# Adjoint of a Wave Function

Why is the adjoint of a function simply it's complex conjugate? Normally with a vector we consider the adjoint to be the transpose (And the conjugate? I don't know why), so does this concept carry through to these functions? Should I imagine the conjugate of a wave function to actually be a column vector? Further what does it mean for a function to be a row vector, versus a column vector. Do they live in a completely separate space? Are our operators always square?

• The product of a function with its complex conjugate is real-valued and non-negative, both of which are properties needed to define a norm. – David H Oct 29 '13 at 0:00
• – Incnis Mrsi Oct 26 '14 at 9:12
• – Incnis Mrsi Nov 22 '14 at 14:28

I'll just first add to Alfred's answer and also point out that in more generalized discussions (theory of distributions and rigged Hilbert spaces) the notion of adjoint is broadened so that in rigged Hilbert space the set of bras is strictly bigger than the set of kets (see the end of my answer).

Further to Alfred's answer, you can think of the function $\psi(x)$ as a continuous generalization of the following process. Imagine you were doing your calculations numerically, and that you were storing the values of your function sampled at discrete points over some interval of interest to your simulation. You could think of your function as a column vector $\Psi$. To form the inner product between two such functions $\psi_1$, $\psi_2$, discretized to column vectors $\Psi_1$, $\Psi_2$, you would do the vector operation $\left<\psi_1, \psi_2\right> = \Psi_1^\dagger \Psi_2$. So there's your transpose. In computer code you would simply contract the indices (the two functions would have the same shape in memory - you wouldn't make "columns" and "rows" different in memory), so hence you can see that the inner product between two functions becomes $\int \psi_1^*(x) \psi_2(x) dx$.

Also, you probably shouldn't think of the transpose as being a fundamental part of the concept of "adjoint". The transpose in the inner product expressed in matrix notation $\left<\psi_1, \psi_2\right> = \Psi_1^\dagger \Psi_2 = {\Psi_1^*}^T \Psi_2= {\Psi_1^T}^* \Psi_2$only arises when we use matrix notation owing to the familiar "rows on left by columns on right" rule in the matrix product. This rule itself is "only" a convention that is adopted to simplify interpretation of matrices (i.e. with their columns defined as the images of the basis vectors under the transformation the matrix stands for) and how we combine matrices to represent composition of the linear transformations they stand for. The transpose is not an essential part of forming the inner product, which is the essence of the idea of "adjoint" and if $T:\mathbf{H}\to\mathbf{H}$ is an operator mapping some vector space $\mathbf{H}$ to itself, the "adjoint" $T^\dagger$ to $T$ is the unique one such that $\left<T^\dagger X,Y\right> = \left<X, T Y\right>, \;\forall X, Y \in \mathbf{H}$. In the case of matrices representing linear operators on finite dimensional vector spaces, this comes to mean complex conjugate transpose with the wonted matrix multiplication product.

To expand on Alfred's comments on one-forms and duals. This is actually why physicist bang on about "Hilbert spaces". A one form is simply a linear functional $\mathcal{L}:\mathbf{H}\to\mathbb{C}$ mapping a function or vector $\psi(x)$, $\Psi$ in some linear space $\mathbf{H}$ (a linear space of e.g. of complex valued functions $f:\mathbb{R}\to\mathbb{C}$ or column vectors), to a complex number. For example, if we consider a linear space of functions, the linear functional could be $\psi(x)\to\int_{-\infty}^\infty \psi_0(u)^* \psi(u)\mathrm{d}u$ for some test function $\psi_0(x)$, or for vectors in a discrete space it could be $\Psi\to \Psi_0^\dagger \Psi$. The point about a Hilbert space is that we have an inner product defined and that:

ALL continuous linear functionals can be represented by an inner product. Whatever your continuous linear functional $\mathcal{L}$, you can always find a test function/vector $\psi_\mathcal{L}$ such that $\mathcal{L}\psi(x) = \left<\psi_\mathcal{L},\psi\right> = \int \psi_\mathcal{L}^*\psi du$.

So in a Hilbert space, vectors and functions always double up as one-forms if you define the appropriate linear functional with a given function / vector as the kernel of the functional. Abstractly, one can think of functions/vectors and one-forms as the "same" through a one-to-one, onto correspondence. So a Hilbert space is, through this abstract correspondence, "the same" as its space of continuous (sometimes "topological") duals.

Now you may know a Hilbert space is more standardly defined as an inner product space wherein all Cauchy sequences converge in the norm defined by the inner product to a point in the space, i.e. the space is complete with respect to the metric defined by the inner product.

This is indeed the right same notion as the one I have just talked about, the self-dualhood being the one often more workable and usable by physicists. The proof that these two notions are logically equivalent is the Riesz Representation Theorem and you should probably read through the Wikipedia page on this topic to get a deeper insight into what's going on here.

However, inevitably one reaches the fairly obvious question about including such things as the Dirac delta in your space of functionals: we want some bras that cannot be represented as Hilbert space members. Here is where the notion of Rigged Hilbert Space comes in - the ingenious process where we kit a dense subset $S\subset H$ of the original Hilbert space $H$ ("rig it") with a stronger topology, so that things like the Dirac delta are included in the topological dual space $S^*$ where $S\subset H\subset S^*$. Good references for this notion are https://physics.stackexchange.com/a/43519/26076 and also the discussions at https://mathoverflow.net/q/43313. In the latter, Todd Trimble's suspicions that that the usual Gel'Fand triple is $S\subset H = \mathbf{L}^2(\mathbb{R}^N)\subset S^*$ with $S$ , $S^∗$ being the Schwartz space and its topological dual, respectively. (The topological dual being wrt to the stronger topology induced by a family of norms, NOT the $\mathbf{L}^2$ norm of the original Hilbert space). $S^∗$ is the space of tempered distributions as discussed in my answer here. One definition of Hilbert space is that it is isomorphic to the set of its continuous linear functionals (its topological dual): it does not include useful things like the Dirac delta which, although a linear functional, is not continuous in the Hilbert space norm. So we kit a dense subspace of $H$ out with a strong topology to ferret out all the useful distributions in $S^∗$. By "ferret out", of course, I'm saying that the topology is strong enough to make things like the Dirac delta into a continuous functional, whereas it isn't continuous with respect to the Hilbert space norm. The Wikipedia article is a little light on here: there's a great deal of detail about nuclear spaces that's glossed over so at the first reading I'd suggest you should take a specific example $S$ = Schwartz space and $S^∗$ = Tempered Distributions and keep this relatively simple (and, for QM most relevant) example exclusively in mind - for QM you won't need anything else. So the set of bras is strictly bigger than the set of kets in this view.

• Ī do not think one should be introduced to Gelfand’t advances in mathematical formalism before learning ordinary Hilbert spaces. – Incnis Mrsi Oct 26 '14 at 9:18
• @IncnisMrsi I agree; it's always hard to know what to put in an answer. But my experience was that as soon as I learnt about Hilbert spaces, I wanted to know about delta functions - clearly linear functionals, but with no Riesz representation. It took a while for the difference between topological and algebraic dual to dawn on me, especially as there is no difference in finite dimensional spaces. I would have liked for the references I read at the time to state this clearly and to say something like "and, when you're ready, we use the notion of rigged Hilbert space for our descriptions". – WetSavannaAnimal Oct 26 '14 at 9:23

Further what does it mean for a function to be a row vector, versus a column vector

Recall that a vector is a geometric object and is distinct from the components of a vector which are just numbers.

A particular component of a vector is given by the contraction of the vector with a basis one-form,or dual vector, to give a number:

$v_x = \vec v \cdot \hat e_x$

The value of the wavefunction at a particular value of x is a number given by the contraction of the state, a ket, with a basis bra:

$\psi(x) = \langle x|\psi \rangle$

So, to be clear:

• the "column vector" ket is $|\psi\rangle$
• the "row vector" bra is $\langle\psi|$

These are abstract objects, they are not numbers.

• the "component" of $|\psi\rangle$ "in the x direction" is $\psi(x)$.

This is a (complex) number.