Question
Show that the variance, defined by $\left\langle Q^2\right\rangle-\langle Q\rangle^2$, is zero only when $\psi$ is an eigenfunction of $Q$. (Let $\psi=\sum_i c_i \psi_i$ where the $\psi_i$ satisfy $Q \psi_i=q_i \psi_i$.
Step 1
The variance is given by the expression: \[ \text{Var}(Q) = \langle Q^2 \rangle - \langle Q \rangle^2 \] where \( \langle Q \rangle = \frac{\langle \psi | Q | \psi \rangle}{\langle \psi | \psi \rangle} \) and \( \langle Q^2 \rangle = \frac{\langle \psi | Q^2 | Show more…
Show all steps
Your feedback will help us improve your experience
Dr. Rajveer Singh and 65 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Let $\hat{Q}$ be an operator with a complete set of orthonormal eigenvectors: $$ \hat{Q}\left|e_{n}\right\rangle=q_{n}\left|e_{n}\right\rangle \quad(n=1,2,3, \ldots) $$ Show that $\hat{Q}$ can be written in terms of its spectral decomposition: $$ \hat{Q}=\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right| $$ Hint: An operator is characterized by its action on all possible vectors, so what you must show is that $$ \hat{Q}|\alpha\rangle=\left\{\sum_{n} q_{n}\left|e_{n}\right\rangle\left\langle e_{n}\right|\right\}|\alpha\rangle, $$ for any vector $|\alpha\rangle$.
Formalism
Dirac Notation
Show that two noncommuting operators cannot have a complete set of common eigenfunctions. Hint: Show that if $\hat{P}$ and $\hat{Q}$ have a complete set of common eigenfunctions, then $[\hat{P} . \hat{Q}] f=0$ for any function in Hilbert space.
The Uncertainty Principle
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD