# 1.4: Exponential Operators

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Throughout our work, we will make use of exponential operators of the form

$\hat {T} = e^{- i \hat {A}} \label{115}$

We will see that these exponential operators act on a wavefunction to move it in time and space, and are therefore also referred to as propagators. Of particular interest to us is the time-evolution operator, $$\hat {U} = e^{- i \hat {H} t / \hbar},$$ which propagates the wavefunction in time. Note the operator $$\hat{T}$$ is a function of an operator, $$f(\hat{A})$$. A function of an operator is defined through its expansion in a Taylor series, for instance

$\hat {T} = e^{- \hat {i} \hat {A}} = \sum _ {n = 0}^{\infty} \dfrac {( - i \hat {A} )^{n}} {n !} = 1 - i \hat {\hat {A}} - \dfrac {\hat {A} \hat {A}} {2} - \cdots \label{116}$

Since we use them so frequently, let’s review the properties of exponential operators that can be established with Equation \ref{116}. If the operator $$\hat {A}$$ is Hermitian, then$$\hat {T} = e^{- i \hat {A}}$$ is unitary, i.e., $$\hat {T}^{\dagger} = \hat {T}^{- 1}.$$ Thus the Hermitian conjugate of $$\hat {T}$$ reverses the action of $$\hat {T}$$. For the time-propagator $$\vec {U}$$, $$\vec {U}^{\dagger}$$ is often referred to as the time-reversal operator.

The eigenstates of the operator $$\hat {A}$$ also are also eigenstates of $$f(\hat {A})$$, and eigenvalues are functions of the eigenvalues of $$\hat {A}$$. Namely, if you know the eigenvalues and eigenvectors of $$\hat {A}$$, i.e., $$\hat {A} \varphi _ {n} = a _ {n} \varphi _ {n},$$you can show by expanding the function

$f ( \hat {A} ) \varphi _ {n} = f \left( a _ {n} \right) \varphi _ {n} \label{117}$

Our most common application of this property will be to exponential operators involving the Hamiltonian. Given the eigenstates $$\varphi _ {n}$$, then $$\hat {H} | \varphi _ {n} \rangle = E _ {n} | \varphi _ {n} \rangle$$ implies

$e^{- i \hat {H} t / \hbar} | \varphi _ {n} \rangle = e^{- i E _ {n} t / \hbar} | \varphi _ {n} \rangle \label{118}$

Just as $$\hat {D} _ {x} ( \lambda )$$ is the time-evolution operator that displaces the wavefunction in time, $$\hat {D} _ {x} = e^{- i \hat {p} _ {x} x / h}$$ is the spatial displacement operator that moves $$\psi$$ along the $$x$$ coordinate. If we define $$\hat {D}_ {x} = e^{- i \hat {p} _ {x} x / h},$$ then the action of is to displace the wavefunction by an amount $$\lambda$$

$| \psi ( x - \lambda ) \rangle = \hat {D} _ {x} ( \lambda ) | \psi (x) \rangle \label{119}$

Also, applying $$\hat {D} _ {x} ( \lambda )$$ to a position operator shifts the operator by $$\lambda$$

$\hat {D} _ {x}^{\dagger} \hat {x} \hat {D} _ {x} = x + \lambda \label{120}$

Thus $$e^{- i \hat {p} _ {x} \lambda / \hbar} | x \rangle$$ is an eigenvector of $$\hat {x}$$ with eigenvalue $$x + \lambda$$ instead of $$x$$. The operator

$$\hat {D} _ {x} = e^{- i \hat {p} _ {x} \lambda / h}$$ is a displacement operator for $$x$$ position coordinates. Similarly, $$\hat {D} _ {y} = e^{- i \hat {p} _ {y} \lambda / h}$$generates displacements in $$y$$ and $$\hat {D_z}$$ in $$z$$. Similar to the time-propagator $$\boldsymbol {U}$$, the displacement operator $$\boldsymbol {D}$$ must be unitary, since the action of $$\hat {D}^{\dagger} \hat {D}$$ must leave the system unchanged. That is if $$\hat {D} _ {x}$$ shifts the system to $$x$$ from $$x_0$$, then $$\hat {D} _ {x}^{\dagger}$$ shifts the system from $$x$$ back to $$x_0$$.

We know intuitively that linear displacements commute. For example, if we wish to shift a particle in two dimensions, x and y, the order of displacement does not matter. We end up at the same position, whether we move along x first or along y, as illustrated in Figure 7. In terms of displacement operators, we can write

\begin{aligned} \left|x_{2}, y_{2}\right\rangle &=e^{-i b p_{y} / \hbar} e^{-i a p_{x} / \hbar}\left|x_{1}, y_{1}\right\rangle \\ &=e^{-i a p_{x} / \hbar} e^{-i b p_{y} / \hbar}\left|x_{1}, y_{1}\right\rangle \end{aligned}

These displacement operators commute, as expected from $$[p_x,p_y] = 0.$$

Similar to the displacement operator, we can define rotation operators that depend on the angular momentum operators, $$L_x$$, $$L_y$$, and $$L_z$$. For instance,

$\hat {R} _ {x} ( \phi ) = e^{- i \phi L _ {x} / h}$

gives a rotation by angle $$\phi$$ about the $$x$$ axis. Unlike linear displacement, rotations about different axes do not commute. For example, consider a state representing a particle displaced along the z axis, $$| z 0 \rangle$$. Now the action of two rotations $$\hat {R} _ {x}$$ and $$\hat {R} _ {y}$$ by an angle of $$\phi = \pi / 2$$ on this particle differs depending on the order of operation, as illustrated in Figure 8. If we rotate first about $$x$$, the operation

$e^{- i \dfrac {\pi} {2} L / h} e^{- i \dfrac {\pi} {2} L _ {x} / h} | z _ {0} \rangle \rightarrow | - y \rangle \label{122}$

leads to the particle on the –y axis, whereas the reverse order

$e^{- i \dfrac {\pi} {2} L _ {x} / \hbar} e^{- i \dfrac {\pi} {2} L _ {y} / \hbar} | z _ {0} \rangle \rightarrow | + x \rangle \label{123}$

leads to the particle on the +x axis. The final state of these two rotations taken in opposite order differ by a rotation about the z axis. Since rotations about different axes do not commute, we expect the angular momentum operators not to commute. Indeed, we know that

$\left[ L _ {x} , L _ {y} \right] = i \hbar L _ {z}$

where the commutator of rotations about the x and y axes is related by a z-axis rotation. As with rotation operators, we will need to be careful with time-propagators to determine whether the order of time-propagation matters. This, in turn, will depend on whether the Hamiltonians at two points in time commute.

Properties of exponential operators

1. If $$\hat{A}$$ and $$\hat{B}$$ do not commute, but $$[ \hat {A} , \hat {B} ]$$ commutes with $$\hat{A}$$ and $$\hat{B}$$, then

$e^{\hat {A} + \hat {B}} = e^{\hat {A}} e^{\hat {B}} e^{- \frac {1} {2} [ \hat {A} , \hat {B} ]} \label{124}$

$e^{\hat {A}} e^{\hat {B}} = e^{\hat {B}} e^{\hat {A}} e^{- [ \hat {B} , \hat {A} ]} \label{125}$

2. More generally, if $$\hat{A}$$ and $$\hat{B}$$ do not commute,

$e^{\hat {A}} e^{\hat {B}} = {\mathrm {exp}} \left[ \hat {A} + \hat {B} + \dfrac {1} {2} [ \hat {A} , \hat {B} ] + \dfrac {1} {12} ( [ \hat {A} , [ \hat {A} , \hat {B} ] ] + [ \hat {A} , [ \hat {B} , \hat {B} ] ] ) + \cdots \right] \label{126}$

3. The Baker–Hausdorff relationship:

$\left. \begin{array} {r l} {\mathrm {e}^{i \hat {G} \lambda} \hat {A} \mathrm {e}^{- i \hat {G} \lambda} = \hat {A} + i \lambda [ \hat {G} , \hat {A} ] + \left( \dfrac {i^{2} \lambda^{2}} {2 !} \right) [ \hat {G} , [ \hat {G} , \hat {A} ] ] + \ldots} \\ {} & {+ \left( \dfrac {i^{n} \lambda^{n}} {n !} \right) [ \hat {G} , [ \hat {G} , [ \hat {G} , \hat {A} ] ] ] \ldots ] + \ldots} \end{array} \right. \label{127}$

where $$λ$$ is a number.

1.4: Exponential Operators is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Andrei Tokmakoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.