State Vectors

A state vector is a nonzero vector in a complex Hilbert space used to represent a pure quantum state. Physicists usually choose a normalized ket $|\psi\rangle$ with $\langle\psi|\psi\rangle=1$, while remembering that all normalized kets differing only by a global phase represent the same physical pure state.

State vectors are powerful but not universal. They represent pure states of the declared system. General mixed states and reduced states require density operators, and ideal continuum eigenkets such as $|x\rangle$ are generalized vectors rather than normalizable state vectors.

Helpful background. Quantum States distinguishes a physical state from its representatives; Bra–Ket Notation and Inner-Product Conventions supply the notation. The only assumed mathematics is complex linear algebra: vector spaces, inner products, bases, and linear combinations.

Kets as Abstract Vectors

Let $\mathcal H$ be a complex Hilbert space. A ket

$$|\psi\rangle\in\mathcal H$$

is an abstract vector: it is defined by the vector-space operations, inner product, and limiting structure of $\mathcal H$, not by one particular list of components. The zero vector is excluded as a state representative because it cannot be normalized and assigns no probability rule.

The site uses the standard physics convention for inner products:

$$\langle\phi|\psi\rangle \text{ is conjugate-linear in }\phi \text{ and linear in }\psi.$$

Thus

$$\begin{aligned} \langle a\phi_1+b\phi_2|\psi\rangle &=a^*\langle\phi_1|\psi\rangle +b^*\langle\phi_2|\psi\rangle,\\ \langle\phi|a\psi_1+b\psi_2\rangle &=a\langle\phi|\psi_1\rangle +b\langle\phi|\psi_2\rangle. \end{aligned}$$

The induced norm is

$$\|\psi\| =\sqrt{\langle\psi|\psi\rangle}.$$

Completeness of the Hilbert space means that every Cauchy sequence in this norm converges to a vector in $\mathcal H$. This analytic condition becomes essential for infinite basis expansions and limits of approximating states. The mathematical details belong to Hilbert Spaces.

Kets, Bras, and Outer Products

The bra $\langle\psi|$ associated with $|\psi\rangle$ is the corresponding continuous linear functional supplied by the inner product. It acts on another ket to produce the complex number

$$\langle\psi|\phi\rangle.$$

In a finite-dimensional orthonormal basis, if

$$[|\psi\rangle]_{\mathcal B} =\begin{pmatrix} \psi_1\\ \vdots\\ \psi_d \end{pmatrix},$$

then the corresponding bra is represented by the conjugate transpose,

$$[\langle\psi|]_{\mathcal B} =\begin{pmatrix} \psi_1^*&\cdots&\psi_d^* \end{pmatrix}.$$

An inner product and an outer product are different objects:

$$\langle\phi|\psi\rangle\in\mathbb C, \qquad |\psi\rangle\langle\phi|:\mathcal H\to\mathcal H.$$

For normalized $|\psi\rangle$, the outer product

$$P_\psi=|\psi\rangle\langle\psi|$$

is the rank-one orthogonal projector onto the ray represented by the ket. It satisfies

$$P_\psi^2=P_\psi, \qquad P_\psi^\dagger=P_\psi, \qquad \operatorname{Tr}P_\psi=1.$$

Normalization and Ray Equivalence

A ket is conventionally normalized by requiring

$$\langle\psi|\psi\rangle=1.$$

Given any nonzero $|\chi\rangle$ with finite norm, a normalized representative is

$$|\psi\rangle =\frac{|\chi\rangle} {\sqrt{\langle\chi|\chi\rangle}}.$$

Normalization fixes the magnitude of the representative but not its global phase. The kets

$$|\psi\rangle \quad\text{and}\quad e^{i\alpha}|\psi\rangle$$

are both normalized and define the same rank-one projector. More generally, all nonzero scalar multiples $z|\psi\rangle$ lie on the same ray. For an unnormalized representative $|\chi\rangle$, the associated pure-state density operator is

$$\rho_\chi =\frac{|\chi\rangle\langle\chi|} {\langle\chi|\chi\rangle}.$$

The physical pure state is the ray, while the unit ket is a convenient representative. Rays and Global Phase owns the physical equivalence relation; Normalization owns ordinary, delta, and box normalization.

Why unit normalization is useful

For an orthogonal projector $P$, the quantity

$$\langle\psi|P|\psi\rangle$$

is a probability only when $|\psi\rangle$ is normalized. With an unnormalized ket, the scale-independent probability is

$$p(P\mid\chi) =\frac{\langle\chi|P|\chi\rangle} {\langle\chi|\chi\rangle}.$$

Normalizing once lets the denominator be suppressed throughout a calculation.

Expansion in an Orthonormal Basis

Let $\{|n\rangle\}$ be a complete orthonormal basis, so

$$\langle m|n\rangle=\delta_{mn}, \qquad \sum_n|n\rangle\langle n|=I.$$

Inserting the identity gives

$$|\psi\rangle =\sum_n c_n|n\rangle, \qquad c_n=\langle n|\psi\rangle.$$

The coefficients $c_n$ are the coordinates of the abstract vector in that basis. Orthonormality and normalization imply Parseval’s identity,

$$\langle\psi|\psi\rangle =\sum_n|c_n|^2 =1.$$

For a finite-dimensional space, the sum has finitely many terms. For a countably infinite orthonormal basis, the partial sums converge to $|\psi\rangle$ in Hilbert-space norm:

$$\lim_{N\to\infty} \left\| |\psi\rangle- \sum_{n=1}^{N}c_n|n\rangle \right\|=0.$$

Norm convergence need not imply pointwise convergence of the corresponding wavefunctions. This distinction matters when basis series are manipulated as ordinary functions.

Nonorthogonal expansions

The simple formula $c_n=\langle n|\psi\rangle$ depends on orthonormality. In a nonorthogonal basis $\{|e_j\rangle\}$, the overlaps $\langle e_j|\psi\rangle$ are generally not the expansion coefficients. One needs the Gram matrix or a dual basis. The standard first-pass state-vector formalism therefore uses orthonormal bases unless stated otherwise.

Coefficients as Probability Amplitudes

For a projective measurement in the orthonormal basis $\{|n\rangle\}$, the projector for outcome $n$ is

$$P_n=|n\rangle\langle n|.$$

The Born probability is

$$p(n\mid\psi) =\langle\psi|P_n|\psi\rangle =|\langle n|\psi\rangle|^2 =|c_n|^2.$$

The coefficient $c_n$ is a probability amplitude, not a probability. Its complex phase can affect probabilities in another measurement basis through interference.

For a transition test onto another normalized state $|\phi\rangle$, the amplitude and probability are

$$\mathcal A_{\phi\leftarrow\psi} =\langle\phi|\psi\rangle, \qquad p(\phi\mid\psi) =|\langle\phi|\psi\rangle|^2.$$

Cauchy–Schwarz ensures

$$0\le|\langle\phi|\psi\rangle|^2\le1$$

for normalized kets. Equality to one holds exactly when the two kets represent the same ray; equality to zero means the states are orthogonal and perfectly distinguishable by a suitable projective measurement.

Probability Amplitudes develops composition, interference, and basis dependence in detail.

Abstract Vector Versus Coordinate Column

After choosing an ordered orthonormal basis $\mathcal B=\{|1\rangle,\ldots,|d\rangle\}$, the ket has coordinate column

$$[|\psi\rangle]_{\mathcal B} =\begin{pmatrix} c_1\\ \vdots\\ c_d \end{pmatrix}.$$

The bracketed expression emphasizes that the column depends on $\mathcal B$. The abstract ket does not. If $S$ is the unitary overlap matrix from one basis to another, the component column changes as

$$\psi_{\mathcal B'}=S\psi_{\mathcal B}$$

in the site’s basis-change convention. Operator matrices transform with the matching rule, leaving inner products and Born probabilities unchanged.

The same ket can therefore be a one-component basis vector in one basis and a many-component superposition in another. A statement about components is not automatically a basis-independent statement about the physical state.

Bases and Representations introduces the coordinate map, and Change of Basis derives the transformation rules.

Linear Combinations and Superposition

If $|\psi_1\rangle$ and $|\psi_2\rangle$ belong to the same Hilbert space, then

$$|\chi\rangle =a|\psi_1\rangle+b|\psi_2\rangle$$

is another vector. It represents a state only if $|\chi\rangle\ne0$, after normalization. Its squared norm is

$$\begin{aligned} \langle\chi|\chi\rangle &=|a|^2\langle\psi_1|\psi_1\rangle +|b|^2\langle\psi_2|\psi_2\rangle\\ &\quad +a^*b\langle\psi_1|\psi_2\rangle +b^*a\langle\psi_2|\psi_1\rangle. \end{aligned}$$

The cross terms vanish only when the component states are orthogonal or when the coefficients make their real contribution vanish. Consequently, $|a|^2+|b|^2=1$ does not normalize an arbitrary superposition of nonorthogonal states.

A coherent linear combination is not a classical random mixture. The pure state

$$\rho_\chi =\frac{|\chi\rangle\langle\chi|} {\langle\chi|\chi\rangle}$$

contains cross terms, whereas a source that chooses $|\psi_1\rangle$ or $|\psi_2\rangle$ randomly gives a convex sum of projectors without those coherences. Superposition and Relative Phase owns this distinction.

Worked Example: Spin-1/2 State

In the ordered $z$ basis $\{|+z\rangle,|-z\rangle\}$, consider

$$|\psi\rangle =\frac{\sqrt3}{2}|+z\rangle +\frac{i}{2}|-z\rangle.$$

Its coordinate column and bra are

$$\psi_z =\begin{pmatrix} \sqrt3/2\\ i/2 \end{pmatrix}, \qquad \psi_z^\dagger =\begin{pmatrix} \sqrt3/2&-i/2 \end{pmatrix}.$$

The norm is

$$\psi_z^\dagger\psi_z =\frac34+\frac14 =1.$$

A measurement of $S_z$ gives

$$p(+z)=\frac34, \qquad p(-z)=\frac14.$$

Now use

$$|\pm x\rangle =\frac{|+z\rangle\pm|-z\rangle}{\sqrt2}.$$

The $+x$ amplitude is

$$\langle+x|\psi\rangle =\frac{\sqrt3+i}{2\sqrt2},$$

so

$$p(+x) =\left| \frac{\sqrt3+i}{2\sqrt2} \right|^2 =\frac12.$$

Similarly $p(-x)=1/2$. The relative phase $i$ does not affect the $z$-basis probabilities, but it changes which transverse spin direction has a biased distribution. For example, a $y$-basis measurement does not give equal probabilities.

Worked Example: A Three-Level System

Let $\{|0\rangle,|1\rangle,|2\rangle\}$ be an orthonormal basis and define

$$|\chi\rangle =2|0\rangle-i|1\rangle+2e^{i\varphi}|2\rangle.$$

Its squared norm is

$$\langle\chi|\chi\rangle =4+1+4 =9,$$

so a normalized representative is

$$|\psi_\varphi\rangle =\frac13 \left( 2|0\rangle-i|1\rangle +2e^{i\varphi}|2\rangle \right).$$

Measurement in the defining basis gives

$$p(0)=\frac49, \qquad p(1)=\frac19, \qquad p(2)=\frac49.$$

These probabilities are independent of $\varphi$, but the state is not. To probe the relative phase between levels $0$ and $2$, use

$$|u_\pm\rangle =\frac{|0\rangle\pm|2\rangle}{\sqrt2}.$$

Then

$$\begin{aligned} p(u_+) &=|\langle u_+|\psi_\varphi\rangle|^2 =\frac49(1+\cos\varphi),\\ p(u_-) &=|\langle u_-|\psi_\varphi\rangle|^2 =\frac49(1-\cos\varphi). \end{aligned}$$

The remaining probability $1/9$ belongs to $|1\rangle$. The example shows why a component phase can be invisible in one basis and observable in another.

Infinite-Dimensional State Vectors

A wave-mechanics state vector can belong to an infinite-dimensional Hilbert space such as $L^2([0,L])$. Choosing the position representation gives the wavefunction

$$\psi(x)=\langle x|\psi\rangle,$$

with norm

$$\|\psi\|^2 =\int_0^L|\psi(x)|^2\,dx.$$

The ket can also be expanded in a countable orthonormal basis of ordinary normalizable states. For the infinite square well,

$$u_n(x) =\sqrt{\frac2L} \sin\!\left(\frac{n\pi x}{L}\right), \qquad n=1,2,\ldots,$$

and

$$|\psi\rangle =\sum_{n=1}^{\infty}c_n|n\rangle, \qquad \psi(x) =\sum_{n=1}^{\infty}c_nu_n(x).$$

The coefficients are

$$c_n =\langle n|\psi\rangle =\int_0^L u_n^*(x)\psi(x)\,dx,$$

and Parseval’s identity gives

$$\int_0^L|\psi(x)|^2\,dx =\sum_{n=1}^{\infty}|c_n|^2.$$

Two-mode well state

As a normalizable example, take

$$|\psi(0)\rangle =\frac{|1\rangle+i|2\rangle}{\sqrt2}.$$

Its position wavefunction is

$$\psi(x,0) =\sqrt{\frac1L} \left[ \sin\!\left(\frac{\pi x}{L}\right) +i\sin\!\left(\frac{2\pi x}{L}\right) \right].$$

An energy measurement gives $E_1$ or $E_2$ with probability $1/2$. Under the well Hamiltonian,

$$|\psi(t)\rangle =\frac1{\sqrt2} \left( e^{-iE_1t/\hbar}|1\rangle +i e^{-iE_2t/\hbar}|2\rangle \right).$$

The energy probabilities stay fixed, while the relative phase evolves and can change the position probability density through interference.

Generalized basis kets are not states

The formal resolution

$$I=\int_{-\infty}^{\infty}|x\rangle\langle x|\,dx$$

is useful, but $|x\rangle$ is not a normalizable vector in $L^2(\mathbb R)$. It obeys delta normalization,

$$\langle x|x'\rangle=\delta(x-x'),$$

and belongs to a generalized spectral framework. A physical localized state is represented by a normalizable wave packet, not an exact position eigenket.

Wavefunctions as Representations and Infinite Square Well own the detailed wave-mechanics constructions.

Time Evolution of State Vectors

For a closed system, a pure state evolves by a unitary operator,

$$|\psi(t)\rangle =U(t,t_0)|\psi(t_0)\rangle.$$

Unitarity preserves normalization and all inner products:

$$\begin{aligned} \langle\psi(t)|\psi(t)\rangle &=\langle\psi(t_0)|U^\dagger U|\psi(t_0)\rangle =1,\\ \langle\phi(t)|\psi(t)\rangle &=\langle\phi(t_0)|\psi(t_0)\rangle. \end{aligned}$$

For a time-independent Hamiltonian,

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

The evolution is linear on kets, which preserves coherent superpositions. A global phase acquired by the entire state does not change its ray, whereas different phases acquired by different energy components alter relative phase and can change later interference.

Unitary Time Evolution owns the dynamical derivation and its assumptions.

Composite-State Vectors

If systems $A$ and $B$ have Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$, a pure joint state is represented in

$$\mathcal H_{AB} =\mathcal H_A\otimes\mathcal H_B.$$

Product kets have the form

$$|\psi\rangle_A\otimes|\phi\rangle_B,$$

often abbreviated $|\psi\rangle_A|\phi\rangle_B$. Not every joint state vector is a product. For example,

$$|\Phi^+\rangle =\frac{|00\rangle+|11\rangle}{\sqrt2}$$

is entangled. It is a pure state vector of the joint system, but neither subsystem has its own pure state vector; each reduced state is $I/2$.

This is another reason to state which system a ket represents. The Composite Systems and Entanglement volume is the current home for composition theory.

When a State Vector Is Insufficient

One ket in $\mathcal H$ does not represent every state of a system. Use a density operator when:

• the preparation is a classical mixture and the label is unavailable;
• the system is a subsystem of an entangled joint state;
• noise or uncontrolled degrees of freedom produce a mixed state;
• an unconditioned measurement averages over outcomes;
• the state is thermal or otherwise statistically mixed.

A density operator is pure exactly when it can be written as a rank-one projector $|\psi\rangle\langle\psi|$. Purifying a mixed state introduces an auxiliary system; the resulting ket belongs to the enlarged Hilbert space, not to the original system alone.

Density Operators is the canonical home for the general state formalism.

Numerical State Vectors

In computation, a state vector is stored as a finite complex array only after a basis and truncation are chosen. A trustworthy numerical representation records:

• the basis ordering and tensor-factor ordering;
• dimensions and any symmetry-sector restriction;
• normalization and global-phase convention when comparisons require one;
• units and quadrature weights for grid representations;
• the truncation rule for an infinite-dimensional space;
• convergence of reported observables as the basis or grid is enlarged.

For a grid wavefunction, Euclidean array normalization $\sum_j|\psi_j|^2=1$ is generally wrong unless the values have been scaled to absorb the integration measure. With direct samples of $\psi(x_j)$ on a uniform grid,

$$\sum_j|\psi(x_j)|^2\,\Delta x \approx1.$$

The array is a discretized representation, not the abstract state vector itself, and convergence must be checked.

A State-Vector Checklist

Before using a ket, ask:

1. Which Hilbert space and physical system does it belong to?
2. Is the state known to be pure, or is a density operator required?
3. Is the ket normalized, nonzero, or intentionally left unnormalized?
4. Which basis or representation defines its components?
5. Does the inner-product convention match the rest of the calculation?
6. Are the basis vectors orthonormal and complete?
7. Which measurement makes the coefficients relevant amplitudes?
8. Are phases global, relative, or convention dependent?
9. In infinite dimensions, is the vector genuinely normalizable or only a generalized eigenket?
10. In numerics, has truncation and measure-weighted normalization been tested?

Common Mistakes

• Calling the zero vector or a non-normalizable function a physical state vector.
• Treating a normalized ket as the unique representative of a pure state and forgetting global phase.
• Normalizing by the sum of components rather than the sum of squared magnitudes.
• Forgetting complex conjugation when constructing a bra.
• Confusing the scalar $\langle\phi|\psi\rangle$ with the operator $|\psi\rangle\langle\phi|$.
• Treating basis coefficients as probabilities before taking squared moduli.
• Assuming $|a|^2+|b|^2=1$ normalizes a superposition of nonorthogonal states.
• Identifying an abstract ket with one basis-dependent coordinate column.
• Saying a state is “in a superposition” without naming the basis.
• Treating an exact $|x\rangle$ or $|p\rangle$ as a normalizable wave packet.
• Using one ket for a mixed subsystem of an entangled state.
• Normalizing a sampled wavefunction without its grid measure.
• Interpreting every two- or three-component ket as a vector in ordinary physical space.

Connections

• Quantum States gives the broader preparation and probability-rule definition.
• Pure States identifies rays, rank-one projectors, and extremal density operators.
• Rays and Global Phase explains which state-vector representatives are physically equivalent.
• Bases and Representations separates abstract kets from coordinate columns.
• Probability Amplitudes develops inner products, interference, and transition probabilities.
• Density Operators extends the state language beyond pure-state vectors.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, Wiley, 1977. See Complement A and Chapters II–III for state vectors and representations.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958. See Chapters I–III for ket vectors, transformations, and representations.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013. See Chapters 2 and 6–8 for Hilbert spaces, Dirac notation, and quantum states.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010. See Sections 2.1–2.2 for finite-dimensional state vectors, composite systems, and density operators.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995. See Chapters 2–4 for preparations, state vectors, and tests.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised ed., Academic Press, 1980. See Chapters II and VII for Hilbert spaces and unitary operators.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020. See Chapter 1 for kets, bases, spin, and measurement amplitudes.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994. See Chapters 1 and 4 for vector spaces, basis expansions, and quantum states.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955. See Chapter II for Hilbert space and Chapter III for states and observables.

Exercises

1. Normalize and form the projector. Let

   $$|\chi\rangle =(1+i)|0\rangle+2|1\rangle.$$

   Find a normalized representative and the pure-state density matrix in the ordered basis $\{|0\rangle,|1\rangle\}$.

Solution

The squared norm is

$$\langle\chi|\chi\rangle =|1+i|^2+|2|^2 =2+4 =6.$$

Thus

$$|\psi\rangle =\frac1{\sqrt6} \left[(1+i)|0\rangle+2|1\rangle\right].$$

The projector is

$$\rho_\psi =\frac16 \begin{pmatrix} 2&2+2i\\ 2-2i&4 \end{pmatrix}.$$

It has trace one and satisfies $\rho_\psi^2=\rho_\psi$.

2. Nonorthogonal superposition. Let $|a\rangle$ and $|b\rangle$ be normalized with $\langle a|b\rangle=s$, where $s$ is real. Normalize $|a\rangle+|b\rangle$ when $s>-1$.

Solution

The squared norm is

$$\begin{aligned} \bigl\|\,|a\rangle+|b\rangle\,\bigr\|^2 &=\langle a|a\rangle+ \langle a|b\rangle+ \langle b|a\rangle+ \langle b|b\rangle\\ &=2(1+s). \end{aligned}$$

Therefore

$$|\psi\rangle =\frac{|a\rangle+|b\rangle} {\sqrt{2(1+s)}}.$$

The condition $s>-1$ excludes the case $|b\rangle=-|a\rangle$, for which the sum is the zero vector.

3. Bra and outer product. In an orthonormal basis, let

   $$|\psi\rangle =\frac{1}{\sqrt3} \begin{pmatrix}1\\i\\1\end{pmatrix}, \qquad |\phi\rangle =\frac{1}{\sqrt2} \begin{pmatrix}1\\0\\-1\end{pmatrix}.$$

   Compute $\langle\phi|\psi\rangle$ and the matrix $|\psi\rangle\langle\phi|$.

Solution

The amplitude is

$$\langle\phi|\psi\rangle =\frac1{\sqrt6}(1-1) =0.$$

The outer product is the operator

$$|\psi\rangle\langle\phi| =\frac1{\sqrt6} \begin{pmatrix} 1&0&-1\\ i&0&-i\\ 1&0&-1 \end{pmatrix}.$$

The first result is a scalar; the second is a rank-one matrix.

4. Qutrit phase test. For the three-level state $|\psi_\varphi\rangle$ on this page, verify that $p(u_+)+p(u_-)+p(1)=1$. Find the distributions at $\varphi=0$ and $\varphi=\pi$.

Solution

Using the formulas above,

$$p(u_+)+p(u_-) =\frac49(1+\cos\varphi) +\frac49(1-\cos\varphi) =\frac89.$$

Adding $p(1)=1/9$ gives one. At $\varphi=0$,

$$(p(u_+),p(u_-),p(1)) =\left(\frac89,0,\frac19\right),$$

while at $\varphi=\pi$,

$$(p(u_+),p(u_-),p(1)) =\left(0,\frac89,\frac19\right).$$

The defining-basis probabilities remain $4/9,1/9,4/9$ in both cases.

5. Change of basis. Let

   $$S=\frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}, \qquad \psi_z=\begin{pmatrix}\alpha\\\beta\end{pmatrix}.$$

   Show that $\psi_x=S\psi_z$ has the same norm and identify its components as the $x$-basis amplitudes.

Solution

The transformed column is

$$\psi_x =\frac1{\sqrt2} \begin{pmatrix} \alpha+\beta\\ \alpha-\beta \end{pmatrix}.$$

These entries are $\langle+x|\psi\rangle$ and $\langle-x|\psi\rangle$. Since $S^\dagger S=I$,

$$\psi_x^\dagger\psi_x =\psi_z^\dagger S^\dagger S\psi_z =\psi_z^\dagger\psi_z.$$

The coordinates change; the abstract ket and its norm do not.

6. Infinite-well mode expansion. For

   $$|\psi\rangle =\frac{|1\rangle+i|2\rangle}{\sqrt2},$$

   verify normalization directly in position space using orthonormality of the well eigenfunctions. What are $\langle H\rangle$ and $\operatorname{Var}(H)$?

Solution

Orthonormality gives

$$\int_0^L|\psi(x)|^2\,dx =\frac12\langle1|1\rangle +\frac12\langle2|2\rangle =1,$$

because the cross terms contain $\langle1|2\rangle=0$. The energy probabilities are equal, so

$$\langle H\rangle =\frac{E_1+E_2}{2},$$

and

$$\operatorname{Var}(H) =\frac{E_1^2+E_2^2}{2} -\left(\frac{E_1+E_2}{2}\right)^2 =\frac{(E_2-E_1)^2}{4}.$$

7. Truncation error. Let $|\psi\rangle=\sum_{n=1}^{\infty}c_n|n\rangle$ be normalized and define the unnormalized truncation $|\chi_N\rangle=\sum_{n=1}^{N}c_n|n\rangle$. Express its squared norm and the norm of the discarded tail. Give the normalized truncation when its norm is nonzero.

Solution

Orthonormality gives

$$\|\chi_N\|^2 =\sum_{n=1}^{N}|c_n|^2.$$

The discarded tail has squared norm

$$\left\| |\psi\rangle-|\chi_N\rangle \right\|^2 =\sum_{n=N+1}^{\infty}|c_n|^2 =1-\|\chi_N\|^2.$$

If $\|\chi_N\|>0$, the normalized truncated state is

$$|\psi_N\rangle =\frac{|\chi_N\rangle}{\|\chi_N\|}.$$

Renormalization does not remove truncation error; convergence of observables still has to be checked.

8. Pure joint state without local state vectors. For

   $$|\Phi^+\rangle =\frac{|00\rangle+|11\rangle}{\sqrt2},$$

   show that no product kets $|a\rangle|b\rangle$ reproduce the state. Then compute the reduced density operator of the first qubit.

Solution

If $|a\rangle=a_0|0\rangle+a_1|1\rangle$ and $|b\rangle=b_0|0\rangle+b_1|1\rangle$, their product has coefficients

$$a_0b_0, \quad a_0b_1, \quad a_1b_0, \quad a_1b_1.$$

Matching the Bell state would require the first and last to be nonzero while both middle coefficients vanish, which is impossible for scalar coefficients. Tracing out the second qubit gives

$$\rho_A =\frac12|0\rangle\langle0| +\frac12|1\rangle\langle1| =\frac I2.$$

The joint system has a pure state vector, while the subsystem is mixed and has no pure state vector of its own.