Consider a system of particles with wave function $\psi$(x) (x can be understood to stand for all degrees of freedom of the system; so, if we have a system of two particles then x should represent {$x_1; y_1; z_1; x_2; y_2; z_2$}). The expectation value of an operator $\hat{A}$ that operates on is defined by : $$\langle\hat{A}\rangle = \int\psi^{*}\hat{A}\psi dx$$

Yup this makes sense to me and there's nothing new here.

If $\psi$ is an eigenfunction of $\hat{A}$ with eigenvalue $a$, then, assuming the wave function to be normalized, we have : $$⟨ \hat{A} ⟩ = a$$

This is where I want to confirm something.

$$\hat{A}\psi = a\psi$$

Hence, $$⟨ \hat{A} ⟩ =\int\psi^{*} a \psi dx$$

Since $a$ is a constant I can take it out :

$$\langle\hat{A}\rangle = a \int\psi^{*} \psi dx$$

We assumed that the wave function was normalized hence $$\int\psi^{*} \psi dx = 1$$

Leaving $$\langle\hat{A}\rangle = a$$

Now consider the rate of change of the expectation value of $\langle\hat{A}\rangle$:

$$\frac{d\langle\hat{A}\rangle}{dt} = \int{\frac{\partial}{\partial t}}(\psi^{*}\hat{A}\psi)dx$$

$$=\int{\frac{\partial \psi^{*}}{\partial t}\hat{A}\psi+\psi^{}\frac{\partial\hat{A}}{\partial t}\psi^{*}}+\frac{\partial \psi}{\partial t}\hat{A}\psi^{*} dx$$

$$=\int{\langle\frac{\partial\hat{A}}{\partial t}\rangle} +\frac{i}{\hbar}\int{[(\hat{H}\psi)^{*}\hat{A}\psi-\psi^{*}\hat{A}\hat{H}\psi]}dx$$

where we have used the Schrodinger equation :

$$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$

The second line is easily obtained via differentiation. The second term in the second line corresponds to the first term in the third line, correct ?

I do not see how this term was obtained. In particular where the $\frac{i}{\hbar}$ originates from :


Please help me.

PS : Take it easy on me. QM in general is quite new to me.

  • $\begingroup$ What is your question? $\endgroup$
    – Kyle Kanos
    Commented Apr 3, 2014 at 1:25
  • $\begingroup$ Give me a few minutes. I'm still editing my post :) $\endgroup$ Commented Apr 3, 2014 at 1:32
  • 1
    $\begingroup$ Okay. You don't need to write your equations as $<\hat{A}>$=$a$, simply use $\langle\hat{A}\rangle=a$ (i.e., surround single equation with $ for an inline equation and $$ for a centered equation) $\endgroup$
    – Kyle Kanos
    Commented Apr 3, 2014 at 1:34
  • $\begingroup$ Alrighty, will have to make some changes to my latest edit then. $\endgroup$ Commented Apr 3, 2014 at 1:54

3 Answers 3


They have used the Schrödinger equation: $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H} \psi$$ And hence: $$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi$$

And since $1/i = -i$: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$

Similarly for $\psi^*$ we take the complex conjugate: $$\frac{\partial \psi^*}{\partial t} = \frac{i}{\hbar}(\hat{H}\psi)^*$$

Plugging these values into the integral gives us the last line you were confused about.


From the Schroedinger equation, $$i\hbar\frac{\partial\psi}{\partial t}=\hat{H}\psi$$ simply divide by $i\hbar$: $$\frac{\partial\psi}{\partial t}=\frac{1}{i\hbar}\hat{H}\psi=-\frac{i}{\hbar}\hat{H}\psi\\$$ where we used $$ \frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i $$

Now note that you've incorrectly applied the derivative. In quantum mechanics, $\hat{A}$ is an operator and acts on objects to the right: $\hat{H}\psi=E\psi$. That is to say, order matters: $\psi^*\hat{A}\psi$ is not necessarily the same as $\psi\hat{A}\psi^*$. As such, the line becomes $$\frac{\partial \psi^*}{\partial t}\hat{A}\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi+\psi^*\hat{A}\frac{\partial \psi}{\partial t} $$ (obviously ignoring the integral). This can be rearranged to give you the line you expected: $$\psi^*\frac{\partial\hat{A}}{\partial t}\psi+\frac{\partial \psi^*}{\partial t}\hat{A}\psi+\psi^*\hat{A}\frac{\partial \psi}{\partial t} $$ So that first term is indeed the middle term of the previous line. The remaining two terms use the Schroedinger equation $$\frac{\partial \psi^*}{\partial t}\hat{A}\psi+\psi^*\hat{A}\frac{\partial \psi}{\partial t} =\left(-\frac i\hbar\hat{H}\psi\right)^*\hat{A}\psi+\psi^*\hat{A}\left(-\frac{i}{\hbar}\hat{H}\psi\right)$$ This should reduce to your solution (though I think the minus signs are different).

  • $\begingroup$ Yeah, I think that was more to do with my poor LaTeX skills. I was having a hard time trying to put everything in, but yes I am aware that the order matters. Thanks though ! $\endgroup$ Commented Apr 3, 2014 at 2:30
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    $\begingroup$ @AriBenCanaan, rather than dividing both sides by i, instead you can simply multiply both sides by i. E.g. iA = B, then iiA = iB, then -A = iB. I think this trick is easier than the divide by i suggested by Kyle and Nuclear, but ultimately it is just a matter of individual style, but it is good to know both. $\endgroup$
    – Kenshin
    Commented Apr 3, 2014 at 2:38
  • $\begingroup$ Yes of course ! That does make sense too. I'm an organic chemist so you can see why QM isn't my strongest ! :P $\endgroup$ Commented Apr 3, 2014 at 2:40

To arrive at the 2nd term of the third line, they have simply employed the SE equation, $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$

Re-arranging the above equation gives: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$

Therefore by replacing $\frac{\partial \psi}{\partial t}$ in the second line with $-\frac{i}{\hbar}\hat{H}\psi$, you will arrive at the 3rd line without too much trouble.


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