Schrödinger equation as a limit of von Neumann equation

Question

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?

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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.

Answer 1

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).

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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.

Answer 2

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, $\partial_t|\Psi(t)\rangle = \partial_t e^{-\frac{i}{\hbar}Ht}|\Psi\rangle = -\frac{i}{\hbar}H|\Psi(t)\rangle$

Answer 3

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, $$(i \partial_t \langle \psi |) | \psi \rangle = - \langle \psi | H | \psi \rangle $$ $$i \partial_t | \psi \rangle = H | \psi \rangle $$