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}<br /> =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}<br /> =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 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

<br /> e^p=e^(p). (e^p)^{\dagger)=1? since p^{dagger}=-p, then (e^p)(e^-p)^{\dagger}=e^0=1<br /> 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]

<br /> e^p=e^(p). (e^p)^{\dagger)=1? since p^{dagger}=-p, then (e^p)(e^-p)^{\dagger}=e^0=1<br /> 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=&gt; (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.