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In my QM book I often see partial derivatives mixed with kets, like

$$ \frac{\partial}{\partial a} |\psi \rangle $$

where $a \in \{x, y, z\}$. Here I'm assuming that $| \psi \rangle \in \mathbb{C}^n$ for some arbitrary $n > 0$ (in particular $n$ can be much larger than $3$).

Question: What is a partial derivative of a ket? Don't partial derivatives operate on functions? What does it mean to operate on a complex n-tuple?

Example: Schrödinger's equation famously involves mixing partial derivatives with kets:

$$i\hbar\partial_t|\Psi\rangle=-\frac{\hbar^2\partial_i^2}{2m}|\Psi\rangle+V(\hat{x})|\Psi\rangle$$

Here there is a partial derivative operator $\partial_i$ as well as a partial time-derivative operator $\partial_t$.

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    $\begingroup$ You’re right! This is bad notation; the good books don’t do this. $\endgroup$
    – knzhou
    Oct 7, 2020 at 15:54
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    $\begingroup$ If you “often see” this and it bothers you, you should toss all the books you’re using in the trash and switch to standard ones. $\endgroup$
    – knzhou
    Oct 7, 2020 at 15:54
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    $\begingroup$ @knzhou why? Would $\partial_t |\psi(t)\rangle$ be better? Vectors can depend on a parameter, and that just means $\lim_{h\to 0} \frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h}$. I would write this as an answer but your comments are making me doubt myself. $\endgroup$ Oct 7, 2020 at 16:10
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    $\begingroup$ @user2723984 The time derivative is fine, the space derivative isn’t. $\endgroup$
    – knzhou
    Oct 7, 2020 at 16:56
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    $\begingroup$ Looks more like eq. (7.24) on p. 285. $\endgroup$
    – Qmechanic
    Oct 7, 2020 at 17:14

3 Answers 3

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It is an abuse of notation: it mixes the Schrodinger notation (i.e. the usual wave function representation) with the abstract Dirac notation (the bra-ket notation). In your example, the Hamiltonian operator is expressed in the Schrodinger notation (i.e. with position operator that is "diagonal" and the momentum operator represented by the gradient), while the "state" is expressed in the Dirac notation (it is an abstract ket).

Long story short:

State at a particular time $t$: it is an abstract container $| w, t \rangle$ for some information $w$ at time $t$ ($w$ is a collection of values and/or procedures that determines the state of your object.. some sort of recipe to "prepare" it).

Wave function relative to the info $w$ at time $t$: think about is as a complex function $\Psi_w(x,y,z,t) = \langle {\bf x}| w, t \rangle $. Usually $w$ is dropped (in fact you are not even able to write it down most of the times, and it is a formal label for the state) and the common notations $\Psi({\bf x},t) = \langle {\bf x}| \Psi, t \rangle$ or $\Psi({\bf x},t) = \langle {\bf x}| \Psi(t) \rangle$ are used. You can think of $\Psi$ (or whatever name you want to use as something that completely labels the state and that is directly used in place of $w$). Note: $| {\bf x}\rangle$ is the abstract eigenstate of the position operator relative to the eigenvalue ${\bf x}$ (in the case of this "position ket" the information $w$ is simply $w={\bf x}$, namely "the particle is in ${\bf x}$"). In other simple cases the label $w$ may be a set of "quantum numbers" or eigenvalues.

Completeness: since (formally) you have that $\sum_{x,y,z}| {\bf x}\rangle \langle {\bf x}|$ is the identity operator,

$$ |\Psi,t\rangle = \sum_{x,y,z} |{\bf x}\rangle \langle {\bf x}| \Psi(t) \rangle = \sum_{x,y,z} \Psi({\bf x},t) |{\bf x}\rangle \, , $$

which means that the ket $|\Psi,t\rangle $ and the associated wave function carry the same info (are both complete and equivalent representations of the state.. the ket one by definition). Moreover,

$$ \partial_t |\Psi,t\rangle = \sum_{x,y,z} (\partial_t \Psi({\bf x},t) ) |{\bf x}\rangle $$

Bad notation: on the other hand, $\nabla |\Psi,t\rangle$ is just abuse of notation. To make sense of this, the gradient must be understood as an "abstract operator", acting on kets, that has the effect of differentiating wave functions once the expression is projected onto the position basis: you can try to use the completeness to obtain the decomposition of the abstract momentum operator (EDIT: see the nice answer of Qmechanic for this point!).

Time VS space: the derivative in time, although formal, is not an abuse of notation if applied to kets. In fact, time is not the eigenvalue of something, position yes. QM treats time and space in fundamentally different ways. Time is just a parameter in QM, so you can think of $|\Psi,t\rangle$ as a curve in the abstract space of states, and this curve has a "tangent" that is formally indicated as $\partial_t |\Psi,t\rangle$.

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    $\begingroup$ By $\mathbf{x}$ do you mean $(x, y, z)$, for $x,y,z \in \mathbb{R}$? $\endgroup$
    – George
    Oct 7, 2020 at 17:23
  • $\begingroup$ Yes. The sum is over an infinite (uncountable) set and is formal (it is more an integral than a sum), but this kind of notation is sometimes used and not too bad. See this: physics.stackexchange.com/q/89958/226902 (here formal integrals are used in place of formal summation.. since everything is formal, use the symbol you like!). $\endgroup$
    – Quillo
    Oct 7, 2020 at 17:32
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The derivative operator $\frac{\partial}{\partial x^j}$ in the Dirac notation is ambiguous because it depends on whether the derivative is supposed to act to the right (on a ket) or to the left (on a bra). See also my Phys.SE answer here. In Ref. 1 the derivative operator came from the Schrödinger representation of the momentum operator $$\begin{align}\hat{p}_j ~=~&\int_{\mathbb{R}^3} \mathrm{d}^3p~ |{\bf p}\rangle ~ p_j ~\langle {\bf p}|\cr ~=~&\int_{\mathbb{R}^3} \mathrm{d}^3x~ |{\bf x}\rangle ~ \frac{\hbar}{i}\frac{\stackrel{\rightarrow}{\partial}}{\partial x^j} ~\langle {\bf x}|\cr ~=~&\int_{\mathbb{R}^3} \mathrm{d}^3x~ |{\bf x}\rangle ~ i\hbar\frac{\stackrel{\leftarrow}{\partial}}{\partial x^j} ~\langle {\bf x}| ,\end{align}\tag{1}$$ so that $$\begin{align}\langle \phi |\hat{p}_j|\psi \rangle ~\stackrel{(1)+(3)}{=}&\int_{\mathbb{R}^3} \mathrm{d}^3x~ \phi^{\ast}({\bf x}) ~ \frac{\hbar}{i}\frac{\partial \psi({\bf x})}{\partial x^j} \cr ~\stackrel{(1)+(3)}{=}&\int_{\mathbb{R}^3} \mathrm{d}^3x~ i\hbar \frac{\partial \phi^{\ast}({\bf x})}{\partial x^j} ~\psi({\bf x}) ,\end{align}\tag{2}$$ where we have defined the wavefunctions $$ \begin{align}\psi({\bf x})~:=~& \langle {\bf x}|\psi \rangle, \cr \phi^{\ast}({\bf x})~:=~& \langle \phi |{\bf x}\rangle.\end{align}\tag{3}$$

References:

  1. J. Schwichtenberg, No-Nonsense Quantum Mechanics: A Student-Friendly Introduction, 2018; eq. (7.24) on p. 285.
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The other two answers are both great and this may be unnecessary but it might be worth explicitly mentioning that in Dirac notation the Schrodinger equation looks like $$i \hbar \partial_t \lvert \Psi \rangle = \hat{H} \lvert \Psi \rangle$$ where the Hamiltonian can be further split up by defining the momentum and potential energy operators as $$ \hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}.$$

The answer from Qmechanic gives a very comprehensive definition of each component of $\hat{p}$ in position basis and following in the same vein as Quillo the potential operator relates to the scalar function $V(x, t)$ as $\langle x \rvert \hat{V} \lvert x \rangle = V(x,t)$ so that (using the identity operator) we have $$V(x,t) \Psi(x,t) = \langle x \rvert \hat{V} \lvert \Psi \rangle$$ and likewise $$\frac{-\hbar^2 \partial_i^2}{2m} \Psi(x,t) = \frac{1}{2m} \langle x \rvert \hat{p}_i^2 \lvert \Psi \rangle.$$ This hopefully will demonstrate the equivalence (in most straightforward cases) of the two formulations when the position basis is used for Dirac notation.

Note: one of the most powerful things about Dirac notation is its generality in not specifying a basis whereas the form of the Schrodinger equation that is most familiar is specifically in the position basis. If instead we chose to use momentum space we could still get a valid form of the Schrodinger equation by doing something like $\langle p \rvert \hat{H} \lvert \Psi \rangle.$ However, the operators $\hat{p}$ and $\hat{V}$ will have different forms in this scenario as they will be expressed in terms of the momentum basis so you can't just jump straight to the familiar form of the equation.

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