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Starting from:

$$ -i\hbar \frac{d|\psi⟩}{dt} = H|\psi⟩ $$

I was able to do some working to prove that $U$ in the corresponding discrete representation

$$ U(t_1,t_2) = exp\frac{-iH(t_2-t_1)}{\hbar} $$

is unitary if and only if $H$ is Hermitian. That is:

$$ U^\dagger(t_1,t_2)U(t_1,t_2) = I \iff H = H^\dagger $$

Cool! But now I'm stuck trying to understand the physical significance of the fact that $H$ is Hermitian. I try to see $H$ as a "velocity function" because it gives the instantaneous change in $|\psi⟩$. That's as far as my intuition goes in terms of understanding the Hamiltonian.

So what's another intuitive way of understanding why the Hamiltonian must be Hermitian? And as a bonus, can the Hamiltonian also be unitary?

EDIT

I think the following "intuitive" description works best for me (the answers led me here):

So we start from my idea that $H$ is like a "velocity function" that takes as input the current state vector $|\psi⟩$ and spits out the infinitesimal change $d|\psi⟩/dt$. Now we consider two things:

1) $H$ being hermitian means it has real eigenvalues (the proof of that is off-topic). So if you apply it to a state vector you're just scaling each of its components by a purely real number. Or in math speak, if you apply $H$ to $|\psi⟩ = \sum_i{c_i}|\psi_i⟩$ (where I've just expressed $|\psi⟩$ as a weighted sum over basis vectors $|\psi_i⟩$), you end up getting $\sum_iR_ic_i|\psi_i⟩$ where $R_i$ are purely real.

2) But then we also consider that $H$ is also multiplied by $i$. So that turns all those purely real coefficients to purely imaginary coefficients. Switching to the polar description of complex numbers, the infinitesimal change to the state vector is just a phase added to each of its components, and no amplitude scaling.

Therefore $H$ is like an instantaneous tangential nudge to the vector. So by integrating, the corresponding discrete operator $U$ is unitary.

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  • $\begingroup$ For any moderators who see this. I couldn't find tags for Hamiltonian, unitary-gate, or postulate-two. I'd think they should make sense to have in this SE. $\endgroup$ – Alexander Soare May 21 at 13:22
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    $\begingroup$ The eigenvalues of a Hermitian operator are real, that is why any observable, not only energy, is a Hermitian operator. $\endgroup$ – kludg May 21 at 13:31
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    $\begingroup$ @kludg That's actually a false reason... Hermitian matrices having real eigenvalues doesn't actually imply that a matrix with real eigenvalues has to be Hermitian $\endgroup$ – Mithrandir24601 May 21 at 14:35
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    $\begingroup$ @Mithrandir24601 this is true reason. Matrices having real eigenvalues and not Hermititan should be rather weird; ex, the spectral theorem does not apply to them. $\endgroup$ – kludg May 21 at 16:00
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    $\begingroup$ @Mithrandir24601 right. I misunderstood you as saying that non-normal observables are also physically relevant. Anyway, I also don't quite agree with saying that having real eigenvalues is fundamentally important, see e.g. discussion here physics.stackexchange.com/a/264439/58382 (which I know is not what you were stating, I'm just pointing it out in regard to the whole discussion) $\endgroup$ – glS May 22 at 17:53
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More Pragmatic Answer after Dialogue with OP in Comments

Rearranging the wave equation in the question to $$\frac{d \vert \Psi \rangle}{dt} = \frac{i}{\hbar} H \vert \Psi \rangle \tag{1}$$ should make it obvious that wavefunction takes the form $$\vert \Psi \rangle = e^{\frac{i}{\hbar} H t}. \tag{2}$$ However, the wavefunction is not only a function of time, but also of position. We can${}^\ast$ think about the time evolution of the system as $$\vert \Psi(t) \rangle = e^{\frac{i}{\hbar}Et} \vert \psi_{r} \rangle, \tag{3}$$ where $\vert \psi_{r} \rangle$ depends on the position within some coordinate system and is independent of time, and $E$ generally represents the energy in the system, which is a scalar value.

We can now consider the time derivative of Eqs. (2) and (3) in the $\vert \psi_r \rangle$ basis,

$$\frac{d \vert \Psi \rangle}{dt} = \frac{i}{\hbar} H \,e^{\frac{i}{\hbar}Et} \vert \psi_{r} \rangle=\frac{i}{\hbar} E \, e^{\frac{i}{\hbar}Et} \vert \psi_r \rangle, \tag{4}$$ and rearrange to get $$-\left( i \, \hbar \, e^{\frac{-i}{\hbar}Et} \right) \, \frac{d \vert \Psi \rangle}{dt} = H \vert \psi_{r} \rangle= E \vert \psi_r \rangle.$$ The second equality here is the eigenvalue equation. It follows directly that if we need energy to be real and positive, which we do, $H$ must have positive real eigenvalues. The eigenvalue equation above also implies that $H$ is diagonalizable. As noted in the comments to your question, these two conditions are sufficient to require that $H$ is Hermitian.

${}^\ast$ After thinking about this some more, I should point out that Eq. (3) amounts to an assumption that at least one eigenvalue of $\vert \Psi \rangle$ can be expressed as a function of time (independent of position). Similarly, Eq. (4) amounts to an assumption that the positional basis $\vert \psi_r \rangle$ is independent of time.

Einstein showed that these assumptions are only low velocity approximations, i.e. that time is not independent of space. This is one way of seeing why the Shrödinger equation is inherently non-relativistic.

Original Less Pragmatic Answer/Geometric Intuition Regarding Hermitian Matrices

The intuition behind Hermitian matrices, is that they are the "real" matrices in $\mathbb{C}^{n \times n}$. The most basic property of any Hermitian matrix ($H$) is that it equals its conjugate transpose $H=H^\dagger$ (in direct analogy to $r \in \mathbb{R}$ where $r = r^\ast$). Equally fundamental, a Hermitian matrix has real eigenvalues and it's eigenvectors form a unitary basis that diagonalizes $H$.

Those are the key mechanical properties, but they probably don't do much for intuition. I'll try to give a sense of the geometric role that Hermitian matrices play in complex spaces in hopes that it may help you intuit the more direct and less transparent explanations that you've probably already seen.

Consider the familiar case of generating a 1-sphere in $\mathbb{C}^1$ by taking the exponent $e^{i \theta}$, with $\theta \in \mathbb{R}$, such that $i\theta$ is a purely imaginary value. In this case, our implicit real basis for the real parameter $\theta$ is $1$, and not very interesting. If we wanted to be pedantic we could call $[1]$ a $1 \times 1$ Hermitian matrix.

In the same manner, we can generate a 3-sphere in $\mathbb{C}^{2 \times 2}$ by $U = e^{\frac{i}{2} \vec{\phi} \, \cdot \, \vec{\sigma}}$, where $\vec \phi$ is a vector in $\mathbb{R}^3$, and $\vec \sigma$ are the three Hermitian Pauli matrices (which are much more interesting than $[1]$). This guarantees that $U$ is an arbitrary element of $SU(2)$, which is isomorphic to the 3-sphere. Setting $\vec \phi = \alpha (\hat \phi_1, \hat \phi_2, \hat \phi_3)$, Taylor expansion gives $$U=e^{\frac{i}{2} \vec{\phi} \, \cdot \, \vec{\sigma}}=\begin{bmatrix} \cos \frac{\alpha}{2} + i \hat \phi_3 \sin \frac{\alpha}{2} & \sin \frac{\alpha}{2} (\hat \phi_2 + i \hat \phi_1) \\ \sin \frac{\alpha}{2} (-\hat \phi_2 + i \hat \phi_1) & \cos \frac{\alpha}{2}- i \hat \phi_3 \sin \frac{\alpha}{2} \end{bmatrix}.$$ The elements of the Pauli basis generate a representation of the (real) Lie algebra $\mathfrak{su_2}$, and are related to the three independent rotations ($R_n$), by $\sigma_n = 2i \frac{\partial R_n}{\partial\alpha} |_{\alpha=0}$. Although $\sigma_2$ makes use of imaginary numbers, the Pauli basis is real in all the ways that matter. For example $\sigma_n^2 = I$ and $(\vec \phi \cdot \vec \sigma)^2 = \vert \vec \phi \vert^2$. Just as in the case of $i \theta$, when we multiply the $\sigma_n$ by $i$, they become purely imaginary, i.e. $(i \sigma_n)^2 = -I$ and $(i \vec \phi \cdot \vec \sigma)$ is isomorphic to the pure quaternions (the imaginary part of the quaternions).

Recalling that $\sigma_n^2 = I$, if we forgo the $i$ in the exponentiation, we get something directly analogous to the 1-dimensional case of exponentiating with the split complex number, $k^2=1$, where $e^{k \beta} = \cosh{\beta} + k \sinh{\beta}$. In the three dimensional case we have, by Taylor expansion again:

$$ V = e^{\frac{1}{2} \vec \phi \, \cdot \, \vec \sigma} = \begin{bmatrix} \cosh \frac{\alpha}{2} + \hat \phi_3 \sinh \frac{\alpha}{2} & \sinh \frac{\alpha}{2} (\hat \phi_1 - i \hat \phi_2) \\ \sinh \frac{\alpha}{2} (\hat \phi_1 + i \hat \phi_2) & \cosh \frac{\alpha}{2} - \hat \phi_3 \sinh \frac{\alpha}{2} \end{bmatrix}, $$ which are Hermitian matrices with unit determinant and Minkowski signature $(+,-,-,-)$. This matrix can be directly identified with a real 4-vector as discussed below. The 1-dimensional projection operators $\frac{1}{2}(1 \pm k)$ are also strikingly similar to the 3-dimensional Hermitian projection operators $\frac{1}{2}(I \pm \hat \phi \cdot \vec \sigma)$.

Pauli used his namesake matrices to formulate the Pauli equation, which is unfortunately non-relativistic since it fails to treat space and time on an equal footing. One of Dirac's great breakthroughs was generalizing the Pauli matrices into the gamma matrices, which enabled him to formulate the Dirac equation, which was one of the great breakthroughs of the past century.

Attempting to explain Dirac spinors at an intuitive level would be quite a task (I can provide some good references if you like), but we're only a breath away from coordinate generating spin matrices, which provide great insight into the value and necessity of Hermitian matrices. This is largely due to Wheeler's great presentation of them in Section 41.3 of Gravitation, which I highly recommend.

Essentially if we take a direct sum $\mathfrak{su}_2$ (which generates the element $U$ above) with $i \mathfrak{su}_2$ (which generates $V$ above) we get the 6-dimensional (complex) Lie algebra $sl_{2 \mathbb{C}}$, which generates the universal cover of the Lorentz group $SL(2,\mathbb{C})$, providing both Lorentzian rotations ($U$), boosts ($V$) and combinations of the two ($L \in SL(2,\mathbb{C}$)). In other words, any $L$ has unique left and right polar decompositions as $UV_R$ or $V_L U$, where $V$ is analogous to a radius, and $U$ to an angle. ($U$ and $V$ commute iff they share the same unit vector $\hat \phi$).

Wheeler calls the matrix $L$, which effects an arbitrary Lorentz transformation, a Lorentzian spin transformation matrix. $L$ acts on an Hermitian "coordinate-generating spin matrix" ($X$), $$X=\begin{bmatrix} t + z & x-iy \\ x+iy & t-z \end{bmatrix},$$ (note the same Hermitian structure and Minkowski signature as $V$) according to $X^\prime = L X L^\dagger$.

The structure $X$ and $L$ ensures that $X^\prime$ remains Hermitian by $$ (X^\prime)^\dagger = (LXL^\dagger)^\dagger = (L^\dagger)^\dagger(X)^\dagger(L)^\dagger = LXL^\dagger = X^\prime $$ and thus $(t,x,y,z) \mapsto (t',x',y',z')$ remains real.

I'm not sure I answered the question, as I didn't say much specifically about Hamiltonians, but in some sense that boils down to Hamiltonians describe real things (specifically kinetic energy + potential energy of a system) in complex spaces, and Hermitian matrices are mathematical representations of real things in complex spaces. Alternative explanations sometimes take forms like (quoting Weyl) "In an infinitesimal unitary rotation of a vector field the velocity $\frac{d \chi}{d \tau}$ is related to $\chi$ by by means of a correspondence whose matrix is i times an Hermitian matrix."

Hopefully the color above about the role Hermitian matrices play in complex spaces helps make sense of the generally unintuitive direct answers to your question.

EDIT: I forgot to answer the bonus question. The Pauli matrices are both unitary and Hermitian. The Hadamard matrix, $$\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}, $$ which is prominent in QIS/QIT, is also unitary and Hermitian.

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  • $\begingroup$ Jonathan, this is really awesome. Although I feel like if I could even understand a third of it, I probably would have never needed to ask my question in the first place! I suppose I will revisit it every now and then on my journey to understanding quantum computing. $\endgroup$ – Alexander Soare May 22 at 9:59
  • $\begingroup$ @AlexanderSoare Sorry, one the challenging things about this site is gauging peoples familiarity with subjects to avoid talking over or under them. Hopefully the last 3 paragraphs are of some use at least. $\endgroup$ – Jonathan Trousdale May 22 at 10:37
  • $\begingroup$ Ah thanks for narrowing my focus. Yes this infinitesimal unitary rotation is tying in nicely with my notion of the Hermitian being like a velocity function. So something about the Hermitianess of the matrix makes sure that the velocity is such that the vector in question does pure rotation without scaling. $\endgroup$ – Alexander Soare May 22 at 10:44
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    $\begingroup$ @AlexanderSoare I may be an outlier, but none of this ever made much sense to me until working through Weyl's book, "The Theory of Groups and Quantum Mechanics." He's the only author I know that starts with robust coverage of Unitary Geometry. It's not any easy book, but Wheeler taught himself QM with that book at age 19, and it was very influential on Wigner as well. $\endgroup$ – Jonathan Trousdale May 22 at 10:49
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    $\begingroup$ @AlexanderSoare Yes, you can think of $H$ as a tangent vector (to some transformation manifold) defining the "direction" in the next instant. When thinking about an infinitesimal transformation you want to think about it as tiny movement away from the identity, which looks like $$R=I + i H \delta t,$$ since higher order terms can be neglected. Finite rotations are then simply $$e^{i H t} = \lim_{N\to\infty} \left(I + \frac{i H t}{N} \right)^N.$$ If $H$ isn't Hermitian (i.e. "real"), then this isn't a pure rotation, and the Hamiltonian is not conserved. $\endgroup$ – Jonathan Trousdale May 22 at 14:17
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I will basically quote directly from Susskind as its a great explanation. First of all being Hermitian tells us that $H$ is an observable, secondly, it has a complete set of eigenvectors and eigenvalues that define the energy levels of the system by spectral decompsition:

$H = \sum E |E\rangle \langle E|$

To derive the condition for $H$ being Hermitian we look at the evolution of a system. Firstly for any time change of a system, the change needs to be unitary thus:

$U^\dagger U = I$,

based on smoothness of time evolution we also require $U$ be of the form that:

$U = I - i\epsilon H$,

thus for $U$ to remain unitary we must have:

$U^\dagger U = \left(I + i\epsilon H^\dagger\right)\left(I - i\epsilon H\right) = I$,

expanding this out we find that:

$H^\dagger - H = 0$,

which is our condition of being Hermitian.

Can $H$ be unitary? For $H$ to have a spectral decomposition the eigenvalues need to be real, and we cannot guarantee this for a unitary. Secondly for $H$ to be Unitary and Hermitian, this means that:

$H = (2P - I)$,

where $P$ is an orthogonal projector. This will have eigenvalues $\pm 1$, thus you will have degenerate energy levels. So yes you can, but you are rather limited.

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  • $\begingroup$ Hmm, I just tried that expansion and theres also a $H^\dagger H$ term in it. Unless I already know that that's purely real I can't really get your last equation. Am I missing something? In any case I'm also trying to see how this gives any more of a physical intuition. $\endgroup$ – Alexander Soare May 21 at 13:48
  • $\begingroup$ You only need to expand up to first order if $\epsilon$, however I've seen you derived $H$ is unitary already so I guess this doesn't help too much. $\endgroup$ – Sam Palmer May 21 at 13:52
  • $\begingroup$ I will add an edit about it being unitary $\endgroup$ – Sam Palmer May 21 at 13:53
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    $\begingroup$ Thanks for that's a great answer. Will leave it open for a while to try and get more $\endgroup$ – Alexander Soare May 21 at 14:20
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Let's say we take for granted that physical states are described by complex vectors $|\psi\rangle\in\mathbb C^n$ defined up to their normalisation and (global) phase.

(Unitaries are the general way to map states to states) We now want to describe how these things evolve in time. The natural way to describe how a vector changes into another vector is via a linear operator (that is, a matrix). Let's call it $U$. We want this $U$ to map states into states. That means $U$ should preserve normalisation and don't care about the global phases. Moreover, crucially, we want such an operation to not destroy nor create information, that is, we want it to be time-reversible (we could and do consider more general situations involving loss of information, using so-called quantum channels, but that's a discussion for another day). Putting these requirements together identifies $U$ as an element of the group of unitary matrices $SU(n)$.

(Unitaries don't have a notion of time) But there is something rather unphysical about the above description: where is time? Indeed, describing the evolution of a state via unitary operators, $|\psi\rangle\mapsto U|\psi\rangle$, corresponds to considering the state at only different "time snapshots". In other words, unitaries are akin to "black boxes": they describe the overall effect of some physical evolution, but don't delve about the details actually making it up.

(Bringing time into the picture: families of unitaries) Real physical interactions act on states by continuously changing them. In other words, there must be a notion of infinitesimal changes, and these changes must depend on the specifics of the situation (and thus in particular on the input state). Given our above conclusions regarding unitaries, we can write quite generally the time evolution of a state $|\psi\rangle$ as the (continuous) set of states $|\psi(t)\rangle=U(t)|\psi\rangle$, where $U(t)$ is a unitary for each $t$, and satisfies a bunch of conditions to make it model a proper time evolution. In particular, we must have $U(t_1)U(t_2)=U(t_1+t_2)$ for all $t_1,t_2$.

(Finally, where Hamiltonians are born) Ok, so we know how to describe time evolutions via families of unitaries. What happens now if the time $t$ is taken to be very small? It is a natural question to ask in "what direction" the state is changing at any given time. Mathematically, this amounts to studying $$\frac1 {dt} (|\psi(dt)\rangle-|\psi(0)\rangle) = \frac{U(dt) - I}{dt}|\psi(0)\rangle.$$ But now look at the kind of expression we got: infinitesimal changes are described by operators of the form $\frac{U(dt)-I}{dt}$ in the limit $dt\to0$, with $t\mapsto U(t)$ such that $U(0)=I$. As it turns out, these objects are skew-Hermitian operators. More precisely, there is always some Hermitian $H$ such that $U(t)=e^{it H}$. And there you have it: if you want to describe infinitesimal evolution of quantum states, you must do so via Hermitian operators called "Hamiltonians".

(Physically, how do Hermitians act?) Hermitians are those matrices which are orthonormally diagonalisable and have real eigenvalues. Their appearing via exponentials means that $e^{iHt}$ are (orthonormally) diagonalisable and have phases as eigenvalues. This means that these are operators which do nothing but add phase shifts between specific components of the input state.

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