Energy Eigenstates

An energy eigenstate is a state of definite energy: an ideal measurement of the Hamiltonian returns one energy value with probability one. For a normalizable pure state,

$$H\lvert\psi\rangle = E\lvert\psi\rangle.$$

Energy eigenstates are central because the Hamiltonian plays two roles. It is the energy observable, whose spectral projectors define energy probabilities, and it is the generator of closed-system time evolution. For a time-independent Hamiltonian, its spectral representation turns dynamics into phase multiplication.

The phrase needs care. A discrete eigenvalue has normalizable Hilbert-space eigenvectors; a continuous spectral value generally has only generalized, delta-normalized eigenvectors. Degeneracy also means that an energy value belongs to an entire eigenspace, not to one privileged vector.

Eigenvalue Equation and Eigenspaces

For a time-independent self-adjoint Hamiltonian $H$, a normalizable energy eigenvector satisfies

$$H\lvert E,\lambda\rangle = E\lvert E,\lambda\rangle.$$

The eigenvalue $E$ lies in the point spectrum of $H$. The label $\lambda$ distinguishes linearly independent vectors with the same energy. If that eigenspace has finite dimension $g_E$, one may choose an orthonormal basis

$$\left\{ \lvert E,\lambda\rangle \right\}_{\lambda=1}^{g_E}.$$

The corresponding spectral projector is

$$P_E = \sum_{\lambda=1}^{g_E} \lvert E,\lambda\rangle \langle E,\lambda\rvert.$$

Although the basis vectors inside a degenerate eigenspace are not unique, $P_E$ is unique. Replacing them by another orthonormal basis related by a unitary matrix within the eigenspace leaves the projector unchanged.

Any vector in the range of $P_E$ is an energy eigenvector with energy $E$:

$$P_E\lvert\psi\rangle = \lvert\psi\rangle \quad\Longrightarrow\quad H\lvert\psi\rangle = E\lvert\psi\rangle.$$

Other compatible observables may be used to choose the degeneracy labels. For example, angular momentum quantum numbers can refine a degenerate energy eigenspace when the corresponding operators commute with $H$. Those extra labels describe a basis choice within the energy subspace; the energy outcome itself is still $E$.

The general linear-algebra definitions are in Eigenvalues and Eigenstates.

Definite Energy

For a normalized pure state $\lvert\psi\rangle$, the following statements are equivalent for a discrete energy $E$:

$$\begin{gathered} H\lvert\psi\rangle=E\lvert\psi\rangle,\\ P_E\lvert\psi\rangle=\lvert\psi\rangle,\\ \langle\psi\vert P_E\vert\psi\rangle=1. \end{gathered}$$

The last line is the Born-rule statement

$$p(E)=1.$$

For every different discrete energy $E'\ne E$,

$$P_{E'}\lvert\psi\rangle=0, \qquad p(E')=0.$$

This is the precise meaning of definite energy. It does not imply definite position, momentum, angular momentum component, or any other observable incompatible with $H$.

Zero energy variance

For a state in the domain of $H^2$, define

$$\langle H\rangle = \langle\psi\vert H\vert\psi\rangle$$

and

$$(\Delta H)^2 = \left\langle \left( H-\langle H\rangle \right)^2 \right\rangle.$$

The variance can be written as a squared norm:

$$(\Delta H)^2 = \left\lVert \left( H-\langle H\rangle \right) \lvert\psi\rangle \right\rVert^2.$$

Therefore

$$\Delta H=0 \quad\Longleftrightarrow\quad H\lvert\psi\rangle = \langle H\rangle\lvert\psi\rangle.$$

A normalized pure state has definite energy exactly when its energy variance vanishes. The corresponding eigenvalue is its expectation value.

Mixed States with Definite Energy

A mixed state can also have a definite energy. The basis-independent condition is that its entire support lie inside one energy eigenspace:

$$\rho = P_E\rho P_E.$$

Then

$$\operatorname{Tr}(\rho P_E)=1,$$

and every other energy outcome has probability zero.

If the eigenspace is degenerate, $\rho$ may be mixed or may contain coherences between different $\lambda$ labels while still having definite energy. For example,

$$\rho = \sum_{\lambda,\lambda'} \rho_{\lambda\lambda'} \lvert E,\lambda\rangle \langle E,\lambda'\rvert$$

has definite energy $E$ whenever the coefficient matrix is positive and has unit trace.

This must be distinguished from a stationary mixture spanning several energies:

$$\rho_{\mathrm{stat}} = \sum_E p_E\rho_E, \qquad \rho_E=P_E\rho_EP_E.$$

Such a state commutes with $H$ and is stationary, but its energy is not definite unless only one $p_E$ is nonzero. Stationarity and definite energy coincide for pure states but not for mixed states.

Expansion in the Energy Representation

Suppose $H$ has a complete discrete orthonormal set of energy eigenvectors. Any state can be expanded as

$$\lvert\psi\rangle = \sum_{n,\lambda} c_{n\lambda} \lvert E_n,\lambda\rangle,$$

with

$$c_{n\lambda} = \langle E_n,\lambda\vert\psi\rangle.$$

Completeness gives

$$I = \sum_n P_{E_n},$$

so the same expansion can be written without choosing a basis inside each degenerate eigenspace:

$$\lvert\psi\rangle = \sum_n P_{E_n}\lvert\psi\rangle.$$

The norm of each projected component determines the energy probability:

$$p(E_n) = \left\lVert P_{E_n}\lvert\psi\rangle \right\rVert^2.$$

If an explicit degeneracy basis is chosen,

$$p(E_n) = \sum_\lambda \left|c_{n\lambda}\right|^2.$$

The projector formula is canonical; the individual coefficients depend on the basis chosen within the eigenspace.

For a Hamiltonian with continuous or mixed spectrum, sums are supplemented or replaced by spectral integrals. In spectral-measure notation,

$$H = \int_{\mathbb R}E\,dP_H(E),$$

and

$$\lvert\psi\rangle = \int_{\mathbb R}dP_H(E)\lvert\psi\rangle.$$

This notation does not pretend that every spectral value has a normalizable eigenket. See Spectral Decomposition and Discrete and Continuous Spectra.

Energy Measurement

For a pure state and a discrete eigenvalue $E_n$,

$$p(E_n) = \langle\psi\vert P_{E_n}\vert\psi\rangle.$$

For a density operator,

$$p(E_n) = \operatorname{Tr}(\rho P_{E_n}).$$

If the outcome $E_n$ is selected in an ideal projective measurement, the Lüders update is

$$\rho \longmapsto \frac{ P_{E_n}\rho P_{E_n} }{ \operatorname{Tr}(\rho P_{E_n}) }.$$

For a degenerate energy, this update projects onto the entire eigenspace. It does not select a particular $\lambda$ unless the apparatus also resolves an additional compatible observable. The canonical state-update discussion is Degenerate Measurements and Lüders Rule.

For a Borel set of energies $\Delta\subset\mathbb R$, the general probability rule is

$$\Pr(E\in\Delta) = \langle\psi\vert P_H(\Delta)\vert\psi\rangle$$

or

$$\Pr(E\in\Delta) = \operatorname{Tr} \left[ \rho P_H(\Delta) \right].$$

This form covers discrete eigenvalues, continuum intervals, and mixed spectra in one statement.

Time Evolution in the Energy Representation

For a time-independent Hamiltonian,

$$U(t,t_0) = \exp\!\left[ -\frac{i}{\hbar}H(t-t_0) \right].$$

Each energy eigenspace acquires one phase:

$$U(t,t_0)P_E = e^{-iE(t-t_0)/\hbar}P_E.$$

Thus a discrete expansion evolves as

$$\lvert\psi(t)\rangle = \sum_{n,\lambda} c_{n\lambda} e^{-iE_n(t-t_0)/\hbar} \lvert E_n,\lambda\rangle.$$

The coefficient magnitudes do not change. Only phases between distinct energies evolve.

In spectral-measure form,

$$\lvert\psi(t)\rangle = \int_{\mathbb R} e^{-iE(t-t_0)/\hbar} dP_H(E)\lvert\psi(t_0)\rangle.$$

Because $U(t,t_0)$ commutes with every spectral projector of $H$,

$$\Pr_t(E\in\Delta) = \Pr_{t_0}(E\in\Delta).$$

Energy probabilities are therefore conserved under a time-independent Hamiltonian, even when the state is not an energy eigenstate. Other measurement probabilities may change because relative energy phases evolve.

The operator construction is detailed in Time-Evolution Operator.

Relation to Stationary States

Every normalizable pure energy eigenstate of a time-independent Hamiltonian evolves as

$$\lvert E,\lambda;t\rangle = e^{-iE(t-t_0)/\hbar} \lvert E,\lambda;t_0\rangle.$$

The phase is global, so the ray and every physical prediction remain fixed. Any coherent superposition inside one degenerate energy eigenspace behaves the same way.

The broader mixed-state criterion is different: every density operator satisfying $[H,\rho]=0$ is stationary, including mixtures with uncertain energy. See Stationary States for the complete distinction.

Bound States and Continuum States

Discrete bound states

For many confining potentials, discrete energy eigenvalues have normalizable eigenfunctions:

$$\int \left|\psi_n(x)\right|^2dx = 1.$$

These vectors belong to the Hilbert space and can be prepared, at least ideally, with a sharp discrete energy.

Generalized continuum states

For a free particle on the line,

$$H=\frac{\hat p^2}{2m}.$$

A momentum generalized eigenket satisfies

$$H\lvert p\rangle = \frac{p^2}{2m}\lvert p\rangle.$$

It is also a generalized energy eigenket, but it is delta-normalized:

$$\langle p'\vert p\rangle = \delta(p-p').$$

It is not a normalizable physical state on the whole line. Normalizable wave packets have a spread of momenta and generally a spread of energies.

Energy degeneracy appears even in one-dimensional free motion. The momenta $p$ and $-p$ have the same energy:

$$E=\frac{p^2}{2m}.$$

An energy-resolved continuum basis must therefore retain a direction or channel label in addition to $E$.

For a normalized state with an absolutely continuous energy distribution, an exact continuum value has probability zero. Probabilities are assigned to intervals:

$$\Pr(E\in[E_1,E_2]) = \int_{E_1}^{E_2} \rho_E(E)\,dE.$$

The density $\rho_E(E)$ depends on normalization conventions and includes degeneracy and Jacobian factors. It is not obtained by simply renaming a momentum probability density.

Mixed spectra

Many Hamiltonians have discrete bound states below one or more continuum thresholds. Their spectral resolution contains both sums and integrals. A normalizable state can have a finite probability of occupying a bound eigenspace and a complementary probability of lying in the continuum.

The canonical free-particle treatment is Free Particle, and continuous Born probabilities are introduced in Born Rule for Continuous Spectra.

Example: Qubit in a Transverse Basis

Consider

$$H = \frac{\hbar\Omega}{2}\sigma_x.$$

The familiar computational-basis vectors $\lvert0\rangle$ and $\lvert1\rangle$ are not energy eigenstates. The energy eigenvectors are

$$\begin{aligned} \lvert+x\rangle &= \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2},\\ \lvert-x\rangle &= \frac{ \lvert0\rangle-\lvert1\rangle }{\sqrt2}, \end{aligned}$$

with

$$\begin{aligned} E_+&=+\frac{\hbar\Omega}{2},\\ E_-&=-\frac{\hbar\Omega}{2}. \end{aligned}$$

Since

$$\lvert0\rangle = \frac{ \lvert+x\rangle+\lvert-x\rangle }{\sqrt2},$$

an energy measurement in $\lvert0\rangle$ gives either energy with probability $1/2$.

Time evolution yields

$$\begin{aligned} \lvert\psi(t)\rangle &= \frac{1}{\sqrt2} \left( e^{-i\Omega t/2}\lvert+x\rangle + e^{i\Omega t/2}\lvert-x\rangle \right)\\ &= \cos\!\left(\frac{\Omega t}{2}\right) \lvert0\rangle - i\sin\!\left(\frac{\Omega t}{2}\right) \lvert1\rangle. \end{aligned}$$

The computational-basis probabilities oscillate, while the two energy probabilities remain $1/2$. This example separates “basis vector” from “energy eigenvector” and “conserved energy distribution” from “stationary state.”

Example: Infinite Square Well

For a particle in an infinite square well on $0<x<L$ and $n=1,2,\ldots$,

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin\!\left(\frac{n\pi x}{L}\right),$$

with energies

$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}.$$

The one-dimensional spectrum is nondegenerate. If

$$\psi(x,0) = \sum_n c_n\psi_n(x),$$

then

$$p(E_n)=\left|c_n\right|^2$$

at every time, while position-space interference terms can vary through relative phases. The full boundary-value problem is Infinite Square Well.

Time-Dependent Hamiltonians

If $H=H(t)$, the instantaneous equation

$$H(t)\lvert n(t)\rangle = E_n(t)\lvert n(t)\rangle$$

defines instantaneous energy eigenvectors. They need not be solutions of the time-dependent Schrödinger equation. Changes in the eigenvectors can drive transitions, and the distribution of instantaneous energy can change.

Simple phase evolution is exact only under additional conditions. Adiabatic following is an approximation with its own hypotheses, while driven and Floquet systems use other structures. See Time-Dependent Hamiltonians.

What Definite Energy Does Not Imply

Claim	Correct statement
The state has definite energy.	An energy measurement returns one spectral value with probability one.
The state is an energy eigenvector.	This is literal for a normalizable pure state in the point spectrum.
The energy is degenerate.	The outcome identifies an eigenspace, not one basis vector within it.
The state is stationary.	True for pure definite-energy states under a time-independent Hamiltonian.
The state has definite values of other observables.	Only if it also lies in their relevant eigenspaces.
The energy distribution is constant.	True for every state under the same time-independent Hamiltonian.
A continuum ket has definite energy.	It is a generalized, delta-normalized state rather than a Hilbert-space vector.

Common Mistakes

• Treating every convenient basis vector as an energy eigenvector.
• Replacing a degenerate energy eigenspace by one arbitrary basis state.
• Forgetting that mixed states supported inside one degenerate eigenspace can have definite energy.
• Calling every stationary mixture a definite-energy state.
• Thinking $\langle H\rangle=E$ alone proves definite energy without checking $\Delta H=0$.
• Treating a continuum generalized eigenket as a normalizable bound state.
• Ignoring degeneracy and Jacobian factors when converting momentum densities to energy densities.
• Assuming exact energy probabilities can change under a time-independent Hamiltonian.
• Assuming an instantaneous eigenvector of $H(t)$ evolves only by a phase.
• Concluding that definite energy implies definite position, momentum, or classical rest.

Cross-Links

• Hamiltonians
• Stationary States
• Time-Evolution Operator
• Eigenvalues and Eigenstates
• Spectral Decomposition
• Variance and Standard Deviation
• Degenerate Measurements and Lüders Rule
• Free Particle
• Infinite Square Well

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 10–11 and 26.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977, vol. 1, chs. 2–3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, chs. 4 and 12.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chs. 1–2.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chs. 5–7.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised ed., Academic Press, 1980, chs. 7–8.

Exercises

1. Prove that a normalized pure state has definite energy if and only if $\Delta H=0$, assuming the state lies in the domain of $H^2$.

Solution

Let

$$\bar E=\langle\psi\vert H\vert\psi\rangle.$$

Then

$$\begin{aligned} (\Delta H)^2 &= \langle\psi\vert \left(H-\bar E\right)^2 \vert\psi\rangle\\ &= \left\lVert \left(H-\bar E\right) \lvert\psi\rangle \right\rVert^2. \end{aligned}$$

The squared norm vanishes exactly when

$$\left(H-\bar E\right) \lvert\psi\rangle=0.$$

Thus $\Delta H=0$ exactly when

$$H\lvert\psi\rangle = \bar E\lvert\psi\rangle,$$

so the state is an energy eigenvector. The reverse implication follows immediately from the same formula.

2. Let $P_E$ project onto a two-dimensional degenerate energy eigenspace. Show that

$$\rho = \frac12 \begin{pmatrix} 1 & \gamma\\ \gamma^* & 1 \end{pmatrix}$$

in an orthonormal basis of that eigenspace has definite energy $E$ whenever it is positive. Find the allowed values of $\gamma$.

Solution

Because the matrix acts entirely within the range of $P_E$,

$$P_E\rho P_E=\rho.$$

Therefore an energy measurement returns $E$ with probability

$$\operatorname{Tr}(\rho P_E) = \operatorname{Tr}\rho = 1.$$

The matrix eigenvalues are

$$r_\pm = \frac12 \left( 1\pm\lvert\gamma\rvert \right).$$

Positivity requires

$$\lvert\gamma\rvert\le1.$$

Nonzero $\gamma$ represents coherence within the degenerate eigenspace and does not create energy uncertainty.

3. For

$$H=\frac{\hbar\Omega}{2}\sigma_x,$$

find the energy eigenvectors and the two energy probabilities in the initial state $\lvert0\rangle$. Compute $\langle H\rangle$ and $\Delta H$.

Solution

The eigenvectors of $\sigma_x$ are

$$\begin{aligned} \lvert+x\rangle &= \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2},\\ \lvert-x\rangle &= \frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2}, \end{aligned}$$

with energies $\pm\hbar\Omega/2$. Since

$$\lvert0\rangle = \frac{ \lvert+x\rangle+\lvert-x\rangle }{\sqrt2},$$

each energy occurs with probability $1/2$. Hence

$$\langle H\rangle=0$$

and

$$\begin{aligned} (\Delta H)^2 &= \frac12 \left(\frac{\hbar\Omega}{2}\right)^2 + \frac12 \left(-\frac{\hbar\Omega}{2}\right)^2\\ &= \left(\frac{\hbar\Omega}{2}\right)^2. \end{aligned}$$

Therefore $\Delta H=\hbar\lvert\Omega\rvert/2$.

4. Show that the projector

$$P_E = \sum_{\lambda=1}^{g_E} \lvert E,\lambda\rangle \langle E,\lambda\rvert$$

is unchanged under a unitary change of orthonormal basis within the eigenspace.

Solution

Let

$$\lvert E,a\rangle' = \sum_\lambda V_{a\lambda} \lvert E,\lambda\rangle,$$

where $V$ is unitary. Let $Q$ denote the projector constructed from the primed basis. Expanding the primed vectors and using $\sum_a V_{a\lambda}V_{a\mu}^*=\delta_{\lambda\mu}$ gives

$$\begin{aligned} Q &= \sum_{\lambda,\mu} \delta_{\lambda\mu} \lvert E,\lambda\rangle \langle E,\mu\rvert\\ &= P_E. \end{aligned}$$

The spectral projector is basis-independent even though its rank-one decomposition is not.

5. Prove that the probability of every energy set $\Delta$ is conserved under a time-independent Hamiltonian.

Solution

The spectral projector $P_H(\Delta)$ is a function of $H$, so it commutes with

$$U(t,t_0) = e^{-iH(t-t_0)/\hbar}.$$

For a pure state,

$$\begin{aligned} \Pr_t(E\in\Delta) &= \langle\psi_0\vert U^\dagger P_H(\Delta)U \vert\psi_0\rangle\\ &= \langle\psi_0\vert P_H(\Delta) \vert\psi_0\rangle. \end{aligned}$$

The same result for a density operator follows from cyclicity of the trace. Conservation of the energy distribution does not require the state itself to be stationary.

6. Replace a time-independent Hamiltonian by $H'=H+E_{\mathrm c}I$. How do its eigenvalues, eigenvectors, energy probabilities, and time-evolution operator change?

Solution

If

$$H\lvert E,\lambda\rangle = E\lvert E,\lambda\rangle,$$

then

$$H'\lvert E,\lambda\rangle = \left(E+E_{\mathrm c}\right) \lvert E,\lambda\rangle.$$

The eigenvectors and spectral projectors are unchanged, while every eigenvalue shifts by $E_{\mathrm c}$. Therefore the probabilities associated with corresponding eigenspaces are unchanged.

The propagator becomes

$$U'(t,t_0) = e^{-iE_{\mathrm c}(t-t_0)/\hbar} U(t,t_0).$$

The extra factor is a common global phase for state vectors evolved under this one Hamiltonian.

7. A normalized free-particle state on the line has momentum-space wavefunction $\phi(p)$ with

$$\int_{-\infty}^{\infty} \left|\phi(p)\right|^2dp = 1.$$

For $E>0$, derive the energy probability density $\rho_E(E)$.

Solution

The two momenta

$$p_E=\sqrt{2mE} \quad\text{and}\quad -p_E$$

have the same energy. Since

$$\left| \frac{dp}{dE} \right| = \frac{m}{p_E},$$

both branches contribute:

$$\rho_E(E) = \frac{m}{p_E} \left[ \left|\phi(p_E)\right|^2 + \left|\phi(-p_E)\right|^2 \right].$$

Indeed,

$$\int_0^\infty \rho_E(E)\,dE = \int_{-\infty}^{\infty} \left|\phi(p)\right|^2dp = 1.$$

The two terms encode the degeneracy, and $m/p_E$ is the Jacobian.

8. In the infinite square well, let

$$\psi(x,0) = \frac{ \psi_1(x)+e^{i\theta}\psi_2(x) }{\sqrt2}.$$

Find the energy probabilities at arbitrary time. Is the state stationary?

Solution

Time evolution gives

$$\begin{aligned} \psi(x,t) &= \frac{1}{\sqrt2} e^{-iE_1t/\hbar}\psi_1(x) \\ &\quad+ \frac{1}{\sqrt2} e^{i\theta} e^{-iE_2t/\hbar}\psi_2(x). \end{aligned}$$

The coefficient magnitudes remain $1/\sqrt2$, so

$$p(E_1)=p(E_2)=\frac12$$

for every $t$. The state is not stationary because $E_1\ne E_2$: the relative phase

$$e^{i\theta} e^{-i(E_2-E_1)t/\hbar}$$

changes with time and can alter position-space interference.