In quantum mechanics, the Schrödinger equation describes how a system changes with time. It does this by relating changes in the state of the system to the energy in the system (given by an operator called the Hamiltonian). Therefore, once the Hamiltonian is known, the time dynamics are in principle known. All that remains is to plug the Hamiltonian into the Schrödinger equation and solve for the system state as a function of time.[1] [2]
Often, however, the Schrödinger equation is difficult to solve (even with a computer). Therefore, physicists have developed mathematical techniques to simplify these problems and clarify what is happening physically. One such technique is to apply a unitary transformation to the Hamiltonian. Doing so can result in a simplified version of the Schrödinger equation which nonetheless has the same solution as the original.
A unitary transformation (or frame change) can be expressed in terms of a time-dependent Hamiltonian
H(t)
U(t)
H\toUH{U\dagger}+i\hbar{
U |
U\dagger} =:\breve{H} (0)
The Schrödinger equation applies to the new Hamiltonian. Solutions to the untransformed and transformed equations are also related by
U
\psi(t)
U\psi(t)
Recall that by the definition of a unitary matrix,
U\daggerU=1
\psi | =- |
i | |
\hbar |
H\psi
we can therefore insert
U\daggerU
H/\hbar
U
U\psi | =- |
i | |
\hbar |
\left(UHU\dagger\right)U\psi (1)
Next, note that by the product rule,
d | \left(U\psi\right)= | |
dt |
U | \psi+U |
\psi |
Inserting another
U\daggerU
U\psi | = |
d | |
dt |
(U\psi) -
U |
U\daggerU\psi (2)
Finally, combining (1) and (2) above results in the desired transformation:
d | |
dt |
(U\psi)=-
i | |
\hbar |
(UH{U\dagger}+i\hbar
U |
{U\dagger}) (U\psi) \left(3\right)
If we adopt the notation
\breve{\psi}:=U\psi
(3)
d | |
dt |
\breve{\psi}=-
i | |
\hbar |
\breve{H}\breve{\psi} \left(4\right)
which can be rewritten in the form of the original Schrödinger equation,
\breve{H}\breve{\psi}= i\hbar{\operatorname{d}\breve{\psi}\over\operatorname{d}t}.
The original wave function can be recovered as
\psi=U\dagger\breve{\psi}
Unitary transformations can be seen as a generalization of the interaction (Dirac) picture. In the latter approach, a Hamiltonian is broken into a time-independent part and a time-dependent part,
H(t)=H0+V(t) (a)
In this case, the Schrödinger equation becomes
\psiI | = - |
i | |
\hbar |
iH0t/\hbar | |
\left(e |
V
-iH0t/\hbar | |
e |
\right)\psiI
\psiI=
iH0t/\hbar | |
e |
\psi
The correspondence to a unitary transformation can be shown by choosing . As a result,
{U\dagger}(t)=\exp\left[{-iH0t}/\hbar\right].
Using the notation from
(0)
\breve{H}=U\left[H0+V(t)\right]U\dagger+i\hbar
U |
U\dagger (b)
First note that since
U
H0
\dagger=H | |
UH | |
0 |
which takes care of the first term in the transformation in
(b)
\breve{H}=H0+UV(t)U\dagger+i\hbar
U |
U\dagger
\begin{align}i\hbar
U |
U\dagger&= i\hbar\left({\operatorname{d}U\over\operatorname{d}t}\right)
-iH0t/\hbar | |
e |
\\ &=i\hbar(iH0/\hbar)
+iH0t/\hbar | |
e |
-iH0t/\hbar | |
e |
\\ &=i\hbar\left({iH0}/\hbar\right)\\ &=-H0,\ \end{align}
which cancels with the other
H0
\breve{H}=UVU\dagger
\psiI |
=-
i | |
\hbar |
UVU\dagger\psiI
When applying a general unitary transformation, however, it is not necessary that
H(t)
U(t)
|g\rangle
|e\rangle
H=\hbar\omega{|{e}\rangle\langle{e}|}
\omega
\omegad
H/\hbar=\omega|e\rangle\langlee|+
i\omegadt | |
\Omega e |
|g\rangle\langlee|+\Omega* e
-i\omegadt | |
|e\rangle\langleg|
for some complex drive strength
\Omega
\omega
\omegad
\Omega
Without a drive, the phase of
|e\rangle
|g\rangle
U=ei\omega
H/\hbar\to\Omega
i(\omegad-\omega)t | |
e |
|g\rangle\langlee|+\Omega*
i(\omega-\omegad)t | |
e |
|e\rangle\langleg|
If the driving frequency is equal to the g-e transition's frequency,
\omegad=\omega
\breve{H}/\hbar= \Omega |g\rangle\langlee|+\Omega* |e\rangle\langleg|
From this it is apparent, even without getting into details, that the dynamics will involve an oscillation between the ground and excited states at frequency
\Omega
As another limiting case, suppose the drive is far off-resonant,
|\omegad-\omega|\gg0
|g\rangle
|e\rangle
|e\rangle
These concepts are illustrated in the table below, where the sphere represents the Bloch sphere, the arrow represents the state of the atom, and the hand represents the drive.
The example above could also have been analyzed in the interaction picture. The following example, however, is more difficult to analyze without the general formulation of unitary transformations. Consider two harmonic oscillators, between which we would like to engineer a beam splitter interaction,
gab\dagger+g*a\daggerb
This was achieved experimentally with two microwave cavity resonators serving as
a
b
In addition to the microwave cavities, the experiment also involved a transmon qubit,
c
\omega1
\omega2
\omega1-\omega2=\omegaa-\omegab
Hdrive/\hbar=\Re\left[\epsilon
i\omega1t | |
1e |
+\epsilon2
i\omega2t | |
e |
\right](c+c\dagger).
In addition, there are many fourth-order terms coupling the modes, but most of them can be neglected. In this experiment, two such terms which will become important are
H4/\hbar=g
i(\omegab-\omegaa)t | |
4(e |
ab\dagger+h.c.)c\daggerc
(H.c. is shorthand for the Hermitian conjugate.) We can apply a displacement transformation,
U=D(-\xi1
-i\omega1t | |
e |
-\xi2
-i\omega2t | |
e |
)
c
Hrm{drive}
c\toc+\xi1
-i\omega1t | |
e |
+\xi2
-i\omega2t | |
e |
H/\hbar=
i(\omegab-\omegaa)t | |
g | |
4(e |
ab\dagger+
i(\omegaa-\omegab)t | |
e |
a\daggerb)(c\dagger
* | |
+\xi | |
1 |
i\omega1t | |
e |
* | |
+\xi | |
2 |
i\omega2t | |
e |
)(c+\xi1
-i\omega1t | |
e |
+\xi2
-i\omega2t | |
e |
)
Expanding this expression and dropping the rapidly rotating terms, we are left with the desired Hamiltonian,
H/\hbar=g4
*\xi | |
\xi | |
2 |
i(\omegab-\omegaa+\omega1-\omega2)t | |
e |
ab\dagger+h.c.=gab\dagger+g*a\daggerb
It is common for the operators involved in unitary transformations to be written as exponentials of operators,
U=eX
X\dagger=-X
Y
UYU\dagger=eXYe-X
[(X)n,Y]\equiv\underbrace{[X,...b[X,[X}n,Y]]...b], [(X)0,Y]\equivY,
we can use a special result of the Baker-Campbell-Hausdorff formula to write this transformation compactly as,
eXYe-X=
infty | |
\sum | |
n=0 |
[(X)n,Y] | |
n! |
,
or, in long form for completeness,
eXYe-X=Y+\left[X,Y\right]+
1 | [X,[X,Y]]+ | |
2! |
1 | |
3! |
[X,[X,[X,Y]]]+ … .