Standard deviation of the Hamiltonian?

  • #1

Summary:

I'm having trouble, or at least am unsure of my interpretation of stationary states and standard deviations of dynamical observables.

Main Question or Discussion Point

I am currently reading Griffiths Introduction to Quantum Mechanics, 2nd Edition. I am aware that, in light of considering potential functions independent of time, the Schrödinger equation has separable solutions and that these solutions are stationary states. I am also aware (If I stand correct) that the stationary states can be put into a linear combination (I'm supposing for now states without the time dependent solution attached, which is the exponential factor obtained from the method of separation of variables) to form a wave function that is also a solution to the time independent SE.

1) Am I correct in interpreting each term of the linear combination (without the time dependence) as a single stationary state of the wave function? Is this still true if we tack on the time dependence?

2) For stationary states, the standard deviation of the Hamiltonian is zero (I believe). However, my state of confusion arised in doing a problem where the standard deviation of the Hamiltonian was NOT zero. Why is this the case?

I would appreciate any help or guidance in putting me on the right track, if it seems that my train of thought is derailing in any manner.
 
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Answers and Replies

  • #2
I should ask rather, instead of a vague "why is this the case," is it true that if the standard deviation of the Hamiltonian is non-zero for a particular state, that state is not a stationary state?
 
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  • #3
vanhees71
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The time-independent Schrödinger equation is simply the eigenvalue problem of the Hamiltonian, i.e., it determines the eigenvalues and the corresponding eigenstates. For any eigenvalue there's a set of eigenvectors, which can be chosen to be orthonormal. All eigenvectors together form then a complete orthonormal basis (if your Hamiltonian has a continuous part in the spectrum, of course you have to take into account that these are generalized eigenvectors in the sense of distributions). Let's denote this complete orthonormal set by ##|E,\alpha \rangle##, where ##E## runs over the eigenvalues of ##E## and for each ##E##, ##\alpha## is a set of parameters specifying the different orthonormal eigenvectors of ##E##.

Now a superposition of such eigenvectors is only another eigenvector, if the superposition is only over the eigenvectors to the same eigenvalue, i.e., if
$$|\psi_E \rangle=\sum_{\alpha} c_{\alpha} |E,\alpha \rangle.$$
A general superposition is not an eigenvector and thus also doesn't solve the time-independent Schrödinger equation.

Of course, it's right that you can solve the time-dependent Schrödinger equation for any initial condition easily by such a general superposition, i.e., (in the Schrödinger picture of the time evolution)
$$|\psi(t) \rangle = \sum_{E,\alpha} \psi_0(E,\alpha) \exp(-E t/\hbar) |E,\alpha \rangle,$$
where
$$\psi_0(E,\alpha)=\langle E,\alpha|\psi_0 \rangle$$
with the given state vector ##|\psi_0 \rangle## at time ##t=0##.

This again shows that ##|\psi(t) \rangle## is a stationary state only, if it's an eigenstate of the Hamiltonian, i.e., if
$$\psi(t)=\exp(\mathrm{i} E t/\hbar) \sum_{\alpha} \psi_0(E,\alpha) |E,\alpha \rangle,$$
i.e., if
$$\psi_0(E',\alpha)=\delta_{E,E'} \psi_0(E,\alpha).$$
Of course, if there are continuous parts in the spectrum and in the parameter(s) ##\alpha##, instead of sums you have to write integrals and instead of ##\delta_{E,E'}## it's a Dirac-##\delta##, ##\delta(E-E')##.
 

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