Adjoint of an operator for particles

[noblegas]

Homework Statement

Let x be the position coordinate for a particle that moves in one dimension and let $$p=-i\hbar*d/dx$$ be the usual momentum operator . State whether each of the following operators is self-adjoint, anti-self-adjoint($$A^{\dagger} =A$$ , unitary($$A^{\dagger}A=1$$, or, if none of the above , what the adjoint is:

xxp, xpx, xpp+ppx, d^2/dx^2 , d^3/dx^3,e^p

Homework Equations

The Attempt at a Solution

I will perform the first operator xxp. $$p=-i\hbar*d/dx$$, therefore xxp=x^2*$$p=-i\hbar*d/dx$$=2*x*$$-\hbar*i$$? So would this operator be a self-adjointed one ?

 

[Preno]

That depends on whether it equals the adjoint. Compute the operator and its adjoint and see if they are equal or not. Remember that $(AB)^\dagger = B^\dagger A^\dagger$.

I will perform the first operator xxp. $$p=-i\hbar*d/dx$$, therefore xxp=x^2*$$p=-i\hbar*d/dx$$=2*x*$$-\hbar*i$$? So would this operator be a self-adjointed one ?

This is chaotic and probably false. Why would you state that $xxp = -i \hbar \frac{\textrm{d}}{\textrm{d}x}$, when p is $-i \hbar \frac{\textrm{d}}{\textrm{d}x}$?

 

[gabbagabbahey]

Homework Statement

Let x be the position coordinate for a particle that moves in one dimension and let $$p=-i\hbar*d/dx$$ be the usual momentum operator . State whether each of the following operators is self-adjoint, anti-self-adjoint($$A^{\dagger} =A$$ , unitary($$A^{\dagger}A=1$$, or, if none of the above , what the adjoint is:

xxp, xpx, xpp+ppx, d^2/dx^2 , d^3/dx^3,e^p

Homework Equations

The Attempt at a Solution

I will perform the first operator xxp. $$p=-i\hbar*d/dx$$, therefore xxp=x^2*$$p=-i\hbar*d/dx$$=2*x*$$-\hbar*i$$? So would this operator be a self-adjointed one ?

First, this is incredibly sloppy notation. Usually one denotes operators with uppercase letters, and an operator's expansion into a certain basis (the x-basis in this case[/itex] is denoted with the symbol $\to$ instead of an equal sign. So, $P\to -i\hbar \frac{d}{dx}$ and $X\to x$ in the x-basis.

Second, the order of multiplication between operators is important (sound familiar?).

$$XXP\to x^2\left(-i\hbar \frac{d}{dx}\right)\neq \left(-i\hbar \frac{d}{dx}\right)x^2$$

To determine the nature of its adjoint, you might consider actually computing the adjoint!...In the x-basis, you have:

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}$$

 

[noblegas]

First, this is incredibly sloppy notation. Usually one denotes operators with uppercase letters, and an operator's expansion into a certain basis (the x-basis in this case[/itex] is denoted with the symbol $\to$ instead of an equal sign. So, $P\to -i\hbar \frac{d}{dx}$ and $X\to x$ in the x-basis.

Second, the order of multiplication between operators is important (sound familiar?).

$$XXP\to x^2\left(-i\hbar \frac{d}{dx}\right)\neq \left(-i\hbar \frac{d}{dx}\right)x^2$$

To determine the nature of its adjoint, you might consider actually computing the adjoint!...In the x-basis, you have:

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}$$

would I need to prove(or disprove) that $$(xxp)^{\dagger}=(xxp),xxp*(xxp)^{\dagger}=1, , and (xxp)^{\dagger}=-(xxp)$$

 

[gabbagabbahey]

Well, it turns out that $XXP$ is anti-self-adjoint...to show that, just calculate $$\left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}$$ and compare it to the x-basis representation of $XXP$...

 

[noblegas]

Well, it turns out that $XXP$ is anti-self-adjoint...to show that, just calculate $$\left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}$$ and compare it to the x-basis representation of $XXP$...

$$xxp^{\dagger}=(-i\hbar*d/dx)^{\dagger}*x^{\dagger}x^{\dagger}=xxp, since x=x^{/dagger}, p=p^{/dagger}$$; If PF doesn't display my Latex code properly, Here is the code I was orginally trying to translate through Latex xxp^{\dagger}=(-i\hbar*d/dx)^{\dagger}*x^{\dagger}x^{\dagger}=xxp, since x=x^{/dagger}, p=p^{/dagger}

 

[gabbagabbahey]

$P\neq P^{\dagger}$...why would you think that they were equal?

And again, order is important (drill that into your head):

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=x^{\dagger}x^{\dagger}\left(-i\hbar \frac{d}{dx}\right)^{\dagger}\neq\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}$$

The $\dagger$ symbol tells you to take the complex conjugate transpose of each quantity...so do just that! The quantities $x$ and $-i\hbar\frac{d}{dx}$ are scalars, so taking their transpose is trivial...but what about their complex conjugates?

 

[noblegas]

$P\neq P^{\dagger}$...why would you think that they were equal?

And again, order is important (drill that into your head):

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=x^{\dagger}x^{\dagger}\left(-i\hbar \frac{d}{dx}\right)^{\dagger}\neq\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}$$

The $\dagger$ symbol tells you to take the complex conjugate transpose of each quantity...so do just that! The quantities $x$ and $-i\hbar\frac{d}{dx}$ are scalars, so taking their transpose is trivial...but what about their complex conjugates?

But according to my textbook$$(AB)^{\dagger} =(B)^{\dagger}A^{\dagger}$$ so therefore $$(xxp)^{\dagger}=p^{\dagger}x^{\dagger}x^{\dagger}$$ so I don't understand why my expression would be in correct and don't understand why you think your expression is correct

 

[gabbagabbahey]

But according to my textbook$$(AB)^{\dagger} =(B)^{\dagger}A^{\dagger}$$ so therefore $$(xxp)^{\dagger}=p^{\dagger}x^{\dagger}x^{\dagger}$$ so I don't understand why my expression would be in correct and don't understand why you think your expression is correct

Yes, sorry , my mistake.

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}$$

but my other two comments still stand.

 

[noblegas]

Yes, sorry , my mistake.

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger}$$

but my other two comments still stand.

I am kinda of confused because my book(Peebles) gives two definitions for a self-adjoint operator. $$Q^{\dagger}=Q$$ and the other definition is $$c^{\dagger}=c*$$ where c* represent the adjoint of ca complex conjugate. For the latter case, the x matrices will not change since they are real numbers and contain no imaginary terms right?

 

[Preno]

An operator adjoint to A in a Hilbert space with the product $\langle \ldots;\ldots \rangle$ is an operator B such that for all vectors $\varphi,\psi$:

$\langle \varphi; A \psi \rangle = \langle B\varphi; \psi \rangle$

(You can try writing this out in x-representation.)

 

[noblegas]

An operator adjoint to A in a Hilbert space with the product $\langle \ldots;\ldots \rangle$ is an operator B such that for all vectors $\varphi,\psi$:

$\langle \varphi; A \psi \rangle = \langle B\varphi; \psi \rangle$

(You can try writing this out in x-representation.)

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger }=\left(i\hbar\frac{d}{dx}\right)x*x=2*i\hbar*x$$ since by definition, $$c=c*$$

 

[gabbagabbahey]

I am kinda of confused because my book(Peebles) gives two definitions for a self-adjoint operator. $$Q^{\dagger}=Q$$ and the other definition is $$c^{\dagger}=c*$$ where c* represent the adjoint of ca complex conjugate. For the latter case, the x matrices will not change since they are real numbers and contain no imaginary terms right?

$Q^{\dagger}=Q$ is the definition of a self-adjoint operator.

$c^{\dagger}=c^{*}$ just tells you how to take the adjoint of a complex scalar $c$. (The fact that $c$ is a scalar tells you $c^{T}=c$, so you don't need to worry about the transpose, just the complex conjugation). $x$ is such a scalar,and is also real valued, so $x^{\dagger}=x^{*}=x$

$$(XXP)^{\dagger}\to \left[x^2\left(-i\hbar \frac{d}{dx}\right)\right]^{\dagger}=\left(-i\hbar \frac{d}{dx}\right)^{\dagger}x^{\dagger}x^{\dagger }=\left(i\hbar\frac{d}{dx}\right)x*x=2*i\hbar*x$$ since by definition, $$c=c*$$

Right, $(XXP)^{\dagger}=2i\hbar x$...does that equal $XXP$? does it equal $-XXP$? Does the product $(XXP)(XXP)^{\dagger}$ equal the identity operator?

 

[noblegas]

$Q^{\dagger}=Q$ is the definition of a self-adjoint operator.

$c^{\dagger}=c^{*}$ just tells you how to take the adjoint of a complex scalar $c$. (The fact that $c$ is a scalar tells you $c^{T}=c$, so you don't need to worry about the transpose, just the complex conjugation). $x$ is such a scalar,and is also real valued, so $x^{\dagger}=x^{*}=x$
Right, $(XXP)^{\dagger}=2i\hbar x$...does that equal $XXP$? does it equal $-XXP$

?
no.

Does the product $(XXP)(XXP)^{\dagger}$ equal the identity operator?

$$XXP(XXP)^{\dagger}=X^2*(-i\hbar*d/dx)2i\hbar x\neq\leftI$$ , i.e. not equal to the identity.

 

[gabbagabbahey]

Right, so $XXP$ is neither self-adjoint, anti-self-adjoint or unitary.

Now move on to the rest of the operators in your question...

 

[noblegas]

Right, so $XXP$ is neither self-adjoint, anti-self-adjoint or unitary.

Now move on to the rest of the operators in your question...

$$(xpp+ppx)^{\dagger}=(xpp)^{\dagger}+{ppx}^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}$$. $$p^{\dagger}=p*,x^{\dagger}=x. (i*\hbar*d/dx)(-i*\hbar)+x(\hbar)^2*d^2/dx^2=0+0$$ Therefore, there it is neither

$$(xpx)^{\dagger}=x^{\dagger}p^{\dagger}x^{\dagger). xpx=x(-i*\hbar*d/dx)x=-xi\hbar.(xpx)^{\dagger}=x(i*\hbar*d/dx)x=xi\hbar$$. Therefore, operator is anti-self-adjoint.

$$e^p=e^(-i*\hbar*d/dx). (e^p)^{\dagger)=e^(i*\hbar*d/dx). (e^p)(e^p)^{\dagger}=e^0=1$$. Therefore, $$e^p$$ is unitary. not sure how to prove how [tex] d^2/dx^2[\tex] and $$d^3/dx^3$$ are adjoint, anti-self-adjoint or unitary.

 

[gabbagabbahey]

$$(xpp+ppx)^{\dagger}=(xpp)^{\dagger}+{ppx}^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}$$. $$p^{\dagger}=p*,x^{\dagger}=x. (i*\hbar*d/dx)(-i*\hbar)+x(\hbar)^2*d^2/dx^2=0+0$$ Therefore, there it is neither

$(xpp+ppx)^{\dagger}=(xpp)^{\dagger}+(ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}$ is correct, but I'm not sure what you're doing after that. (Your $\LaTeX$ is very messy!)

Just use $x^{\dagger}=x$ (since $x$ is a real valued scalar) and $p^{\dagger}=-p$ (Since $\left[-i\hbar\frac{d}{dx}\right]^{\dagger}=i\hbar\frac{d}{dx}$ )

That gives you,

$$(xpp+ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}=(-p)(-p)x+x(-p)(-p)=ppx+xpp$$

 

[noblegas]

$(xpp+ppx)^{\dagger}=(xpp)^{\dagger}+(ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}$ is correct, but I'm not sure what you're doing after that. (Your $\LaTeX$ is very messy!)

Just use $x^{\dagger}=x$ (since $x$ is a real valued scalar) and $p^{\dagger}=-p$ (Since $\left[-i\hbar\frac{d}{dx}\right]^{\dagger}=i\hbar\frac{d}{dx}$ )

That gives you,

$$(xpp+ppx)^{\dagger}=p^{\dagger}p^{\dagger}x^{\dagger}+x^{\dagger}p^{\dagger}p^{\dagger}=(-p)(-p)x+x(-p)(-p)=ppx+xpp$$

$$e^p=e^(p). (e^p)^{\dagger)=1? since p^{dagger}=-p, then (e^p)(e^-p)^{\dagger}=e^0=1$$ Therefore , $$e^p$$ is unitary
Don't know how o show whether or not $$d^2/dx^2$$ or $$d^3/dx^3$$ is adjoint, self-adjoint ,unitary , or neither

 

[gabbagabbahey]

$$e^p=e^(p). (e^p)^{\dagger)=1? since p^{dagger}=-p, then (e^p)(e^-p)^{\dagger}=e^0=1$$ Therefore , $$e^p$$ is unitary

Let's see,

$$(e^p)^{\dagger}=\left(\sum_{n=0}^{\infty}\frac{p^n}{n!}\right)^{\dagger}=\sum_{n=0}^{\infty}\frac{(p^n)^{\dagger}}{n!}=\sum_{n=0}^{\infty}\frac{(p^{\dagger})^n}{n!}=\sum_{n=0}^{\infty}\frac{(-p)^n}{n!}=e^{-p}$$

So yes, I would say $e^{p}$ is unitary.

Don't know how to show whether or not $$d^2/dx^2$$ or $$d^3/dx^3$$ is adjoint, self-adjoint ,unitary , or neither

I'd start by defining the operator $D\leftrightarrow\frac{d}{dx}$, then $\frac{d^2}{dx^2}\leftrightarrow D^2=DD$, then just apply the same reasoning as before.

 

[noblegas]

Let's see,

$$(e^p)^{\dagger}=\left(\sum_{n=0}^{\infty}\frac{p^n}{n!}\right)^{\dagger}=\sum_{n=0}^{\infty}\frac{(p^n)^{\dagger}}{n!}=\sum_{n=0}^{\infty}\frac{(p^{\dagger})^n}{n!}=\sum_{n=0}^{\infty}\frac{(-p)^n}{n!}=e^{-p}$$

So yes, I would say $e^{p}$ is unitary.
I'd start by defining the operator $D\leftrightarrow\frac{d}{dx}$, then $\frac{d^2}{dx^2}\leftrightarrow D^2=DD$, then just apply the same reasoning as before.

i.e. show that $$D^{\dagger}=D$$ and $$(D^2)^{\dagger}=D^{\dagger}D^{\dagger}$$?

 

[gabbagabbahey]

i.e. show that $$D^{\dagger}=D$$ and $$(D^2)^{\dagger}=D^{\dagger}D^{\dagger}$$?

Yes.

 

[noblegas]

Yes.

$$D=d/dx=> (D)^{\dagger}=(d/dx)^{\dagger} =d/dx$$

$$(D^2)^{\dagger}=(DD)^{\dagger}=D^{\dagger}D^{\dagger} =(d/dx)^{\dagger}(d/dx)^{\dagger},$$. Since d/dx has no imaginary conjugate$$(d/dx)^{\dagger}(d/dx)^{\dagger}=(d/dx)(d/dx)=d^2/dx^2$$

 

[gabbagabbahey]

Right, so $(D^2)^{\dagger}=D^2$ and hence, $\frac{d^2}{dx^2}$ is self-adjoint.