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?

# [Physics] Adjoint of a Wave Function

quantum mechanics

#### Related Solutions

Let's go step by step:

1) A particle (or a system in general) can be found in a given state $|\psi\rangle$, which mathematically is defined as a vector of an Hilbert state $\mathcal{H}$. This is to ensure that the concept of scalar product (necessary to define ortonormal basis) and norm (necessary to define the probability amplitude) are well defined. When you then decide to expand your wavefunction on the position basis you get something like: $$ |\psi\rangle = \mathbb{1}|\psi\rangle = \int dr |r\rangle\langle r|\psi\rangle = \int dr\, \Psi(r)|r\rangle $$

where I defined $\Psi(r): \mathbb{R}^3\rightarrow\mathbb{C}$ as the wavefunction of the state $|\psi\rangle$ represented at the position $r$. If you want your wavefunction at position $x$, all you have to do is multiply by $\langle x|$ the above expression: $$ \langle x|\psi\rangle = \int dr \,\Psi(r)\langle x|r\rangle = \int dr \,\Psi(r)\delta(x-r) = \Psi(x) $$

and there you go.

2) Suppose now that you want to perform a measurement to characterize your system. Mathematically an operator $\hat O$ is associated to the measurement process, which you suppose has a complete orthonormal set of eigenvalues $\{|\psi_n\rangle \}_{n=0}^\infty$. You can then expand the wavefunction associated to your system as in the formula you provided. Now, since you choose a set of basis functions, what specifies the wavefunction is the coefficients $c_n$ of the expansion. Therefore, if you perturbe the system, your wavefunction will have another set of coefficients $c'_n$ onto the same basis. Since the probability amplitude is defined as the square modulus of the wavefunction, you'll get (exercise!) $$ |\Psi(x,t)|^2 = \sum_n |c_n|^2 $$ therefore a different set of coefficients will give you a different probability distribution.

3) If you are in the ground state of the system, the result of the measurement will be $$ \hat O|\psi\rangle = |\psi_0\rangle $$ every time you perform the measurement (see the Von Neumann's postulate). This means that in the coefficient expansion only $|c_0|=1$, while or the other coefficients are zero. It's really wrong to say that the basis terms must be zero, the basis is just the basis and it's associated to the measurement operator, it's the coefficients that matter for the wavefunction. Also "making the wavefunction depend on one basis" doesn't make much sense. Of course the representation of the wavefunction depends on the basis of choice, but you choose the basis with respect to the measurement you want to perform on it, i.e. with respect to the operator you're considering.

You're final equation is false. It is **not** true that,
$$
\left\langle ... n _i ... \right| a = \sqrt{ n _i } \left\langle ... n _i - 1 ... \right|
$$
The correct result is
$$
\left\langle ... n _i ... \right| a = \left\langle ... n _i + 1 ... \right| \sqrt{ n _i +1}
$$
as you have just proven.

A non-Hermitian operator doesn't act in the same way to the left as it does to the right. The rules of working with bra's and ket's are if you are acting backwards with the operator then you need to use the Hermitian adjoint (see for example Wikipedia). Otherwise you end up with inconsistencies as you did above. More formally, for any operator $ A$ and states $ \left| \alpha \right\rangle $ and $ \left| \beta \right\rangle $ you can write, \begin{equation} \left\langle \alpha \right| A \left| \beta \right\rangle \equiv \left\langle \alpha |A \beta \right\rangle = \langle A ^\dagger \alpha | \beta \rangle \end{equation}

and only for a Hermitian operator can you omit the dagger.

## Best Answer

Recall that a vector is a geometric object and is distinct from the

componentsof a vector which are just numbers.A particular

componentof a vector is given by the contraction of the vector with a basis one-form,or dual vector, to give anumber:$v_x = \vec v \cdot \hat e_x$

The value of the wavefunction at a particular value of

xis a number given by the contraction of the state, aket, with a basisbra:$\psi(x) = \langle x|\psi \rangle$

So, to be clear:

These are abstract objects, they are not numbers.

xdirection" is $\psi(x)$.This is a (complex) number.