Inner-Product Conventions

The site uses the standard physics convention for complex inner products:

$$\langle\phi|\psi\rangle \quad\text{is conjugate-linear in }\phi \quad\text{and linear in }\psi.$$

This choice determines where coefficients are conjugated, how bras act, and how the adjoint is written. It does not change convention-independent predictions. A source using the common mathematics convention describes the same Hilbert-space geometry after its arguments are translated consistently.

This page owns the sitewide slot convention and its translation rules. State Vectors owns the physical use of kets and amplitudes; Self-Adjoint Operators owns the full domain theory.

The Physics Slot Convention

For vectors $|\phi\rangle,|\psi\rangle,|\chi\rangle$ and scalars $a,b\in \mathbb C$, linearity in the ket slot means

$$\langle\phi|\bigl(a\psi+b\chi\bigr)\rangle =a\langle\phi|\psi\rangle +b\langle\phi|\chi\rangle.$$

Conjugate-linearity in the bra slot means

$$\langle a\phi+b\chi|\psi\rangle =a^*\langle\phi|\psi\rangle +b^*\langle\chi|\psi\rangle.$$

The remaining inner-product axioms are

$$\langle\phi|\psi\rangle =\langle\psi|\phi\rangle^*, \qquad \langle\psi|\psi\rangle\ge 0,$$

with equality in the second relation only for the zero vector. In particular, $\langle\psi|\psi\rangle$ is real even though a general overlap is complex.

The bra $\langle\phi|$ is therefore a linear functional of the ket on its right. Scaling a ket and then taking its bra gives

$$\bigl(a|\phi\rangle\bigr)^\dagger =a^*\langle\phi|.$$

This conjugation is the source of many sign and phase errors in component calculations.

Coordinates and Continuous Representations

In an orthonormal basis of $\mathbb C^n$, write

$$\phi= \begin{pmatrix} \phi_1\\ \vdots\\ \phi_n \end{pmatrix}, \qquad \psi= \begin{pmatrix} \psi_1\\ \vdots\\ \psi_n \end{pmatrix}.$$

Then

$$\langle\phi|\psi\rangle =\phi^\dagger\psi =\sum_{j=1}^{n}\phi_j^*\psi_j.$$

The formula changes with a nonorthonormal coordinate basis. If the Gram matrix is $G_{jk}=\langle e_j|e_k\rangle$, coordinate columns obey

$$\langle\phi|\psi\rangle =\phi^\dagger G\psi,$$

not generally $\phi^\dagger\psi$. The abstract inner product did not change; only its coordinate representation did.

For wavefunctions in $L^2(X,d\mu)$, the same convention reads

$$\langle\phi|\psi\rangle =\int_X \phi(x)^*\psi(x)\,d\mu(x).$$

The measure is part of the Hilbert-space specification. In spherical coordinates, for example, the radial measure is $r^2dr$ rather than $dr$. Likewise, a sampled wavefunction on a uniform grid approximates the norm by

$$\langle\psi|\psi\rangle \approx \sum_j |\psi(x_j)|^2\,\Delta x,$$

unless the quadrature weights have already been absorbed into the stored array. Complex conjugation and measure weighting are separate requirements.

Norms, Angles, and Transition Amplitudes

The induced norm is

$$\|\psi\| =\sqrt{\langle\psi|\psi\rangle}.$$

A pure-state representative is normalized when $\langle\psi|\psi\rangle=1$. For normalized $|\phi\rangle$ and $|\psi\rangle$, Cauchy–Schwarz gives

$$0\le |\langle\phi|\psi\rangle|^2\le 1.$$

The overlap is a transition amplitude; its squared modulus is invariant under the two slot conventions and under independent global rephasings of the two representatives. Its complex phase can still enter an interference experiment when this amplitude is combined coherently with another amplitude.

For nonzero, not necessarily normalized vectors, the scale-independent transition probability is

$$P(\phi\leftarrow\psi) =\frac{|\langle\phi|\psi\rangle|^2} {\langle\phi|\phi\rangle\langle\psi|\psi\rangle}.$$

The denominator is essential: an inner product alone is not a probability unless the relevant state representatives are normalized.

Bras, Adjoints, and Matrix Elements

For a densely defined operator $A$, the physics convention defines its adjoint through

$$\langle\phi|A\psi\rangle =\langle A^\dagger\phi|\psi\rangle.$$

In finite dimensions, $A^\dagger=(A^*)^T$ in an orthonormal basis. The matrix element between two kets is

$$A_{\phi\psi} =\langle\phi|A|\psi\rangle.$$

It is conjugate-linear in $|\phi\rangle$ and linear in $|\psi\rangle$. For a self-adjoint $A$ and a vector in its domain,

$$\langle\psi|A|\psi\rangle =\langle\psi|A|\psi\rangle^*,$$

so the expectation value is real.

For an unbounded operator, the displayed adjoint identity also determines a domain. More precisely, $|\phi\rangle$ belongs to $D(A^\dagger)$ when there is a vector $|\eta\rangle$ such that

$$\langle\phi|A\psi\rangle =\langle\eta|\psi\rangle \quad\text{for every }|\psi\rangle\in D(A),$$

and then $A^\dagger|\phi\rangle=|\eta\rangle$. A formal integration by parts does not by itself prove $A=A^\dagger$; boundary conditions and domains must also agree.

Translating the Mathematics Convention

Many mathematics texts use an inner product $(\phi,\psi)_{\mathrm m}$ that is linear in the first argument and conjugate-linear in the second. The exact translation is

$$\boxed{ \langle\phi|\psi\rangle_{\mathrm p} =(\psi,\phi)_{\mathrm m} }$$

or, equivalently,

$$(\phi,\psi)_{\mathrm m} =\langle\psi|\phi\rangle_{\mathrm p} =\langle\phi|\psi\rangle_{\mathrm p}^*.$$

The subscripts label conventions, not different physical inner products. After reversing the arguments, both descriptions give the same norm:

$$(\psi,\psi)_{\mathrm m} =\langle\psi|\psi\rangle_{\mathrm p}.$$

The adjoint operator is also the same geometric object. A mathematics text may write

$$(A\psi,\phi)_{\mathrm m} =(\psi,A^*\phi)_{\mathrm m}.$$

Reversing the arguments gives

$$\langle\phi|A\psi\rangle_{\mathrm p} =\langle A^*\phi|\psi\rangle_{\mathrm p},$$

so that text’s $A^*$ is the operator denoted $A^\dagger$ here. Do not confuse the operator symbol $A^*$ in that convention with entrywise complex conjugation of a matrix.

Worked Two-Component Check

Take

$$|\phi\rangle =\frac{1}{\sqrt2} \begin{pmatrix}1\\i\end{pmatrix}, \qquad |\psi\rangle =\frac{1}{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix}.$$

The physics overlap is

$$\langle\phi|\psi\rangle =\frac12 \begin{pmatrix}1&-i\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix} =\frac{1-i}{2}.$$

Reversing the order gives the conjugate,

$$\langle\psi|\phi\rangle =\frac{1+i}{2}.$$

Now let

$$A=\sigma_y =\begin{pmatrix}0&-i\\i&0\end{pmatrix}.$$

Because $A^\dagger=A$ and $A|\phi\rangle=|\phi\rangle$,

$$\langle\phi|A\psi\rangle =\frac{1-i}{2} =\langle A^\dagger\phi|\psi\rangle.$$

This cross-matrix-element calculation checks the conjugated bra and the adjoint placement without changing conventions midway. It is not a reality test: a self-adjoint operator can have complex off-diagonal matrix elements. The corresponding diagonal expectation value is

$$\langle\phi|A\phi\rangle =\langle\phi|\phi\rangle =1\in\mathbb R,$$

which is the appropriate reality check for this normalized state.

Convention-Invariant Audit

When importing an inner-product formula, check the following in order:

1. Which slot is linear?
2. Does the source’s adjoint symbol mean operator adjoint or entrywise complex conjugation?
3. Is the coordinate basis orthonormal, or is a Gram matrix required?
4. Which measure or quadrature weights define the continuous or numerical inner product?
5. Are all vectors in the domains required by the displayed operator matrix elements?
6. After translation, do norms, absolute transition amplitudes, expectation values, and spectra agree?

A disagreement that survives these invariant checks may be physical. A disagreement that disappears after argument reversal or measure restoration was conventional.

Common Mistakes

Forgetting the bra-slot conjugation. Pulling a scalar out of a bra changes $a$ to $a^*$. This is why a ket column becomes a conjugate-transposed row.

Equating reversed overlaps. In general $\langle\phi|\psi\rangle\ne\langle\psi|\phi\rangle$; they are complex conjugates. They coincide only when the overlap happens to be real.

Normalizing by a component sum. Normalization uses $\sum_j|\psi_j|^2$ in an orthonormal finite basis, or the corresponding measure-weighted expression. The sum $\sum_j\psi_j$ has no general normalization meaning.

Dropping the coordinate metric. In a nonorthonormal basis, the inner product is $\phi^\dagger G\psi$. Treating it as $\phi^\dagger\psi$ changes the geometry rather than merely changing notation.

Using a formal differential adjoint as an operator adjoint. Integration by parts identifies a candidate differential expression. Self-adjointness also requires equality of domains.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, Chapters I–II.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, Chapters 2–3.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised ed., Academic Press, 1980, Chapters II and VIII.

Exercises

1. Slot linearity. Let $|\phi\rangle=(1,i)^T$ and $|\psi\rangle=(2,1)^T$. Compute $\langle\phi|\psi\rangle$ and $\langle\psi|\phi\rangle$.

Solution

Using the physics convention,

$$\langle\phi|\psi\rangle =1^*2+i^*1 =2-i,$$

whereas

$$\langle\psi|\phi\rangle =2^*1+1^*i =2+i =\langle\phi|\psi\rangle^*.$$

2. A nonorthonormal basis. Suppose

   $$G=\begin{pmatrix}1&s\\s^*&1\end{pmatrix}, \qquad |s|<1.$$

   For coordinate columns $\phi=(1,0)^T$ and $\psi=(a,b)^T$, compute $\langle\phi|\psi\rangle$.

Solution

The coordinate rule is $\phi^\dagger G\psi$, so

$$\langle\phi|\psi\rangle =\begin{pmatrix}1&0\end{pmatrix} \begin{pmatrix}1&s\\s^*&1\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} =a+sb.$$

The overlap is not generally the first coordinate $a$ because the basis is not orthonormal.

3. Translate conventions. A mathematics text, linear in its first slot, reports $(\phi,\psi)_{\mathrm m}=3+2i$. Find $\langle\phi|\psi\rangle_{\mathrm p}$ and $\langle\psi|\phi\rangle_{\mathrm p}$.

Solution

The translation gives

$$\langle\psi|\phi\rangle_{\mathrm p} =(\phi,\psi)_{\mathrm m} =3+2i.$$

Therefore

$$\langle\phi|\psi\rangle_{\mathrm p} =3-2i.$$

Both conventions assign the same absolute overlap.

4. Measure-weighted normalization. Samples of a wavefunction on a uniform grid have values $(1,2i,1)$ and spacing $\Delta x=0.2$. Find the scale factor $N>0$ such that $N(1,2i,1)$ is normalized by the rectangular quadrature rule.

Solution

The unscaled discrete approximation is

$$\sum_j|\psi_j|^2\Delta x =(1+4+1)(0.2) =1.2.$$

Thus $N^2(1.2)=1$, so

$$N=\frac{1}{\sqrt{1.2}} =\sqrt{\frac56}.$$