Expanding a bracket of derivatives

[etotheipi]

Homework Statement

Show that $(\frac{d}{dx} +x)(-\frac{d}{dx} + x)z = -\frac{d^{2}z}{dx^{2}}+x^{2}z + z$

Relevant Equations

N/A

I am wondering why the two methods below give different answers. If I multiply z through the second bracket I get $$(\frac{d}{dx} +x)(-\frac{dz}{dx} + xz)$$which, on expansion, yields $$-\frac{d}{dx}\frac{dz}{dx} -x\frac{dz}{dx} + \frac{d(xz)}{dx} + x^{2}z = -\frac{d^{2}z}{dx^{2}} + x^{2}z + z$$ via the product rule on the third term, as required. However, if instead I expand the brackets first before multiplying through by z, I get$$(-\frac{d}{dx}\frac{d}{dx} -x\frac{d}{dx} + x\frac{d}{dx} + x^{2})z = -\frac{d^{2}z}{dx^{2}} + x^{2}z$$ I know the error has something to do with misusing the \frac{d}{dx} operator, but I can't pinpoint it.

 

[fresh_42]

What is / does $z$? Is it left multiplication with a variable called $z$? What is / does $z.f(x)$?

I would apply the entire thing to a function $f$ and work with usual derivatives. And note that they are not commutative!

 

[WWGD]

Yes, please define and/or give refs for the operator.

 

[etotheipi]

What is / does $z$? Is it left multiplication with a variable called $z$? What is / does $z.f(x)$?

I would apply the entire thing to a function $f$ and work with usual derivatives. And note that they are not commutative!

Yes, please define and/or give refs for the operator.

Sorry I should have added, that z(x) is some function of x, though no other information is given apart from that and the problem statement.

 

[fresh_42]

Then apply the operators to $z(x)$ and write $\dfrac{d}{dx}z(x)=z'(x)$. But do not multiply. Chances are you make a mistake since multiplication isn't commutative.
$$
(\frac{d}{dx} +x)(-\frac{d}{dx} + x)z =\left(\frac{d}{dx} +x\right)\left[\left((-\frac{d}{dx} + x)z\right)\right]
$$

 

[PeroK]

However, if instead I expand the brackets first before multiplying through by z, I get$$(-\frac{d}{dx}\frac{d}{dx} -x\frac{d}{dx} + x\frac{d}{dx} + x^{2})z = -\frac{d^{2}z}{dx^{2}} + x^{2}z$$ I know the error has something to do with misusing the \frac{d}{dx} operator, but I can't pinpoint it.

The operators $x$ and $\frac{d}{dx}$ do not commute. That should be:
$$(-\frac{d}{dx}\frac{d}{dx} -x\frac{d}{dx} + \frac{d}{dx}x + x^{2})z $$
And the two terms in the middle do not cancel.

By the way, just out of interest, the fact that $x$ and $\frac{d}{dx}$ do not commute is actually at the heart of Quantum Mechanics.

 

[etotheipi]

Then apply the operators to $z(x)$ and write $\dfrac{d}{dx}z(x)=z'(x)$. But do not multiply. Chances are you make a mistake since multiplication isn't commutative.
$$
(\frac{d}{dx} +x)(-\frac{d}{dx} + x)z =\left(\frac{d}{dx} +x\right)\left[\left((-\frac{d}{dx} + x)z\right)\right]
$$

The operators $x$ and $\frac{d}{dx}$ do not commute. That should be:
$$(-\frac{d}{dx}\frac{d}{dx} -x\frac{d}{dx} + \frac{d}{dx}x + x^{2})z $$
And the two terms in the middle do not cancel.

By the way, just out of interest, the fact that $x$ and $\frac{d}{dx}$ do not commute is actually at the heart of Quantum Mechanics.

Thank you, I think I now understand. My mistake was exactly that, assuming that each term obtained by expanding the brackets could commute when in fact the order of the different terms must be maintained as per the brackets.