I am currently self-learning quantum mechanics with A Textbook of Quantum Mechanics Second Edition by P. M. Matthews and K. Venkatesan.

In Example 3.15 on p. 105, the author proved $\left<T\right>=\frac12\left<\textbf x\cdot\nabla V\right>$ in any stationary state, where $T$ is the kinetic energy operator, $\textbf x$ is the position vector, $V(\textbf x)$ is the potential and $\left<\cdot\right>$ is the expectation value. The proof goes like this:

  1. For any operator $A$ not explicitly dependent on $t$, $\frac{\partial}{\partial t}\left<A\right>=0$ in a stationary state.

  2. Combining Schrodinger's equation $i\hbar\frac{\partial}{\partial t}\psi=H\psi$ and the definition of $\frac{d}{dt} \left<A\right>$, one gets $$\frac{\partial}{\partial t} \left<A\right>=\left<\frac1{i\hbar}[A,H]\right>$$ where $[A,H]$ is the commutator $AH-HA$.

  3. Hence one concludes that $\left<[A,H]\right>=0$.

  4. Then the author considered a strange operator $$A=\textbf x\cdot\textbf p=-i\hbar\sum_{i}x_i\partial_{x_i}.$$

  5. After going through some algebra one arrives at $[A,H]=i\hbar(2T-\mathbf x\cdot\nabla V)$ and thus the desired conclusion follows.

What I am particularly uncomfortable with is step 4: how could one come up with such a choice of operator, given that it has no obvious connections to the statement that we are proving? So, how could one give a more natural proof, ideally starting with $$\left<T\right>=\int \psi^*\left(-\frac{\hbar^2}{2m}\nabla^2\right)\psi \,d^3x~?$$

Aside of this, how could I understand $\left<T\right>=\frac12\left<\textbf x\cdot\nabla V\right>$ intuitively?

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    $\begingroup$ Intuitively, the Virial theorem is just a restatement of the principle of least action as shown here (which makes it quite a fundamental idea!) $\endgroup$ Aug 10, 2020 at 3:57

2 Answers 2


$\mathbf{x\cdot p}$ is not so unnatural, it is the exponent in the Fourier transform and it is from this point of view that this proof of the Virial theorem's relationship to the principle of least action comes out.

In Lie theory, we often consider symmetries which "generate" a continuous set we care about, which often take the form of

$$T = \exp(g)$$

where $g$ is called the generator of $T$. For example, the momentum operator is the generator of spatial translations:

$$\exp\left(\frac{i pL}{\hbar}\right)f(x) = f(x+L)$$

In this view, notice that in the Fourier transform exponential takes the form

$$\exp\left(\frac{i \mathbf{x\cdot p}}{\hbar}\right)$$

where $\mathbf{x\cdot p}$ takes the role of the generator, normalized by $\hbar$. Heuristically, thinking of $\hbar$ as a unit of action, $\mathbf{x\cdot p}$ is a generator with dimensions of action. This leads to the tie in with the Virial Theorem, being the minimizer of action.


Intuitively, the expectation value of the kinetic energy, $\left<T\right>$, in a stationary state has a constant value. Now, $\left<\textbf x\cdot\nabla V\right>$ can intuitively be seen as the expectation value of the product $x\cdot\nabla V$, which can be written as $xF=xma=mv^2$. So the putting the $\frac1 2$ in front of this, we arrive at the expectation value of $\frac 1 2 mv^2$, which is constant in a stationary state.


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