How would I derive the Schrödinger Equation as a limit of the von Neumann equation?

The quantum Liouville equation (von Neumann equation) is given by $$i \hbar \: \partial_t \rho = [ H, \rho ] \quad .$$ Similarly, the Schrödinger equation reads $$ i \hbar \: \partial_t |\psi\rangle = H |\psi \rangle \quad .$$ I'm assuming that the Schrödinger equation is a limiting case of the Liouville (mixed states to pure state), but how would I write that mathematically?

Here is my attempt... My guess is that it goes something like $$\rho = \sum_i p_i | \psi_i \rangle \langle \psi_i | \quad .$$ Insert that into the Quantum Liouville equation: \begin{align} i \hbar \: \partial_t \rho &= \sum_i p_i [ H, | \psi_i \rangle \langle \psi_i |]\\ i \hbar \: \partial_t \rho &= \sum_i p_i (H | \psi_i \rangle \langle \psi_i | - | \psi_i \rangle \langle \psi_i | H )\\ i \hbar \: \partial_t \rho &= \sum_i p_i E_i ( | \psi_i \rangle \langle \psi_i | - | \psi_i \rangle \langle \psi_i | ) = 0 . \end{align} Which obviously doesn't work. Can anyone help?

For reference, I'm looking notation from here on Wikipedia.

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    $\begingroup$ What limit do you mean? Do you mean to resemble the Schrödinger equation of $\rho$ is pure? $\endgroup$ Jan 11 at 22:28
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    $\begingroup$ "This answer seems a bit non-sensical to me." You only plugged in your expression for rho in terms of psi on the RHS... why don't you also plug it in on the LHS and see where you get... I mean the top equation, not the bottom equation, which, yes, doesn't really make a lot of sense at first look... $\endgroup$
    – hft
    Jan 11 at 22:35
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    $\begingroup$ To go from a mixed state to a pure state, take all p's zero except for one. What you wrote is nonsense, as guessed. $\endgroup$ Jan 11 at 22:35
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    $\begingroup$ $H|\psi \rangle = E| \psi \rangle$, you wrote $H|\psi \rangle = E$, so no wonder you get nonsense. Done properly, you will get $\partial_t \rho=0$, as expected for $\rho$ diagonal in the basis of $H$. But why would you derive SE from NE in the first place? SE is more fundamental and NE naturally follows from it. $\endgroup$
    – John
    Jan 11 at 22:49

3 Answers 3


The quantum Liouville equation (von Neumann equation) is given by $$i \hbar \: \partial_t \rho = [ H, \rho ] \quad .$$ Similarly, the Schrödinger equation reads $$ i \hbar \: \partial_t |\psi\rangle = H |\psi \rangle \quad .$$

I'm assuming that the Schrödinger equation is a limiting case of the Liouville (mixed states to pure state), but how would I write that mathematically?

I would not describe the situation this way, but I would say they are certainly related. If you wanted my opinion on which is primal, I would say the Schrodinger equation.

Anyways, take your definition of $$ \rho = \sum_i p_i |\Psi_i\rangle\langle\Psi_i| $$ and take the "limit" where all $p_i\to 0$ except for one of them $p_0 \to 1$. This is called a "pure state," and pure states are the main purview of the Schrodinger equation, so we will be able to derive a relationship.

For the pure state we have: $$ \rho_{\text{pure}} = |\Psi_0\rangle\langle\Psi_0|\;. $$

Setting $\hbar=1$, the Schrodinger equation for our state $|\Psi_0\rangle$ is: $$ i\partial_i|\Psi_0\rangle = H|\Psi_0\rangle\;,\tag{A} $$ and its Hermitian conjugate is: $$ -i\partial_i\langle\Psi_0| = \langle\Psi_0|H.\tag{B} $$

Therefore, the Liouville-von-Neumann equation follows from the Schrodinger equation since: $$ i\partial_t\rho_{\text{pure}}=i\partial_t(|\Psi_0\rangle\langle\Psi_0|) $$ $$ = i\left(H|\Psi_0\rangle\langle\Psi_0| - |\Psi_0\rangle\langle\Psi_0|H\right)\tag{C} $$ $$ =i[H,\rho_{\text{pure}}]\;, $$ where the third line (tagged "C") follows by plugging in the the two forms of the Schrodinger equation in Eq. (A) and Eq. (B).

Update (writing up TF's comments as a answer to the converse question):

Assuming the Hamiltonian is time-independent, I can formally solve the LvN equation by writing: $$ \rho(t) = U(t)\rho(0)U^\dagger(t)\;, $$ where $U(t) = e^{-iHt}$.

Therefore, for a pure state: $$ \rho(t) = |\Psi(t)\rangle\langle\Psi(t)| = U(t)|\Psi(0)\rangle\langle\Psi(0)|U^\dagger(t)\;, $$ where we now identify $$ |\Psi(t)\rangle = U(t)|\Psi(0)\rangle\;. $$

Now, we see that $$ \partial_t|\Psi(t)\rangle = (\partial_t U(t))|\Psi(0)\rangle =-iH|\Psi(t)\rangle\;, $$ which is the Schrodinger equation.

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    $\begingroup$ I think the OP is interested in the converse statement, i.e. deriving the Schrödinger equation from the von Neumann equation. Which, I think, is not much harder at all to derive. $\endgroup$ Jan 11 at 22:54
  • $\begingroup$ Yeah, I think you are probably right, but I'm not sure if that is possible. At least, I don't immediately know how to do it... Do you know if it is even possible? @TobiasFünke $\endgroup$
    – hft
    Jan 11 at 22:56
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    $\begingroup$ Well, I guess something like this (here for time-independent Hamiltonian, but I think this can be easily generalized): The general solution of the von Neumman equation reads $\rho(t) = U(t) \rho_0 U^\dagger(t)$ with $U(t)=e^{-iHt}$, modulo typos. Here we are interested in the case where $\rho_0$ and thus $\rho(t)$ is pure, i.e. $\rho_0=|\psi\rangle\langle\psi|$ for some normalized $\psi$. Now it is easy to see that $|\psi(t)\rangle = U(t)|\psi\rangle$ solves the Schrödinger equation subject to the corresponding initial condition. $\endgroup$ Jan 11 at 22:59

The structure of this question has changed over the course of the answers. I think it is worth pointing out first the error in your derivation, then addressing what appears to be your real question in the comments to your answer, why can we assume that the time evolution of the density matrix is given by $\rho(t)=U(t)\rho(t=0)U^\dagger(t)$?

However, if I have misinterpreted please let me know.

Error in your derivation:

You implicitly assume the density matrix is diagonal in the basis of energy eigenstates. A general density matrix will have off-diagonal terms and is composed of sums of any states $|\psi\rangle\langle\psi|$. As an example in the spin-1/2 case, $\rho = \frac{1}{2}(|s_z+\rangle\langle s_z+|+|s_y+\rangle\langle s_y+|)$ has off-diagonal terms in the $s_z$ basis. If you only considered combinations of $|s_z+\rangle\langle s_z+|$ and $|s_z-\rangle\langle s_z-|$, then you would see trivial time evolution in a magnetic field in the $\hat{z}$ direction.

Take a general state $|\Psi\rangle=\sum_j c_j|\psi_j\rangle$, where $|\psi_j\rangle$ are eigenstates of the Hamiltonian. Now the density matrix is of the form, $\sum_i p_i|\Psi_i\rangle\langle\Psi_i|$, expand it in terms of definite energy eigenstates, $ \sum_i p_i|\Psi_i\rangle\langle\Psi_i|=\sum_i p_i\sum_{jk}c^i_j c^{i*}_k |\psi_j\rangle\langle\psi_k|$ allowing for off-diagonal terms. Now step through your derivation,

\begin{align*} \partial_t \rho &=-\frac{i}{\hbar}\lbrack H, \rho \rbrack\\ &=-\frac{i}{\hbar}\lbrack H, \sum_i p_i\sum_{jk}c^i_j c^{i*}_k |\psi_j\rangle\langle\psi_k| \rbrack\\ &=-\frac{i}{\hbar}\sum_i p_i\sum_{jk}c^i_j c^{i*}_k\lbrack H, |\psi_j\rangle\langle\psi_k| \rbrack\\ &=-\frac{i}{\hbar}\sum_i p_i\sum_{jk}c^i_j c^{i*}_k\bigg( H |\psi_j\rangle\langle\psi_k| - |\psi_j\rangle\langle\psi_k|H \bigg)\\ &=-\frac{i}{\hbar}\sum_i p_i\sum_{jk}c^i_j c^{i*}_k(E_j-E_k)|\psi_j\rangle\langle\psi_k| \end{align*}

Which clearly vanishes for diagonal elements (the only ones you considered), but has non-trivial time evolution for the off-diagonal terms.

The missing steps:

Now briefly let's forget about representing the density matrix as an operator on some Hilbert space of states, let's just take it as a linear mapping of observables to expectation values (along with a few other traits that don't enter in here). I want to argue that the form of the time evolution must be $\rho(t)=U(t)\rho U^\dagger(t)$.

I'll first give an argument that is more algebraic and intuitive (I think), then I'll follow it up with plug and chug through LvN equation to rearrange things into the Schrodinger equation.

first argument

We can take a general mixed state as a convex combination of pure states, i.e. $\rho^M=(1-a)\rho_1 + a\rho_2$. Now the expectation value of an observable in the mixed state is related to that in the pure state by, $\langle A\rangle_{\rho^M}=\rho^M(A)=(1-a)\rho_1(A)+a\rho_2(A)=(1-a)\langle A\rangle_{\rho_1}+a\langle A\rangle_{\rho_2}$

My point in stepping through this, is that if we consider this true at two different times we see that the mixed states time evolution is determined by the pure states, and that the coefficients do not evolve in time. i.e. $\rho^M(t)=(1-a)\rho_1(t) + a\rho_2(t)$

Now we can choose to represent this in some Hilbert space in which case our pure states become projection operators $\rho_i=|\Psi_i\rangle\langle\Psi_i|$. The pure states stay pure states, so $\rho\rightarrow\rho(t)$, meaning $|\Psi\rangle\langle\Psi|\rightarrow|\Psi(t)\rangle\langle\Psi(t)|$ (since it remains a projection operator). Now the time evolution has to be mapping a normalized basis state $|\Psi\rangle$ into another normalized basis state $|\Psi(t)\rangle$, which is the definition of a unitary operator.

Now take the unitary operator as $U(t)=e^{-iKt}$ where $K$ is some currently unknown Hermitian operator. This of course means that the basis states satisfy $\partial_t|\Psi\rangle = -iK|\psi\rangle$. If we consider a pure state we see that it must satisfy, $\partial_t \rho = -i\lbrack K, \rho\rbrack$. And finally taking the LvN equation as given, we see $K=\frac{1}{\hbar}H$, and $U(t)=e^{-i\frac{1}{\hbar} Ht}$

second argument

Now I want to show that the LvN equation also directly implies unitary time evolution, ignoring my arguments above. Note that the LvN equation actually determines all orders of time derivatives of the density matrix. The second derivative given as, \begin{align} \partial_t \dot{\rho}&=\partial_t \bigg(-\frac{i}{\hbar}\lbrack H, \rho \rbrack \bigg)\\ &=-\frac{i}{\hbar}\lbrack H, \partial_t\rho \rbrack \\ &=-\frac{i}{\hbar}\lbrack H, -\frac{i}{\hbar}\lbrack H,\rho \rbrack\rbrack \\ &=-\frac{1}{\hbar^2}\lbrack H, \lbrack H,\rho \rbrack\rbrack \\ \end{align}

We assume the time evolution of $\rho$ is analytic, then a Taylor expansion gives,

\begin{align} \rho(t) &= \rho|{t=0} + \partial_t\rho|_{t=0}+ \frac{1}{2}\partial^2_t\rho|_{t=0} + ...\\ &= \rho -\frac{i}{\hbar}\lbrack H, \rho \rbrack t -\frac{1}{2\hbar^2}\lbrack H, \lbrack H,\rho \rbrack\rbrack t^2 + ...\\ &= \rho -\frac{i}{\hbar}(H\rho-\rho H)t-\frac{1}{2\hbar^2}(H^2\rho - H\rho H - H\rho H + \rho H^2)t^2 + ...\\ &= (1 - \frac{i}{\hbar}Ht - \frac{1}{2\hbar^2}H^2t^2 + ...)\rho(1+\frac{i}{\hbar}Ht-\frac{1}{2\hbar^2}H^2t^2+..)\\ &= e^{-\frac{i}{\hbar}Ht}\rho e^{+\frac{i}{\hbar}Ht}\\ &= U(t)\rho U^\dagger(t) \end{align} giving the unitary time evolution. The difficult step is from step 3 to 4, I would do it in reverse, take step 4, expand out the terms, only keep terms up to order $t^2$ then rearrange into step 3.

Now of course we can follow through the motions, take $\rho = \sum_i p_i |\Psi_i\rangle\langle\Psi_i|$, then $\rho(t) = \sum_i p_i U(t)|\Psi_i\rangle\langle\Psi_i| U^\dagger(t) = \sum_i p_i|\Psi_i(t)\rangle\langle\Psi_i(t)|$. Finally, $\partial_t|\Psi(t)\rangle = \partial_t e^{-\frac{i}{\hbar}Ht}|\Psi\rangle = -\frac{i}{\hbar}H|\Psi(t)\rangle$

  • $\begingroup$ I'll focus on your first argument, if I understand correctly, you introduce a norm preserving mapping and parameterize the mapping with variable t. Since the mapping is norm preserving, then the update operator must be a Lie group element which has an associated Lie algebra and manifold derivative. Finally, given that the pure state comprises rho in the LvN eqn, you can infer the Lie group element must be the Hamiltonian. This seems answer both my original question and my subsequent iterations with Tobias. $\endgroup$
    – ThomasTuna
    Aug 2 at 3:31
  • $\begingroup$ I'll focus on the second argument, you start from the LvN equation, and rather than assume $U = e^{i H t / \hbar}$ you derive that the solution must have a certain form. $\rho(t) = U(t) \rho U^\dagger(t)$ where $U(t)$ must be $U = e^{i H t / \hbar}$. This leads you to expression $\partial_t e^{iHt/\hbar} | \Psi \rangle = \frac{i}{\hbar}H e^{iHt/\hbar} | \Psi \rangle$ which is Schodinger's equation. This argument also seems answer both my original question and my subsequent iterations with Tobias. $\endgroup$
    – ThomasTuna
    Aug 2 at 3:47
  • $\begingroup$ Yes that is how I would summarize both arguments. $\endgroup$
    – TEH
    Aug 2 at 3:55

Here was an answer provided to me by a friend that I am also happy with. It resembles the one from this textbook.

$$i \partial_t \rho = H \rho - \rho H $$ Assume $\rho = | \psi \rangle \langle \psi | $. $$i \partial_t | \psi \rangle \langle \psi |= H | \psi \rangle \langle \psi | - | \psi \rangle \langle \psi | H $$ product rule $$i (\partial_t | \psi \rangle ) \langle \psi | + i | \psi \rangle (\partial_t \langle \psi |) = H | \psi \rangle \langle \psi | - | \psi \rangle \langle \psi | H $$ dot with $| \psi \rangle$ and assume normalized $\langle \psi | \psi \rangle =1 $ $$i (\partial_t | \psi \rangle ) + i | \psi \rangle (\partial_t \langle \psi |)| \psi \rangle = H | \psi \rangle - | \psi \rangle \langle \psi | H | \psi \rangle$$ This equation separates into two Schrodinger eqn's, $$(i \partial_t \langle \psi |) | \psi \rangle = - \langle \psi | H | \psi \rangle $$ $$i \partial_t | \psi \rangle = H | \psi \rangle $$

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    $\begingroup$ Wait... you are using the Schrodinger equation in your derivation... So, I'm not sure this answers your question to show that the Schrödinger equation is a limiting case of the LvN equation... But if you are OK with postulating the schrodinger equation to begin with then, I guess, yes this works... $\endgroup$
    – hft
    Jan 11 at 23:24
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    $\begingroup$ Maybe there's something I'm missing, but in assuming that $U(t) = e^{i Ht}$ and $| \psi(t) \rangle = U(t) | \psi(0) \rangle$ you also have assumed (implicitly) that $| \psi(t) \rangle$ satisfies a Lie Manifold diff equ of the form $\partial_t y = A y$. Thus to me, your solution implicitly assumes 'the schrodinger equation' the same as my friend's does explicitly. See "Geometric Numerical Integration" Hairer, Wanner, Lubich (2002) some section on Lie Groups would probably have the proof. $\endgroup$
    – ThomasTuna
    Jan 11 at 23:41
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    $\begingroup$ I found it, see "IV.7 Methods Based on the Magnus Series Expansion" equ 7.1 and 7.2 $\endgroup$
    – ThomasTuna
    Jan 11 at 23:47
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    $\begingroup$ I put a direct derivation from your assumption to the Schrodinger equation here. $$e^{ i \epsilon H} \psi(t) = \psi(t+\epsilon)$$ $$(1 + i \epsilon H) \psi(t) = \psi(t+\epsilon)$$ $$H \psi(t) = i (\psi(t) - \psi(t+\epsilon))/\epsilon$$ $$H \psi(t) = -i \partial_t \psi$$ $\endgroup$
    – ThomasTuna
    Jan 12 at 1:33
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    $\begingroup$ No, there is no such assumption in the answer of @hft . Rather, this is derived. The procedure is: Show that the general solution of the von Neumann equation is of the form shown there; then simply show that the such defined vector $|\psi(t)\rangle$ obeys the Schrödinger equation. Your proof, OTOH, is meaningless, as you assume the thing you want to prove. Moreover, the book you cite here goes the other way around, i.e. it derives the vN eq. from the SE, so it proves a different thing; which is also done in the other answer. But again: You assume that $\psi$ obeys the SE just to derive that... $\endgroup$ Jan 12 at 7:40

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