Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say you have energy eigenstates

\begin{align} \begin{split} |+\rangle= \frac{1}{\sqrt{2}}|1{\rangle}+\frac{1}{\sqrt{2}}|2 \rangle \end{split} \end{align}

\begin{align} \begin{split} |-\rangle= \frac{1}{\sqrt{2}}|1{\rangle}-\frac{1}{\sqrt{2}}|2 \rangle \end{split} \end{align}


\begin{align} \begin{split} |\psi(0)\rangle= \alpha{_+}|+{\rangle}+\alpha{_-}|- \rangle \end{split} \end{align}


\begin{align} \begin{split} \alpha{_{+}} = {\langle} + | {\psi{(0)}} {\rangle} \end{split} \end{align}

\begin{align} \begin{split} \alpha{_{-}} = {\langle} - | {\psi{(0)}} {\rangle} \end{split} \end{align}

I know that you can find the coefficients $\alpha_+$ and $\alpha_-$ if you have $|\psi(0)\rangle$ already, but I am struggling conceptually with what this means in relation to the Heisenberg uncertainty principle and problem solving for this type of thing in general.

I am also unsure how you find the eigenstates. Though I know mathematically how to get the eigenvalues and eigenvectors from a matrix.

share|cite|improve this question
up vote 6 down vote accepted

An energy eigenstate is just an eigenstate of the Hamiltonian. So, given a particular Hamiltonian operator $H$, the energy eigenstates $\lvert n\rangle$ satisfy

$$H\lvert n\rangle = E_n\lvert n\rangle$$

where $E_n$ is just a number.

The reason energy eigenstates are useful is that according to the Schroedinger equation, they remain unchanged (except for a phase factor) over time. Suppose $\lvert\psi(0)\rangle$ is the initial state of some system with a Hamiltonian $H$. If $\lvert\psi(0)\rangle$ is the $n$th eigenstate of $H$, namely if $\lvert\psi(0)\rangle = \lvert n\rangle$, the system's state at a later time $t$ will be

$$\lvert\psi(t)\rangle = e^{iE_nt}\lvert n\rangle = e^{iE_nt}\lvert \psi(0)\rangle$$

And since the Schroedinger equation is linear, if the initial state is a linear combination of energy eigenstates, $\lvert\psi(0)\rangle = \sum_n \alpha_n\lvert n\rangle$, the same holds for each of the eigenstates in the sum. Essentially you can distribute the time evolution over the sum. Accordingly, this lets you easily write down an expression for the state of the system at time $t$:

$$\lvert\psi(t)\rangle = \sum_n\alpha_n e^{iE_nt}\lvert n\rangle$$

So if you can express the initial state as a sum of coefficients times the energy eigenstates, it makes it pretty trivial to express the state at any later time. That's where the inner products come in. It's often the case that eigenstates of $H$ form a complete orthonormal basis, and when you have an orthonormal basis, the way you decompose an arbitrary state into that basis is by taking inner products, $\alpha_n = \langle n\vert\phi(0)\rangle$.

None of this has anything to do with the uncertainty principle.

share|cite|improve this answer
What if you don't have a particular Hamiltonian operator? How can you identify the energy eigenstates then? Is is sufficient that you know the probability of the energy at that state? I think it might be from what you have written but I am not sure. – Magpie Oct 18 '12 at 0:04
You can't. "Energy eigenstate" just means "eigenstate of the Hamiltonian." So there is no such thing as an energy eigenstate without a Hamiltonian, just as there is no such thing as an eigenvector without a matrix. – David Z Oct 18 '12 at 0:10
Yes but you just said $$H\lvert n\rangle = E_n\lvert n\rangle$$ what about the right hand side then? Surely this is just the probability of being in an energy state n? – Magpie Oct 18 '12 at 0:12
No. That's a quantum state, not a probability. The right side of that equation doesn't represent anything in particular. – David Z Oct 18 '12 at 0:17
Then how do you derive the Hamiltonian? – Magpie Oct 18 '12 at 1:12

protected by Community Feb 16 '15 at 21:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.