How to determine the product of two Hermitian operators is Hermitian

186
2
Let's say we have operator X that is Hermitian and we have operator P that is Hermitian. Is the following true:

[X,P]=ihbar

This is the commutator of X and P.
This particular result is known as the canonical commutation relation.
Expanding:
[X,P]=XP-PX=ihbar
This result indicates that XP[itex]\neq[/itex]PX because XP-PX[itex]\neq[/itex]0
Because XP[itex]\neq[/itex]PX, XP is not a Hermitian operator.
Likewise, because PX[itex]\neq[/itex]XP, PX is not a Hermitian operator.

So to summarize:

The commutator implies multiplication of operators
Multiplication of Hermitian operators does not always produce another Hermitian operator.
If two Hermitian operators do not commute, then their product is not Hermitian.

Any mistakes here?
 

UltrafastPED

Science Advisor
Gold Member
1,910
214
The product of Hermitian operators A,B is Hermitian only if the two operators commute: AB=BA.
This follows directly from the definition of Hermitian: H*=H. Then using the properties of the conjugate transpose:

(AB)*= B*A* = BA which is not equal to AB unless they commute.

The relationship [X,P]=ihbar holds when X,P form a conjugate pair from Lagrangian mechanics - if X is the position operator, then P is the momentum operator conjugate to X.

For example, see http://en.wikipedia.org/wiki/Canonical_coordinates
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,949
531
Actually. XP=PX+const*1 tells you 3 things directly:

1. If the representation is made on a complex separable Hilbert space, this space is infinite dimensional.
2. At least one of the 2 operators is unbounded.
3. The Stone-von Neumann uniqueness theorem.

From these 3 things one can infer that when XP and PX are defined on a common dense everywhere invariant domain of L^2 (R) such as the Schwartz space S(R), X,P are essentially self-adjoint and the operators XP and PX are not even symmetric.
 

UltrafastPED

Science Advisor
Gold Member
1,910
214
Actually. XP=PX+const*1 tells you 3 things directly:

1. If the representation is made on a complex separable Hilbert space, this space is infinite dimensional.
2. At least one of the 2 operators is unbounded.
3. The Stone-von Neumann uniqueness theorem.

From these 3 things one can infer that when XP and PX are defined on a common dense everywhere invariant domain of L^2 (R) such as the Schwartz space S(R), X,P are essentially self-adjoint and the operators XP and PX are not even symmetric.
I didn't understand any of that ... not how it is connected to the OP, not too clear on 1,2, 3, and not at all on the conclusions.


And I had QM as an undergrad and in graduate school ...
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,949
531
1 is a trivial result.
2 Is a result of Wintner (1947) and Wielandt (1948).
3 is a statement of Stone rigorously proved by von Neumann for the first time.
 

George Jones

Staff Emeritus
Science Advisor
Gold Member
7,223
771
I didn't understand any of that ... not how it is connected to the OP, not too clear on 1,2, 3, and not at all on the conclusions.


And I had QM as an undergrad and in graduate school ...
I think this, at least partially, shows the difference between North American and European programs. I also saw none of this in undergrad or grad quantum mechanics (in Canada).

I think your post #2 is the appropriate answer to the OP.

For my own interest, I took loads of pure math courses not required for my physics degrees.

Since all linear operators on finite-dimensional vector spaces are bounded, 1. follows from 2., and 2. is proved here:

Set [itex]\hbar = 1[/itex] and assume

[tex]AB - BA = iI.[/tex]

Multiplying the commutation relation by [itex]B[/itex] and reaaranging gives

[tex]
\begin{equation*}
\begin{split}
AB - BA &= iI \\
AB^2 - BAB &= iB \\
AB^2 - B \left( BA + iI \right) &= iB \\
AB^2 - B^2 A &= 2iB.
\end{split}
\end{equation*}
[/tex]

By induction,

[tex]AB^n - B^n A = niB^{n-1}[/tex]

for every positive integer [itex]n[/itex]. Consequently,

[tex]
\begin{equation*}
\begin{split}
n\left\| B \right\|^{n-1} &=\left\| AB^n - B^n A \right\| \\
&\leq 2\left\| A \right\| \left\| B \right\|^n\\
n &\leq 2\left\| A \right\| \left\| B \right\| .
\end{split}
\end{equation*}
[/tex]

Because this is true for every [itex]n[/itex], at least one of [itex]A[/itex] and [itex]B[/itex] must be unbounded. Say it is [itex]A[/itex]. Then, by the Hellinger-Toeplitz theorem, if [itex]A[/itex] is self-adjoint, the domain of physical observable [itex]A[/itex] cannot be all of Hilbert space!
 

UltrafastPED

Science Advisor
Gold Member
1,910
214
None of this seems relevant to the OP ... no matter how erudite it is.
 
186
2
UltrafastPED, thanks for your post. You are right, that is more along the lines of what I was asking. Thanks also George Jones and dexter, it's interesting to hear about concepts beyond what I have studied so far
 

Related Threads for: How to determine the product of two Hermitian operators is Hermitian

Replies
8
Views
964
Replies
2
Views
5K
  • Posted
Replies
16
Views
1K
  • Posted
Replies
4
Views
2K
  • Posted
Replies
2
Views
1K
  • Posted
Replies
1
Views
1K
  • Posted
Replies
5
Views
2K
Replies
4
Views
18K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top