk gives the broader map of physical and mathematical objects.

Translate an inner product by testing a scalar

The convention used here is

$$\langle u,v\rangle=\sum_j\overline{u_j}v_j,$$

linear in the second slot. Some mathematics texts use a bracket $(u,v)_{\rm m}$ linear in the first slot. To describe the same Hilbert geometry, set

$$(u,v)_{\rm m}=\langle v,u\rangle.$$

Test the convention with $iu$: $\langle iu,v\rangle=-i\langle u,v\rangle$, while $(iu,v)_{\rm m}=i(u,v)_{\rm m}$. This catches conjugation errors before they propagate into amplitudes.

For example, the Dirac rank-one operator $|u\rangle\langle v|$ acts as

$$w\longmapsto u\,\langle v,w\rangle =u\,(w,v)_{\rm m}.$$

Replacing the last bracket by $(v,w)_{\rm m}$ would make the map antilinear in $w$. That would change its mathematical type, not merely its notation.

The adjoint is characterized in the present convention by $\langle u,Av\rangle=\langle A^*u,v\rangle$ on the relevant domains. The same adjoint may be written $A^\dagger$ elsewhere. For an unbounded operator, matching the symbol does not establish matching adjoint domains.

Check a translated formula on a rank-one input

Suppose a source represents operators as vectors by stacking matrix columns. Denote this convention by $\operatorname{vec}$. A frequently used identity is

$$\operatorname{vec}(AXB^*) =(\overline B\otimes A)\operatorname{vec}(X).$$

The conjugation on $B$ is essential. To check it, use $X=|u\rangle\langle v|=uv^*$, for which

$$\operatorname{vec}(uv^*)=\overline v\otimes u.$$

The transformed matrix is $(Au)(Bv)^*$, whose vectorization is $\overline{Bv}\otimes Au$. This is precisely the stated tensor-product action. Rank-one matrices span the finite-dimensional matrix space, so the check establishes the identity there.

A source using row stacking will arrange the factors differently. A source using a conjugate Hilbert space may encode the same conjugation structurally instead of writing a bar. Neither formula should be imported into a calculation until the vectorization and tensor-factor order have been fixed.

Restore normalization when translating momentum and wave number

On the line, the momentum-normalized transform is

$$\widetilde\psi(p) =\frac1{\sqrt{2\pi\hbar}} \int_{\mathbb R}e^{-ipx/\hbar}\psi(x)\,dx,$$

initially for sufficiently regular integrable functions and extended unitarily to $L^2$. A wave-number convention instead uses

$$\widehat\psi(k) =\frac1{\sqrt{2\pi}} \int_{\mathbb R}e^{-ikx}\psi(x)\,dx.$$

Setting $p=\hbar k$ changes the measure as well as the exponential. The correct relation is

$$\widetilde\psi(p)=\hbar^{-1/2}\widehat\psi(p/\hbar), \qquad |\widetilde\psi(p)|^2\,dp =|\widehat\psi(k)|^2\,dk.$$

Thus normalization and probabilities survive the change. Merely substituting $k=p/\hbar$ without the prefactor does not preserve the state norm. The same warning applies to delta-normalized generalized eigenvectors: their normalization depends on the measure used to label them.

The rigorous meaning of continuous bras and kets belongs to Continuous-Basis Notation. A symbol such as $\langle x|\psi\rangle$ need not mean an inner product with a Hilbert-space vector.

Translate a generator together with its parameter

Stone’s theorem is often written for $U(s)=e^{isA}$, whereas physical dynamics uses $U(t)=e^{-itH/\hbar}$. If $s=t$, then $A=-H/\hbar$. If the author instead uses $s=-t/\hbar$, the generator is $H$. The parameter and its units determine which statement is being made.

Check the derivative on its proper domain:

$$\left.\frac{d}{dt}U(t)\psi\right|_{t=0} =-\frac{i}{\hbar}H\psi, \qquad \psi\in D(H).$$

This one line verifies the sign, units, and domain simultaneously. It also prevents a change of convention from being mistaken for a different dynamical law. Stone’s Theorem supplies the underlying equivalence between continuity and self-adjoint generation.

Exercise: translate a transition amplitude

Let $u=(1,i)/\sqrt2$ and $v=(1,1)/\sqrt2$. Compute $\langle u,v\rangle$ in the second-slot-linear convention and identify the first-slot-linear expression representing the same amplitude. Check the transition probability.

Solution

The amplitude is $\langle u,v\rangle=(1-i)/2$. In the other convention the same amplitude is $(v,u)_{\rm m}$, while $(u,v)_{\rm m}=(1+i)/2$ is its conjugate. Both conjugate amplitudes have squared modulus $1/2$, so a probability-only check can miss a conjugation error. An interference calculation or a linearity test can expose it.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — Hilbert-space conventions, Fourier transforms, and generators.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, Chapter 1 — operator-vector correspondence, adjoints, and tensor-product identities.