I am currently trying to follow Leonard Susskind's "Theoretical Minimum" lecture series on quantum mechanics. (I know a bit of linear algebra and calculus, so far it seems definitely enough to follow this course, though I have no university physics education.)

In general, I find these lectures focus a bit too much on the math and not really on the physical motivation behind it, but so be it (if there are other courses aimed at those with reasonable math skills that focus more on physical meaning, let me know!). However, that's only indirectly related to what my question is about.

In Lecture 4, just after the 40-minute mark, Susskind sets out to derive an expression for what he earlier labelled the time-development operator $U$:

$$|\psi(t)\rangle = U(t)|\psi(0)\rangle$$

He starts out as such:

$$U(\epsilon) = I + \epsilon H$$

which makes sense because the change in time will have to be small, i.e. on the order of a small $\epsilon$. However, he then goes ahead and changes this into:

$$U(\epsilon) = I - i\epsilon H$$

which of course is still fine, because we still don't know what $H$ is supposed to be. Now my problem lies with the fact that Susskind then proceeds to derive an expression for $H$ and, subsequently, the Schrödinger equation in which it figures, from the above equation. The $i$ never gets lost and ends up in that equation.

Could we not just as easily have left the $i$ out, or put a 6 or whatever there? Why put $i$ there? I finished the entire lecture hoping Susskind would get back to this, but he never does, unfortunately. (Which is, I guess, another symptom of this course, with which I'm otherwise quite happy, occasionally lacking in physical motivation.)

  1. For those of you who know this lecture, or similar styles of teaching: am I missing something here?

  2. Alternatively, a general answer as to why there is an $i$ in the Hamiltonian and the Schrödinger equation?

  • 8
    $\begingroup$ "if there are other courses aimed at those with reasonable math skills that focus more on physical meaning, let me know!" Landau & Lifshitz, set of books called Course of theoretical physics. $\endgroup$ Commented Feb 19, 2016 at 19:27
  • $\begingroup$ "if there are other courses aimed at those with reasonable math skills that focus more on physical meaning, let me know!" Try this: ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013 $\endgroup$
    – Shing
    Commented Mar 13, 2018 at 1:32

5 Answers 5


Let $U$ be an unitary operator. Write $$ U=\mathbb I+\epsilon A $$ for some $\epsilon\in\mathbb C$, and some operator $A$.

Unitarity means $U^\dagger U=\mathbb I$, i.e., $$ U^\dagger U=(\mathbb I+\epsilon^* A^\dagger)(\mathbb I+\epsilon A)=I+\epsilon^*A^\dagger+\epsilon A+\mathcal O(\epsilon^2) $$

Therefore, if $U^\dagger U=\mathbb I$, we must have $$ \epsilon^*A^\dagger+\epsilon A=0 $$

How can we archive this? We can always redefine both $\epsilon$ and $A$ so that $\epsilon$ is real. If you do this, we get $A^\dagger=-A$, i.e., $A$ is anti-hermitian. In principle, this is perfectly valid, but we can do better.

If we choose $\epsilon$ imaginary, we get $A^\dagger=A$. We like this option better, because we like hermitian operators. If $A$ is to be identified with the Hamiltonian, we better have $\epsilon$ imaginary, because otherwise $H$ cannot be hermitian (i.e., observable).

Note that Susskind writes $U=\mathbb I-i\epsilon H$ instead of $U=\mathbb I+i\epsilon H$. This negative sign is just convention, it is what everybody does, but in principle it could be a $+$ sign. This sign doesn't affect the physics (but we must be consistent with our choice). This is similar to certain ODE's in classical mechanics (driven harmonic oscillator, RLC circuits, etc), where we use the ansatz $x(t)=\mathrm e^{-i\omega t}$, with a minus sign for historical reasons.

So, we include the factor $i$ in $U$, and we end up with the Schrödinger equation. Had we not included the $i$, we would have got $$ \frac{\partial\psi}{\partial t}=\nabla^2\psi $$ where I take $\hbar=2m=1$ and $V=0$ to simplify the analysis (this doesn't change the conclusions). Note that this is the heat equation. The general solution of the heat equation is

$$ \psi(x,t)=\int\mathrm dy\ \frac{\psi(y,0)}{\sqrt{4\pi t}}\mathrm e^{-(x-y)^2/4t} $$

No matter what $\psi(y,0)$ is, this solution is non-propagating, non-oscillatory and decays in time. Therefore, the "particles" described by the heat equation don't move, and they slowly disappear! (for example, "stationary" solutions are of the form $\psi(x,t)=\mathrm e^{-Et}\phi(x)$, which goes to zero as $t\to \infty$).

Also, if it were not for the $i$ in Schrödinger equation, $\int\mathrm dx\ |\psi(x,t)|^2$ wouldn't be time independent, so the total probability would change in time, and this makes no sense. Therefore, the $i$ in Schrödinger equation makes the Born interpretation of the wave-function possible!

Some things you might want to check out:

  • Continuous Groups, Lie Groups, and Lie Algebras, for example in http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter7.pdf

  • Wigner's theorem: symmetries are realised through linear/unitary operators, or through antilinear/antiunitary operators.

  • Translation operator: in quantum mechanics (and in classical mechanics as well, in a sense), space/time translations are represented through unitary operators, where the generators of such operations are the energy/momentum operators.

  • Spectral theorem: Hermitian operators have real eigenvalues.

  • The Maximum Principle of the heat equation: if $\psi(x,t)$ solves the heat equation and $t_2>t_1$, then $\psi (t_2,x)\le \psi(t_1,y)\ \forall x,y$, which means $\psi$ "decreases" in time (therefore, probability "is destroyed", or "particles disappear").

  • Schrödinger versus heat equations

  • 9
    $\begingroup$ It bears mention that mathematicians often bundle the $i$ into $H$, and they're quite happy to have anti-hermitian hamiltonians. That's pretty nuts to a physicist - we like our $H$s hermitian - but it's all just conventions. $\endgroup$ Commented Feb 19, 2016 at 14:10
  • $\begingroup$ Thanks a lot; I actually understand this now (great feeling). Definitely staying tuned and looking forward to the additional details :) $\endgroup$
    – EelkeSpaak
    Commented Feb 19, 2016 at 14:45
  • $\begingroup$ @EelkeSpaak I'm glad I could help! If there is some point of my answer that is not clear enough, please say so and I'll try to explain it better :) $\endgroup$ Commented Feb 19, 2016 at 15:44
  • 2
    $\begingroup$ 1. Please use self-adjoint instead of Hermitian. In the infinite-dimensional case, there is a difference, and the spectral theorem is for self-adjoint operators, not merely Hermitian ones. 2. The infinitesimal expansion you do at the beginning is the physicist's version of showing Stone's theorem - every one-parameter unitary group is generated by a self-adjoint operator times $\mathrm{i}$. $\endgroup$
    – ACuriousMind
    Commented Feb 19, 2016 at 16:49
  • $\begingroup$ @ACuriousMind yes, you are completely right [but note that Susskind says hermitian as well :) ] $\endgroup$ Commented Feb 19, 2016 at 17:41

Generally, the $i$ doesn't have to be there: we can just define the time evolution to be $e^{tK}$, but then $K$ will be anti-hermitian. It really makes no difference. I guess it's more convenient to deal with hermitian operators, so people just factored out an $i$: $$K=iH.$$ As for why Hermitian operators are nicer to deal with, note that Dirac's beloved bra-ket sandwiches with this operator $K$, all over QM, would be slightly more ambiguous: we'd get nonsensical expressions like $$\left<\psi\right|K\left|\psi\right>=\left<\psi\right|-K\left|\psi\right>$$ And so we'd get minus signs, depending on whether we're acting on the left or on the right. It's simply easier to have a hermitian operator around.


My loosey-goosey intuition for why the $i$ appears in Schrödinger's equation comes down to the logarithm of a unitary matrix being $i$ times a Hermitian matrix, and differential equations tending to exponentiate things.

All quantum operations are unitary (besides measurement, depending on your favorite interpretation). Unitary matrices have a lot of nice properties. In particular, their eigendecomposition has perpendicular eigenvectors with eigenvalues from the complex unit circle. Symbolically:

$$\begin{align} U=&\sum_{k} e^{i\theta_k} \left|v_k\right\rangle \left\langle v_k \right| \end{align}$$

Where each $\theta_k$ is between $0$ and $2 \pi$, and the $v_k$'s are all mutually perpendicular.

If you want to turn a unitary matrix into a continuous action over time, you need to interpolate it. You could do that either by raising it to a fractional power like $U_t = U^t$, or by scaling its logarithm in an exponential like $U_t = e^{t \ln(U)}$. We're going to use the logarithm approach because it's more elegant, and more useful for differential equations (*cough* *cough*).

Because we have an eigendecomposition for $U$ in terms of $e^\text{whatever}$, the logarithm has a very nice form. The phase factors' angles drop down:

$$\begin{align} \ln(U)&= \ln\sum_{k} e^{i\theta_k} \left|v_k\right\rangle \left\langle v_k \right| \\&= \sum_{k} \ln(e^{i\theta_k}) \left|v_k\right\rangle \left\langle v_k \right| \\&= \sum_{k} i \theta_k \left|v_k\right\rangle \left\langle v_k \right| \\&= i \sum_{k} \theta_k \left|v_k\right\rangle \left\langle v_k \right| \\&= i H \end{align}$$

Here we've defined $H$ in terms of an eigendecomposition with perpendicular vectors scaled by real eigenvalues. Therefore $H$ is Hermitian. So the logarithm of a unitary matrix is $i$ times a Hermitian matrix, and our interpolation-by-exponentiating-a-scaled-logarithm becomes:

$$U_t = e^{t \ln(U)} = e^{i t H}$$

If you were paying attention, you saw where the $i$ came from.

  • It initially appeared because unitary matrices only apply phase factors to their eigenvectors, so their eigenvalues are like $e^{i \theta_k}$.
  • When we computed the logarithm, to interpolate the unitary matrix's effect, the $i$ dropped out of the exponential and escaped the sum.
  • The remaining sum was a Hermitian matrix.

Since we want our differential equation to have unitary solutions, and differential equations tend to exponentiate things, we'd better put $i$ times a Hermitian matrix in there.

  • $\begingroup$ The proper formalization of "the logarithm of a unitary matrix is anti-Hermitian" (which is not really a good thing to say because the logarithm is a rather badly behaved function) is Stone's theorem - every one-parameter unitary group is generated by a self-adjoint operator. $\endgroup$
    – ACuriousMind
    Commented Feb 19, 2016 at 16:47
  • 1
    $\begingroup$ @ACuriousMind Yes, that's a concise way to put it. (And I did gloss over the fact that the logarithm is actually an infinite family of matrices, with arbitrary offsets of $2 \pi n$ added to each eigenvalue.) $\endgroup$ Commented Feb 19, 2016 at 16:58
  1. The $i$ is derived from the Heisenberg Uncertainty Equation.

  2. As for the importance of the $i$ being there,,it shows mathematically "that a very real part of the future depends on an imaginary part of the present."

  • $\begingroup$ I used derive as evolved. Knew a mathematician would point that out. And evolved isn't right term either. I had trouble explaining my equations to get to theirs to my physics professors' too. Progression is better, Brownian Motion, EMT, HUE when applied to photon form of light leads to Schrodinger,,,but correct not derived by standard formulation $\endgroup$ Commented Feb 21, 2016 at 13:00

You can derive the Schrodinger equation with the knowledge $E=\hbar \omega$ and $p=\hbar k$. Take a general wave and act on it with the gradient or a time derivative and you get the energy and momentum operators:

$ \psi(\vec x, t) = \psi_o e^{i(\vec k x - \omega t)} \\ \nabla \psi = ik\psi \rightarrow \nabla = ip/\hbar \rightarrow \hat p = -i\hbar \nabla \\ \dfrac{\partial \psi}{\partial t} = -i\omega \psi \rightarrow \dfrac{\partial}{\partial t} = -iE/\hbar \rightarrow \hat E = i\hbar \dfrac{\partial}{\partial t} $

Then put the energy and momentum operators in the Hamiltonian.

$ H = T+V \\ \hat E = \dfrac{\hat p^2}{2m}+\hat V \\ i\hbar \dfrac{\partial}{\partial t}\psi = \dfrac{-\hbar^2 \nabla^2\psi}{2m}+V\psi $


  • $\begingroup$ The OP needs derivation from first principles dude! $\endgroup$
    – Karthik
    Commented Oct 30, 2021 at 22:11

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