# Standard deviation of the Hamiltonian?

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• Kyle Nemeth
In summary, 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 Schrödinger equation.
Kyle Nemeth
TL;DR Summary
I'm having trouble, or at least am unsure of my interpretation of stationary states and standard deviations of dynamical observables.
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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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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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')##.

## What is the standard deviation of the Hamiltonian?

The standard deviation of the Hamiltonian is a measure of the spread of values around the average energy value of a quantum system. It is calculated by taking the square root of the variance of the Hamiltonian, which is the average of the squared differences between each energy value and the average energy value.

## Why is the standard deviation of the Hamiltonian important?

The standard deviation of the Hamiltonian is important because it provides information about the uncertainty or variability of the energy values in a quantum system. It can help scientists understand the stability and behavior of the system, and can also be used to calculate other important quantities such as the heat capacity and thermodynamic properties.

## How is the standard deviation of the Hamiltonian related to the uncertainty principle?

The standard deviation of the Hamiltonian is related to the uncertainty principle in that it represents the uncertainty or variability of the energy values in a quantum system. According to the uncertainty principle, there is a fundamental limit to how precisely we can know both the position and momentum of a particle, and this uncertainty is reflected in the standard deviation of the Hamiltonian.

## Can the standard deviation of the Hamiltonian be negative?

No, the standard deviation of the Hamiltonian cannot be negative. It is a measure of the spread of values around the average energy value, and therefore must be a positive value. A negative standard deviation would imply that the energy values are all below the average, which is not possible in a quantum system.

## How is the standard deviation of the Hamiltonian calculated?

The standard deviation of the Hamiltonian is calculated by taking the square root of the variance of the Hamiltonian. The variance is calculated by taking the average of the squared differences between each energy value and the average energy value. This calculation can be done using mathematical equations or with specialized software programs.

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