My question is: Is there any general mechanism to find the adjoint of an operator without using any basis? For example, how do we find the adjoint of $\widehat{x}$ or $\widehat{p}$ operator, without using any basis? Also, kindly tell the same for the operator $a=\widehat{x}-\iota\widehat{p}$.

• Have you looked at the definition of the adjoint of an operator? – SRS May 22 '18 at 4:01
• @SRS Yes. For the cases when we express the operators in terms of matrices (finite or infinite), the adjoint is defined as the transpose of complex conjugate of the given matrix. But, if we haven't expressed it in the form of a matrix, then how do we take the adjoint of the operator, that is my question. – quirkyquark May 22 '18 at 4:15
• Also, as we know in Q-Mech, $\widehat{x}$ and $\widehat{p}$ are the two basic operators, once we find that they are self-adjoint, how do we find the adjint of operators like $a$, as I have mentioned in the question? – quirkyquark May 22 '18 at 4:19
• You have partially answered your own question by stating the relationship $\hat{x}$ and $\hat{p}$ have to their adjoints. Now think about what the adjoint of an object like $i\hat{p}$ is – Triatticus May 22 '18 at 4:51

One should use the definition of adjoint operator, and the knowledge about the action of the operator itself.

Let $\mathscr{H}$ be a (separable) Hilbert space with scalar product $\langle\,\cdot\,,\,\cdot\,\rangle$. Given a densely defined operator $A$ with (dense) domain $D(A)\subseteq \mathscr{H}$, its adjoint $A^*$ is the (closed) operator with domain

$$D(A^*)=\bigl\{\varphi\in\mathscr{H}, \exists \eta\in\mathscr{H}, \forall \psi\in D(A), \langle \psi,\eta\rangle=\langle A\psi,\varphi\rangle\bigr\}\;,$$

and action on $\varphi\in D(A^*)$ defined by $A^*\varphi=\eta$, where $\eta$ is the same vector appearing above in the definition of the domain. In other words, the action of the adjoint $A^*$ is defined, on its domain of definition, by

$$\langle \psi, A^*\varphi\rangle=\langle A\psi,\varphi\rangle\; .$$

Once you have an explicit form of the operator, it is not difficult to find the adjoint applying the definition. For example, let $\mathscr{H}=L^2(\mathbb{R})$, and let $A=\hat{x}$ (multiplication by $x$), with domain the rapidly decreasing test functions $\mathscr{S}(\mathbb{R})$ and action: $$\langle \hat{x}\psi,\varphi\rangle_2= \int_{\mathbb{R}} x\bar{\psi}(x)\varphi(x)\mathrm{d}x\; .$$ Clearly, the adjoint action of $\hat{x}^*\lvert_{\mathscr{S}}=\hat{x}$, however the domain of the adjoint is larger in this case, $D(\hat{x}^*)\supset \mathscr{S}(\mathbb{R}^d)$, with the inclusion being strict. The case of the momentum operator $\hat{p}=-i\partial_x$ (with the same domain of definition) is rather similar, but it involves an integration by parts. The domain of $\hat{p}^*$ in this case is a well-known functional space, the non-homogeneous Sobolev space $H^1(\mathbb{R})\subset L^2(\mathbb{R})$. In fact, if we start from the operator $\hat{p}$ but with domain of definition $D(\hat{p})=H^1(\mathbb{R})$, we find that $D(\hat{p}^*)=H^1(\mathbb{R})$ and $\hat{p}^*=\hat{p}$. In this case the operator is self-adjoint (while when it is defined only on $\mathscr{S}$ is just symmetric, that means $D(\hat{p})\subset D(\hat{p}^*)$ and $\hat{p}^*\lvert_{D(\hat{p})}=\hat{p}$). The above position operator is only symmetric as well, but it also has a unique self-adjoint extension given precisely by $\hat{x}^*$. There are however densely defined symmetric operators that admit more than one self-adjoint extension, or none at all.

To find the adjoint action of linear combinations of operators is formally easy if you know the adjoint of the components, as it is seen from the definition; whether the formal action is valid on some domain is however much trickier in general (it could be only true for the vector $0$). In particular, let $\{A_j\}_{j=1}^N$, $D=\bigcap_{j=1}^n D(A_j)$, and $A=\sum_{j=1}^N z_j A_j$ on $D$ (the $z_j$ are complx numbers). Then the adjoint action is, on $D'=\bigcap_{j=1}^n D(A_j^*)$, $$A^*\lvert_{D'}=\sum_{j=1}^N \bar{z}_j A^*_j\; .$$ In the explicit case the OP mentions, it is true that $a^*\lvert_{\mathscr{S}}= \hat{x}+i\hat{p}$ (it is actually true on a slightly more general domain), however $D(a^*)\supset \mathscr{S}$ and there may be vectors $\psi$ for which it is not possible to write $a^*\psi=(\hat{x}+i\hat{p})\psi$, for the right hand side could fail to make sense (but the left hand side still does if we use the definition of adjoint). Unbounded operators are tricky!

Let me remark that, aside from the definition, abstractly only few information can be given about the adjoint if we do not have any information about the explicit action of the operator on the given Hilbert space. One such information is, e.g., that the adjoint $A^*$ is always a closed operator if the original operator is densely defined (but possibly not closed itself), and that in this case if also $D(A^*)$ is dense then $A^{**}$ is the closure of $A$, i.e. the smallest closed operator containing $A$.

Suppose $\mathcal{H}$ is a Hilbert space, and suppose $A : \mathcal{D}(A)\subset \mathcal{H}\rightarrow\mathcal{H}$ is a linear operator on a dense subspace $\mathcal{D}(A)$ of $\mathcal{H}$. Then $y\in\mathcal{D}(A^*)$ iff there exists a constant $M$ such that $$|\langle Ax,y\rangle| \le M\|x\|,\;\; \forall x\in\mathcal{D}(A).$$ In this case, $\phi_y(x)=\langle Ax,y\rangle$ has a unique extension to a bounded linear functional on $\mathcal{H}$, which, by the Riesz representation theorem, gives the existence of a unique vector $A^*y$ such that $\phi_y(x) = \langle x,A^*y\rangle$. So $\langle Ax,y\rangle = \langle x,A^*y\rangle$ for all $x\in\mathcal{D}(A)$.

In particular, you know that $x$ and $p$ are self adjoint, meaning that $x = x^{\dagger}$ and $p = p^{\dagger}$.

If you wanted to calculate the adjoint of an operator like $a$, you would need to complex-conjugate and transpose everything in it. You know how to do that for $x$ and $p$, so what about the $-i$ in there? It gets complex-conjugated only (you can't transpose it because it's a number!).

So $a^{\dagger} = x^{\dagger} + ip^{\dagger} = x + ip$. Note that $a$ is not self-adjoint, because $a \neq a^{\dagger}$.

Is there any general mechanism to find the adjoint of an operator without using any basis?

If $A$ is an operator in a Hilbert space, $A^{\dagger}$ makes sense even when you are not talking about a matrix explicitly, because the $\dagger$ operation is something you perform on operators in general. Matrices are just representations of operators in a certain basis so the $\dagger$ operation takes a certain form to operate on them (as noted, it is complex-conjugating and then transposing).

For operator objects in general, $A^{\dagger}$ means that the operator must act to the left on a bra like $⟨ψ|A^†$ (as opposed to $A$ that must act to the right on kets like $A|ψ⟩$).

To know what form $A^{\dagger}$ will take explicitly you need to know what effect it has on a set of chosen states, meaning you need to choose a basis.

• I get what u say. Just one more thing. How do we find the adjoint of $\hat{x}$ and $\hat{p}$? – quirkyquark May 22 '18 at 7:01
• Let |n⟩ be a momentum eigenstate, and $\hat{p}|n⟩ = p_n |n⟩$ with $p_n$ real. Then, $⟨n|p|n⟩ = p_n = p_{n}^{*} = (⟨n|p|n⟩)^{\dagger} = ⟨n|p^{\dagger}|n⟩$. It follows that $p = p^{\dagger}$. This is a way to get the result you wanted. The fact that the eigenvalues of $x$ and $p$ are real forces them to be self-adjoint. – Luke May 22 '18 at 7:15
• Remember that $p_{n}^{*}=(⟨n|p|n⟩)^{\dagger}$ because $⟨n|p|n⟩$ is just a number, so $⟨n|p|n⟩^* = ⟨n|p|n⟩^{\dagger}$. – Luke May 22 '18 at 7:23

As you may already be aware, the Hilbert space is a linear vector space. The wave function or the quantum states are represented as a vector in the ket-space, or a column vector.

For convenience and to define an inner product, the Hilbert space is extended to a dual-space a.k.a the bra-space, or the complex-conjugate transpose of the column vector. For a state $|\psi\rangle$ in the ket-space, we have a state $\langle \psi |$ in the dual space so that $$\langle \psi | \psi \rangle = ||\psi|| ,$$ where $||...||$ represents the norm of the vector.

For any quantum state, irrespective of the basis used to represent this, the action of an operator $\hat{A}$ on the state $|\psi\rangle$ can be written as: $$|\psi'\rangle = \hat{A} |\psi \rangle.$$ The adjoint of the operator $\hat{A}$ is defined as an operator $\hat{A}^\dagger$ such that $$\langle \psi'| = \langle \psi | \hat{A}^\dagger ,$$ that is to say that $\hat{A}^\dagger$ is the dual of $\hat{A}$ in the same sense as the bra state $\langle \psi|$ is the dual of the ket state $|\psi\rangle$.

Mathematically, for obvious reasons, this adjoint turns out to be the complex-conjugate transpose. However, if you were to define the dual space in a different manner, the mathematical definition of adjoint would have to be altered accordingly.