We consider a particle in an infinite potential well in one dimension. The particle is describe by the arbitrary wave function $\psi(x)$.

For our particle, we have to show that: \begin{equation} \langle{E}\rangle\geq{E}_{1} \end{equation}

This is a problem I found in the Introductory quantum mechanics of Richard L. Liboff. I tried to solve the problem this way.

\begin{equation} \langle{Q(x,p)}\rangle=\int\Psi^{*}Q\Big(x,\frac{\hbar}{i}\frac{\partial}{\partial{x}}\Big)\Psi{d}x \end{equation}

Because we can write the energy as $E=\frac{p^2}{2m}$ we have that:

\begin{equation} \langle{E}\rangle=\int\Psi^{*}\hat{E}\Psi{d}x=-\frac{\hbar^2}{2m}\int\Psi^{*}\frac{\partial^2\Psi}{\partial{x}^2}dx \end{equation}

But if we use the time-independant Schrödinger equation, we have (because the potential energy is zero):

\begin{equation} \langle{E}\rangle=\int\Psi^{*}E\Psi{d}x=E\int\Psi^{*}\Psi{d}x=E \end{equation}

This is what I get, and I think it's logic because the energy conservation. My problem is to talk about the inequality. Do we have $\langle{E}\rangle\geq{E}_{1}$ because $E$ is the summation of all the energy $E_{n}$ associated with a stationary state $\psi_{n}(x)$? My idea is that, because we don't have negative energy, if we only take one term on a summation we have a smaller value than the summation of all $E_{n}$. The only way to b equal mean that $\psi(x)=\psi_{1}(x)$.

I'm not sure if I'm doing mistakes or not or if my mathematics are rigorous or if there is a little something I don't understand.

  • $\begingroup$ Make sure you understand the question. An expectation value is the value the weighted averages approach as you have a larger and larger sample. So you measure the energy on an ensemble of identically prepared systems and get certain results with certain frequencies. The results are the eigenvalues of the operator, and the relative frequencies are the ratio of the square of the projections. Since the things you are weighting together are greater than or equal to E1 ... $\endgroup$
    – Timaeus
    Sep 24, 2015 at 15:58
  • $\begingroup$ Yes I've seen it that way after I notice that the question asked to answer with an arbitrary wave function. But if I understand correctly, the only possible value to get are the eigenvalues of the operator? $\endgroup$
    – WernerSpin
    Sep 24, 2015 at 18:48
  • $\begingroup$ for a strong measurement in the span jf the point spectrum You always get an eigenvalue $\endgroup$
    – Timaeus
    Sep 24, 2015 at 19:00

2 Answers 2


An arbitrary state can be written as a linear combination of the energy normalized eigenstates ${{\Psi }_{n}}$:

$\Psi =\sum\limits_{n}{{{c}_{n}}{{\Psi }_{n}}}$

and the mean energy would be calculated as:

$\left\langle E \right\rangle :=\left\langle \Psi ,H\Psi \right\rangle =\sum\limits_{n,m}{c_{n}^{*}{{c}_{m}}\left\langle {{\Psi }_{n}},H{{\Psi }_{m}} \right\rangle }=\sum\limits_{n}{{{\left| {{c}_{n}} \right|}^{2}}{{E}_{n}}}$

where we have used the energy eigenvalues equation: $H{{\Psi }_{n}}={{E}_{n}}{{\Psi }_{n}}$ and the orthogonallity condition: $\left\langle {{\Psi }_{n}},{{\Psi }_{m}} \right\rangle ={{\delta }_{nm}}$. Since ${{E}_{n}}\ge {{E}_{1}}>0\ ,\ \forall n\in \mathbb{N}$ we deduce that:

$\left\langle E \right\rangle =\sum\limits_{n}{{{\left| {{c}_{n}} \right|}^{2}}{{E}_{n}}}\ge {{E}_{1}}\sum\limits_{n}{{{\left| {{c}_{n}} \right|}^{2}}}={{E}_{1}}$

where we have used the normalization condition:

$\left\langle \Psi ,\Psi \right\rangle =1\Rightarrow \sum\limits_{n}{{{\left| {{c}_{n}} \right|}^{2}}}=1$


For a infinite quantum well from $x=0$ to $x=a$, the Hamiltonian (energy operator) and the wave function are: $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$$ $$\psi(x)=A\sin kx$$ where A is defined by the normalization condition. And $k=\frac{\sqrt{2mE}}{\hbar}$.

From the boundary condition, we can find out $E_n$.

For the energy expectation value: $$\langle{E}\rangle=\int\limits_{0}^{a}\psi H\psi dx$$ From those, you can compare $\langle{E}\rangle$ and $E_1$.

  • $\begingroup$ The question asked you to consider an arbitrary state, not just an energy eigenstate. $\endgroup$
    – Timaeus
    Sep 24, 2015 at 16:00

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