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Hi I have a question that might be a bit trivial. I have just completed learning a section on the bra-ket notation. There is a statement that the following is prohibited $$\hat{A}\langle\psi|, ~|\psi\rangle |\phi\rangle,~\langle \psi| \langle\phi|, |\psi \rangle \hat{A}.$$ I understand that this can be seen when dealing with the finite-dimensional case where we can consider $\hat{A}$ as a matrix and the above then reduces to cases which are prohibited in matrix multiplication. In a set of notes I am using it proves the conservation of probability (conservation of normalization) by first stating that

$$\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)| \Psi(t) \rangle = \left(\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)|\right)|\Psi(t) \rangle + \langle \Psi(t)|\left(\frac{\mathrm d | \Psi(t) \rangle}{\mathrm dt}\right)$$

It seems that $\frac{\mathrm d}{\mathrm dt}$ does not satisfy the definition of an operator used in the context above so we can have that $\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)|$ on the right-hand side makes sense. Would I be right in saying that this is because $\frac{\mathrm d}{\mathrm dt}$ does not satisfy the operator definition in QM which is a mapping that simply takes a ket to a ket and a bra to a bra? I know that the inner-product result used (even though I thought that this was only for finite dimensional inner product spaces) is $$\frac{\mathrm d}{\mathrm dt}\langle f,g \rangle = \langle f(t), g'(t) \rangle + \langle f'(t),g(t) \rangle.$$

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  • $\begingroup$ I've found bra-ket notation is not uniform across texts/authors. What source are you using? $\endgroup$ – anon01 Jun 3 '16 at 13:16
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Bra-ket notation is just a useful short hand for some well-defined objects in functional analysis (or linear algebra if you work in finite dimensions).

To understand what is allowed and what isn't, you would better know what those concepts are, so let's quickly recap:

  • A ket $|\psi\rangle$ is just a vector in some Hilbert space $\mathcal{H}$.
  • A bra $\langle \psi |$ is an object in the dual space of that Hilbert space (usually denoted $\mathcal{H}^*$), which is the space of all linear functionals on the Hilbert space. This implies that $\langle \psi |$ is a linear function, where the variable runs over all possible $|\phi\rangle\in \mathcal{H}$ and the result is a number, i.e. $\langle \psi|$ is a function $\mathcal{H}\to \mathbb{C}$.
  • The dual space of a Hilbert space is itself a vector space, hence it makes sense to add bras and multiply them with scalars.
  • Now, given the scalar product of the Hilbert space $\langle \cdot, \cdot \rangle$, you can easily construct such functionals: For every $|\phi\rangle\in \mathcal{H}$, the map $|\psi\rangle \mapsto \langle |\phi\rangle,|\psi\rangle\rangle$ is a linear functional. Identifying this linear functional with $\langle \phi |$ lets you write $\langle \phi |\psi \rangle$, which hints at the fact that you used the scalar product.
  • According to some theorem, that's all there is to the dual space, hence the bra-ket notation captures the Hilbert space, its dual and the natural duality between the two (the inner product).
  • An operator is now a linear function $A:\mathcal{H}\to \mathcal{H}$. This means that it takes vectors to vectors.
  • This explains why what you write as "prohibited" is at first glance ill-defined. However, beware: some of the "prohibited" notions are actually all over other texts. For instance, $|\psi\rangle|\phi\rangle$ is often used as a shorthand for $|\psi\rangle \otimes |\phi\rangle$, but clearly, this has to be defined!

Now back to your question. The main problem is, as you say, that the differentiation is not an operator in the sense above. Why? Because it doesn't really act on your Hilbert space. $|\psi(0)\rangle\in \mathcal{H}$. Also $|\psi(t)\rangle\in \mathcal{H}$ for any fixed $t$, but the Hilbert space doesn't know anything about $t$. $|\psi(t)\rangle$ is in fact a function from some time interval $[a,b]$ to the Hilbert space $\mathcal{H}$. And it's on that parameter that the differential acts.

You can also see this with matrices: Given $\mathcal{H}=\mathbb{R}^2$ as your Hilbert space, we have for example a vector $v\in \mathcal{H}$ which is just a column vector. Also, an operator is a two-by-two matrix (that's the linear functions $\mathbb{R}^2\to\mathbb{R}^2$. Let's pick an operator and call it $A$. Then we can write:

$$ v=\begin{pmatrix}{} v_1 \\ v_2 \end{pmatrix}\quad \mathrm{and}\quad A=\begin{pmatrix}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$

But what about the $t$? Well, clearly, this means that all coordinates are dependent on $t$. You can even have operators that depend on $t$ (see for instance the Heisenberg picture):

$$ v(t)=\begin{pmatrix}{} v_1(t) \\ v_2(t) \end{pmatrix}\quad \mathrm{and}\quad A(t)=\begin{pmatrix}{} a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end{pmatrix} $$

Now if the time differentiation was an operator, you'd have to be able to write it as a two-by-two matrix. Can you? Obviously not. And the reason is that time is just a parameter. It doesn't really belong to the Hilbert space structure (it's not a dimension or anything).

From this picture, you can also see why what you wrote down is valid. How would you differentiate a scalar product of $v,w\in \mathbb{R}^2$? Well, let's write it out:

$$ \frac{\mathrm{d}}{\mathrm{d}t} (v(t)\cdot w(t)) = \frac{\mathrm{d}}{\mathrm{d}t}(v_1(t)w_1(t)+v_2(t)w_2(t))$$

Now you just have a bunch of scalar functions and you can use the usual rules of differentiation and obtain the result.

Can you make time an operator? That's an entirely different question (the answer is "not really").

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Others have pointed out that time is not an operator in the Schrodinger equation (I'll link to posts when I find them), but this isn't the end of the story.

For example, how should one denote the monentum operator acting on a bra? $\partial_x \psi(x,t)$ could be denoted as $\langle \psi | \hat p^\dagger$ (preferred) but I've also seen variants like $\langle \hat p \psi |$ (I find this horrendous). Just keep in mind that you define conjugate operators to act on the bra, even if they appear to the right.

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