2

Consider a (free and massless) real scalar field $\phi(x)$ with Hamiltonian $$ H := \int d^3\mathbf{x}\; \bigg[ \frac{1}{2} \pi^2(\mathbf{x}) + |\nabla\phi(\mathbf{x})|^2 \bigg] $$ where $\pi(\mathbf{x})$ is the conjugate momentum field (where we have $[\phi(\mathbf{x}), \pi(\mathbf{y}) ] = i \delta^{(3)}(\mathbf{x} - \mathbf{y})$). We work in the Schrodinger picture here.

The density matrix for the field $\rho(t)$ satisfies the von Neumann equation $$ \frac{d \rho(t)}{d t} = - i [ H, \rho(t) ] \ . $$ Let $| \phi_a \rangle$ be the eigenkets of the field operator $\phi(\mathbf{x})$ with eigenvalue $\phi_a(\mathbf{x})$. This means $$ \hat{\phi}(\mathbf{x}) | \phi_a \rangle = \phi_a(\mathbf{x}) | \phi_a \rangle \ . $$ From here we will derive an equation of motion for the components of the density matrix in the field basis, where $$ \rho_{\phi_1 \phi_2}(t) := \langle \phi_1 | \rho(t) | \phi_2 \rangle \ . $$ By using the fact that for any state $| \chi \rangle$ we have $$ \langle \phi_1 | \pi(\mathbf{x}) | \chi \rangle = - i \frac{\delta}{\delta \phi_1(\mathbf{x})} \langle \phi_1 | \chi \rangle \ , $$ it is straightforward to use the von Neumann equation to derive the equation for the off-diagonal components $\phi_1 \neq \phi_2$ that $$ \frac{d \rho_{\phi_1\phi_2}}{dt} = - \frac{i}{2} \int d^3\mathbf{x} \left( - \frac{\delta^2}{\delta \phi_1(\mathbf{x})^2} + |\nabla \phi_1(\mathbf{x})|^2 + \frac{\delta^2}{\delta \phi_2(\mathbf{x})^2} - |\nabla \phi_2(\mathbf{x})|^2 \right) \rho_{\phi_1\phi_2}(t) \ . $$ However strangely for diagonal components $\phi_1 = \phi_2$ I am finding that you get $$ \frac{d \rho_{\phi_1\phi_1}}{dt} = 0 \ . $$ What is the meaning of this? This seems to suggest that the diagonal components are always constant, no matter what bizarre state the field starts out in. This somehow seems wrong.

QuantumEyedea
  • 3,659
  • 2
    It might well be instructive to write the NR QM analog for $\langle x|\rho|x'\rangle$ first, in your first paragraph. – Cosmas Zachos Jan 19 '21 at 01:19
  • @CosmasZachos I get the equation $\frac{d}{dt}\rho_{xx'}=i\left(\frac{1}{2m}( \partial_x^2 - \partial_{x'}^2 ) - \frac{m\omega^2}{2} (x^2 - x^{\prime 2}) \right)\rho_{xx'} $ where $\rho_{xx'} := \langle x | \rho(t) | x' \rangle$. When setting $x=x'$ I find $\frac{d}{dt} \rho_{xx} = 0$. I still don't understand. For example, taking an initial state which is a superposition of energy eigenstates $\rho(0) = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)(\langle 0 | + \langle 1 |)$ I would expect the cross terms to have time-dependence once evolved in $t$. Why would $\rho_{xx}(t)$ be static then? – QuantumEyedea Jan 19 '21 at 16:24
  • Somewhat linked. – Cosmas Zachos Jan 19 '21 at 17:07

2 Answers2

2

To answer your comment on the simplest NR QM pure state illustration you chose, $$ \rho (t)= \frac{1}{2} (e^{-it\omega/2}|0\rangle +e^{-it3\omega/2}|1\rangle)(e^{it\omega/2}\langle 0| +e^{it3\omega/2}\langle 1 |), $$ express it in the coordinate basis, yielding Hermite functions, $$ \rho_{xx'}(t)= \frac{1}{2} \left ( \psi_0(x) +e^{-it\omega} \psi_1(x)\right )~~\left ( \psi^*_0(x')+e^{it\omega} \psi_1^* (x') \right ), $$ so $x=x'$ collapses to a real function, a probability density.
Its trace, if you will, in this basis, is the $\int dx \rho=1$, naturally yielding unity. The Landau-Lifshitz NR QM text covers this nicely.

Your point in the comment that this collapsed real function is time dependent $$ \frac{1}{2} ( |\psi_0(x)|^2 + |\psi_1(x)|^2+ \cos (\omega t) \psi_0(x) \psi_1(x) ), $$ is well put. It oscillates with time, as it should. You took the $x'\to x$ limit incorrectly.

The vanishing of the r.h.side of this equation as $x'\to x$ is not sound, given differential operators... there are singular terms omitted. Here is a formal wisecrack possibly evoking this: $$ (\partial_x - \partial_{x'}) e^{ic x} e^{-ic x'}= 2ic e^{ic (x-x')}. $$ But, according to your $x'\to x$ limit the left hand side vanishes, whereas the right hand side not, $=2ic$. Your vanishing quantity is a mirage.

But... you really have seen this before in the diagonal sector, with the probability current! $\rho(x,x)=\rho(x)= |\psi(x)|^2$, the probability density, which is not time invariant at all: probability flows! $$ \partial_t \rho(x) = \partial_x\left ( \frac{i\hbar}{2m} (\psi^*(x)\partial_x \psi(x) - \psi(x) \partial_x \psi^*(x) ) \right )\\ = \frac{i\hbar}{2m} \left ( \psi^*(x)\partial_x^2 \psi(x) - \psi(x) \partial_x^2 \psi^*(x) \right ) \neq 0 ! $$ For your flip-flop wave function, you may confirm on both sides that it is the oscillating quantity $$ -\omega\sin(\omega t) \psi_0 (x) \psi_1(x) . $$

  • Takeaway: the diagonal components are not static. The limit $(\partial_x^2-\partial^2_{x'})\rho_{xx'}\neq 0$, when taken carefully. First the derivatives operate on "half" of $\rho_{xx'}$, with the time variation encoded in the imaginary part, and only then is $x'\to x$ taken.

  • It is the probability current inflow. So the premise of the question, "why", is off: should be missing.

Cosmas Zachos
  • 62,595
  • Setting $x=x'$, I use the fact that $\psi_{0,1}(x)$ are real functions and get the formula $\rho_{xx'}(t) = \frac{1}{2} |\psi_0(x)|^2 + \frac{1}{2} |\psi_1(x)|^2 + \cos(\omega t) \psi_0(x) \psi_1(x)$, which has some time-dependance. I must be missing something though? – QuantumEyedea Jan 19 '21 at 17:07
  • 1
    @CosmasZachos sorry, had a mental blip. Have deleted the comment although in that case I do get the $cos(\omega t)$ that OP gets. – jacob1729 Jan 19 '21 at 17:21
  • @CosmasZachos Interesting. However, I think it makes sense that the time derivative is real --- my proof for this is to consider the $xx$-component of the RHS of the von Neumann eq: $ - i \langle x | [ H, \rho ] | x\rangle = -i \big( \langle x | H\rho| x\rangle - \langle x | H\rho| x\rangle^{\ast} \big) = 2 \mathrm{Im}[ \langle x | H\rho| x\rangle ]$, which is real. However what's bizarre is that it seems $\mathrm{Im}[ \langle x | H\rho| x\rangle ] = 0$ if I naively write this out in terms of the operators in $H$....I think your last paragraph must hold the answer. – QuantumEyedea Jan 19 '21 at 20:10
  • 1
    Right; on the one hand. On the other hand, your explicit expression in the comment is i times a real operator acting on a real function, in the diagonal limit! – Cosmas Zachos Jan 19 '21 at 20:14
  • @CosmasZachos Your connection to the probability current density makes perfect sense, and I am now convinced that $\lim_{x \to x'} ( \partial_x^2 - \partial_{x'}^2 ) \rho(x,x') \neq 0$. But how does one compute this limit in the general case that $\rho$ is not pure? – QuantumEyedea Jan 21 '21 at 16:55
  • 1
    As I indicate in the penultimate bullet, you first perform the derivatives, and then take the limit $x'\to x$, pure state or not. The potential does not matter, as there are no limit subtleties there, just as the probability current lacks potential terms. – Cosmas Zachos Jan 21 '21 at 17:10
  • @CosmasZachos That makes sense. What is puzzling about the potential terms not contributing is its implication on correlation functions. For example the operator $\langle x^2 \rangle = \mathrm{Tr}[x^2\rho(t)] = \int dx\ x^2 \rho_{xx}$. The fact that no potential terms can contribute to the dynamics of $\rho_{xx}$ would imply that $\partial_t \rho_{xx}$ is the same no matter what complicated potential you include for the system, and so $\partial_t \langle x^2 \rangle$ is the same no matter what potential term you include. This doesn't seem right somehow... – QuantumEyedea Jan 22 '21 at 19:48
  • 1
    Νοno, wrong conclusion: You don't see the effects of the potential directly in the probability current, but, as above, you see them in the wavefunction! By Ehrenfest's theorem, the expectation value variation $\langle [\hat p^2,\hat x^2]\rangle$ appears, the same, but it is not the same for any and all wavefunctions, and is certainly non vanishing for the above example! – Cosmas Zachos Jan 22 '21 at 20:22
2

It is not true that $\dot\rho(\phi_1,\phi_1)=0$: the object $\rho(\phi_1,\phi_1)$ does not exist.

It doesn't make sense to talk about $\rho$ for $\phi_1=\phi_2$. The object $\rho$ is a distribution (as a function of $\phi$, not $x$) so it is meaningless to "evaluate" it on a given function.

Indeed, $$ \langle\phi_1|\phi_2\rangle=\delta(\phi_1-\phi_2) $$ where $\delta$ is the Dirac delta in the space of functionals (i.e., with respect to functional integration, not regular integration). See this PSE post for some (formal) applications of this formula.

The density matrix $\rho(\phi_1,\phi_2)$ only makes sense under the integral sign: $$ \int_{\mathcal F^2}\rho(\phi_1,\phi_2)F[\phi_1,\phi_2]\,\mathrm d\phi_1\mathrm d\phi_1 $$ where $\mathcal F$ denotes some reasonable function space and $F$ is a healthy functional on it.

AccidentalFourierTransform
  • 53,248
  • 20
  • 131
  • 253
  • Interesting. But this is confusing for me because $\rho_{\phi_1\phi_2}$ with $phi_1 = \phi_2 = \phi$ seems to appear in traces with $\hat{\phi}$ operators. For example, $\mathrm{Tr}[\phi(x) \phi(y) \rho ] = \int d\phi; \phi(x) \phi(y) \langle \phi | \rho | \phi \rangle $. This is just a correlation function though right? – QuantumEyedea Jan 21 '21 at 14:10
  • 1
    @QuantumEyedea The correct expression to look at is $\int \phi_1(x)\phi_2(y)\langle \phi_1|\rho|\phi_2\rangle,\mathrm d\phi\mathrm d\phi_2$. The density matrix is a bi-distribution: a distribution with respect to two arguments. – AccidentalFourierTransform Jan 21 '21 at 15:25
  • I don't understand why $\int\phi_1(x)\phi_2(x)\langle\phi_1|\rho|\phi_2\rangle d\phi_1d\phi_2$ is a correlator? Taking $\mathrm{Tr}[\phi(x)\phi(y)\rho]$ as a definition for the correlator then I can manipulate this into $\int\langle\phi_2| \phi(x)\phi(y)\rho|\phi_2 \rangle d\phi_2 = \int\ \phi_1(x) \phi_2(y) \langle \phi_2 | \phi_1 \rangle\langle\phi_1|\rho |\phi_2\rangle d\phi_1 d\phi_2$. This doesn't match your expression: is $\mathrm{Tr}[\phi(x) \phi(y) \rho]$ not the right definition? – QuantumEyedea Jan 21 '21 at 15:42