I'm a bit confused as to when ordinary and partial derivatives are used in the Dirac formalism.

In the Schrödinger equation, for instance, Griffiths [3.85] uses ordinary derivatives:

$$ i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \lvert \mathcal{S} \rangle = H \lvert \mathcal{S} \rangle, $$

and so do Schumacher and Westmoreland [5.23]:

$$ H \lvert \psi(t) \rangle = i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \lvert \psi(t) \rangle, $$

and they even denote the explicit time dependence of the ket. Shankar (section 4.1) does the same. This makes sense to me, since the ket is not a function of e.g. space. However, though I haven't closely studied Sakurai, he seems to disagree and writes [2.1.27]:

$$ i \hbar \frac{\partial}{\partial t} \lvert \alpha, t_0; t \rangle = H \lvert \alpha, t_0; t \rangle. $$

Similar is found in e.g. these lecture notes from MIT (PDF). Is there any good reason for using different kinds of derivatives that I'm not seeing? Or any other justification for different notations?

I also had a second confusion, in this case regarding operators. Sakurai defines an observable in the Heisenberg picture (subscript $H$) in terms of the same observable in the Schrödinger picture (subscript $S$) [2.2.10]:

$$ A_H \equiv U^\dagger(t) A_S U(t), $$

where $U$ is the time evolution operator. To derive Heisenberg's equation of motion, I myself would just take the total derivative with respect to time:

$$ \frac{\mathrm{d} A_H}{\mathrm{d} t} = \frac{\mathrm{d} U^\dagger}{\mathrm{d} t} A_S U + U^\dagger \frac{\mathrm{d} A_S}{\mathrm{d} t} U + U^\dagger A_S \frac{\mathrm{d} U}{\mathrm{d} t}. $$

Now Sakurai has instead partial derivatives on the right-hand side [2.2.15]. But why is this, since we're taking the total derivative? He claims (implicitly) that the middle term cancels for observables $A_S$ that do not depend explicitly on time, which I can see if he takes the partial derivative (and not a priori for total derivatives), but I do not understand why we do that.

  • $\begingroup$ The second subquestion (v2) is a duplicate of this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 8, 2017 at 19:30

2 Answers 2


The usage of total and partial derivatives in Physics is not always very rigorous or consistent. Sometimes, authors only write $\frac{\mathrm d}{\mathrm dt}$ if they want to emphasize that there is an implicit time-dependence which is to be included in the derivative, and default to $\frac{\partial}{\partial t}$ otherwise. Sometimes, authors only write $\frac{\partial}{\partial t}$ if they want to emphasize that there is an implicit time-dependence which is to be ignored in the derivative, and default to $\frac{\mathrm d}{\mathrm dt}$ otherwise. Sometimes, authors just write $\partial_t$ because it is less effort than $\frac{\mathrm d}{\mathrm dt}$. Sometimes, authors don't really care and just go by gut feeling.

I'll go through your specific questions:

  • In the Schrödinger equation $H|\psi(t)\rangle = \mathrm i\hbar \partial_t |\psi(t)\rangle = \mathrm i\hbar \frac{\mathrm d}{\mathrm dt} |\psi(t)\rangle$ because the function $|\psi(t)\rangle$ does not depend on any other parameters anyway. (From my experience, most people write a partial derivative here, but either way is correct.)

  • In your example from Sakurai, maybe they want to stress that $\alpha$ and $t_0$ do not depend on time.

  • About the operators: In my (second) edition of Sakurai, there is a total derivative on the left hand side. And there should be, because we want to distinguish the implicit time-dependence that comes from the propagators and the possible explicit time-dependence of $A_S$. This is sometimes written as $$ \frac{\mathrm d A_H}{\mathrm dt} = \frac{\mathrm i}{\hbar} [H, A_H] + \frac{\partial A_H}{\partial t} . $$ Edited to add: People like to write it this way because it shows the analogy with canonical mechanics. There, the distinction between total and partial derivatives is pretty clear.

  • Finally note that for the operator $A_S$ in the Schrödinger picture, there is no implicit time-dependence. So it doesn't matter whether you write a total or partial derivative.

  • $\begingroup$ Oops! I meant to write that Sakurai has partial derivatives on the right-hand side. But your last point answered my question anyway. One more thing, actually. In Heisenberg's equation of motion, I've seen the last term, the explicit time dependence of the operator, written a slightly different way (PDF, page 21): $(\partial A_S / \partial t)_H$. So the time derivative in $S$ transformed to $H$. I'm having trouble seeing how that is equivalent to what you have written. $\endgroup$
    – Danny
    Nov 8, 2017 at 22:35
  • 1
    $\begingroup$ Both terms are supposed to mean $U \frac{\partial A_S}{\partial t} U^\dagger$. The way you write it makes it a bit clearer, but the way I put it is also common notation. With these things you often have to infer meaning from the context -- in the equation I wrote, the partial derivative means "only consider the intrinsic time dependence of the observable, not the time dependence added by the Heisenberg picture". $\endgroup$
    – Noiralef
    Nov 8, 2017 at 23:24

Upvoted the question, because it is a question in physics which is very well-suited for a purely mathematical answer. Which is:

The proper notation for the time derivative to be used in the generic/abstract Schrödinger equation is the full derivative, i.e. $\frac d {dt}$, regardless whether one uses the (mathematically challenging) bra/ket notation, or whether one uses the traditional physicists' notation of Schrödinger wavefunctions. However, the partial derivative can be used under certain mathematical assumptions linked to below.

Explanation: The Hamiltonian in the so-called Schrödinger picture is a proper Hilbert space operator-valued mapping $\mathbb R \ni t\mapsto H (t)\equiv H \in \mathcal L(\mathcal H),\forall t, H = H^\dagger$. The Schrödinger states (state representatives) are proper Hilbert space vector-valued mappings (called quantum trajectories) $\mathbb R \ni t\mapsto \psi (t)\in \mathcal H $, strongly continous in the real parameter $t$ called time. Therefore, it is an assumption (axiom of QM) that the mapping

$$\mathbb{R}\ni t\mapsto \text{s-lim}_{h\to 0} \frac{1}{h} \left[\psi (t+h) - \psi (t)\right]\text{exists} $$ and is denoted by $\frac{d\psi(t)}{dt}$. This limit, under the Schrödinger equation seen as a constraint, properly defines the set of admissible quantum trajectories as a subset of the maximal domain (domain of self-adjointness) of the Hamiltonian operator.

One now could now question if the Hamiltonian is a function of the fundamental operators x,p (in one dimension), by a projection into the Hilbert space $L^2 (\mathbb{R})$, thus it could be seen as a several variable complex function $H = H(x,p,t)$, can one write down the Schrödinger equation using partial derivatives in the form:

$$\frac{\partial \psi (x,p,t)}{\partial t} = \frac{1}{i\hbar} H(x,p) \psi (x,p,t),$$

in other words, what would be the connection to the rigorous formulation above? Valter Moretti here Time derivative of the state vector as expressed in abstract Hilbert space vs. as a wavefunction showed under what exact mathematical assumptions, the $\frac{d}{dt}$ defined above by the strong limit can be transformed to the $\frac{\partial}{\partial t}$.


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