0
$\begingroup$

I found two ways to define a wave function in Heisenberg picture,

  1. $| \psi(t) \rangle_\mathrm{H}=\mathrm e^{\mathrm i H t/\hbar} | \psi(t) \rangle_\mathrm{S}$ which further gives $|\psi(t) \rangle_\mathrm{H}=\psi(t_0) \rangle_\mathrm{H}$, $t_0$ is the initial time.

2 $| \psi(t) \rangle_\mathrm{H}=U(t,t_0)^{+} | \psi(t) \rangle_\mathrm{S}$ where $U(t,t_0)^{+}=\mathrm e^{\mathrm i H( t-t_0)/\hbar}$

here $U(t,t_0)$ is the time-evolution operator

e.g. page 5 of http://bohr.physics.berkeley.edu/classes/221/1112/notes/tevolut.pdf

for this case, we will have $|\psi(t) \rangle_\mathrm{H}=\psi(t_0) \rangle_\mathrm{S}$,

so my questiones is does the difference between these two definitions lies in whether we choose $t_0=0$? Then in case 1, why don't we just write $|\psi(t) \rangle_\mathrm{H}=\psi(0) \rangle_\mathrm{H}$ instead.

$\endgroup$
2
$\begingroup$

In the Heisenberg picture, states are defined not to evolve under the Hamiltonian (instead, the operators are the objects that evolve), so the notation $|\psi(t)\rangle_H$ doesn't make any sense. Usually what is written is $|\psi\rangle_H=|\psi(t_0)\rangle_S$.

The particular time $t_0$ is chosen so that the states (which evolve under the Hamiltonian in the Schrodinger picture but do not in the Heisenberg picture) and operators (which evolve under the Hamiltonian in the Heisenberg picture but do not in the Schrodinger picture) are the same in both pictures at $t_0$. The particular value of $t_0$ is dependent on the particular system you're studying, and generally is taken to be the moment at which the system "starts to evolve" under the Hamiltonian. This is commonly taken to be $t_0=0$, but this is mostly a matter of convention.

$\endgroup$
1
$\begingroup$

The answer given previously is correct, in the Heisenberg picture the states are suposed to remain unchanged in time, while the operators do evolve in time. However the relations you present are valid in the Schrödinger picture where the time evolution of states is. $$|\psi (t)⟩ = \hat{U}(t,t_0)|\psi (t_0)⟩$$ So we need to find the $U$ operator, there are different ways to do that, an easy one is to consider the Schrödinger equation for $|\psi(t)⟩$ $$i\hbar \frac{\partial}{\partial t}|\psi(t)⟩=\hat{H}|\psi(t)⟩\\ i\hbar \frac{\partial}{\partial t}\hat{U}(t,t_0)|\psi (t_0)⟩=\hat{H}\hat{U}(t,t_0)|\psi (t_0)⟩$$

Because $|\psi (t_0)⟩$ is fixed, we get an equation for $\hat{U}(t,t_0)$ $$i\hbar \frac{\partial}{\partial t}\hat{U}(t,t_0)=\hat{H}\hat{U}(t,t_0)$$

Whose solution is given by $$\hat{U}(t,t_0)=\exp{\left[-\left(\frac{i}{\hbar}\right) \int_{t_0}^t dt'\hat{H}(t') \right]}$$

Because in general the hamiltonian is a function of time. In the case that the we are dealing with time independent hamltonians, the time evolution operator is simply $$\hat{U}(t,t_0)=\exp{\left[-\left(\frac{i}{\hbar}\right)\hat{H}(t-t_0)\right]}$$

When the hamiltonian depends on time we end up with the so called Dyson Series

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.