Quick Quantum Mechanics Q about basis

If it's just a partial sum then you need to show that the missing terms are zero. In summary, the expression Ʃ(j)<k|j><j|i> equals <k|i> if j sums over all elements of the orthonormal basis. If it is a partial sum, the missing terms need to be shown to be zero.
  • #1
bon
559
0

Homework Statement



Suppose we have a system and that {|a>, |b>, ...} is a complete and orthonormal basis for the system

Am i right in thinking Ʃ(j) <k|j><j|i> = <k|i> = 0 unless k=i?

In other words, does the LHS expression equal the middle one because Ʃ(j) |j><j| is just the insertion of the identity and we can put it in anywhere?

Homework Equations





The Attempt at a Solution



I've explained my attempt above.

Thanks!
 
Physics news on Phys.org
  • #2
bon said:

Homework Statement



Suppose we have a system and that {|a>, |b>, ...} is a complete and orthonormal basis for the system

Am i right in thinking Ʃ(j) <k|j><j|i> = <k|i> = 0 unless k=i?

In other words, does the LHS expression equal the middle one because Ʃ(j) |j><j| is just the insertion of the identity and we can put it in anywhere?

Homework Equations





The Attempt at a Solution



I've explained my attempt above.

Thanks!

You need to say what |k> and |i> are. If they are elements of the orthonormal basis then if k≠i, <k|i>=0, just because the basis is orthonormal and that's what the 'normal' part means. No need to insert the identity anywhere.
 
  • #3
Dick said:
You need to say what |k> and |i> are. If they are elements of the orthonormal basis then if k≠i, <k|i>=0, just because the basis is orthonormal and that's what the 'normal' part means. No need to insert the identity anywhere.

Thanks, sorry I wasn't clear. |k> and |i> are elements of the orthonormal basis. And I know that this means <k|i>=0 if k doesn't equal i. It's just that (as part of a larger calculation) I have arrived at the expression

Ʃ(j)<k|j><j|i> (where |j> is also an element of the orthnormal basis) and just wanted to check this equals <k|i>. Am I correct in thinking it does?

Thanks again
 
  • #4
bon said:
Thanks, sorry I wasn't clear. |k> and |i> are elements of the orthonormal basis. And I know that this means <k|i>=0 if k doesn't equal i. It's just that (as part of a larger calculation) I have arrived at the expression

Ʃ(j)<k|j><j|i> (where |j> is also an element of the orthnormal basis) and just wanted to check this equals <k|i>. Am I correct in thinking it does?

Thanks again

Sure, if j sums over ALL of the elements of the orthonormal basis.
 
  • #5


Yes, you are correct in your thinking. The expression on the left-hand side represents the inner product of |k> and |i>, which is equal to 0 unless k=i (since the basis is orthonormal). This can be seen by inserting the identity operator Ʃ(j) |j><j|, which essentially acts as a "placeholder" for the basis vectors. This allows us to rearrange the expression and see that it is equivalent to the middle one. This is a fundamental property of orthonormal bases in quantum mechanics. Great job on your understanding!
 

What is a basis in quantum mechanics?

In quantum mechanics, a basis is a set of states or vectors that can be used to express or represent any other state or vector within the system. It serves as a foundation for describing the state of a quantum system and allows for mathematical calculations and predictions.

Why is the choice of basis important in quantum mechanics?

The choice of basis is important in quantum mechanics because it affects the ease and accuracy of calculations and predictions. Different bases may be more suitable for describing certain physical systems, and choosing the appropriate basis can simplify the mathematical representation of the system.

How is a basis determined in quantum mechanics?

In quantum mechanics, the basis is determined by the physical properties of the system being studied. For example, in a system with discrete energy levels, the basis may be determined by the energy states of the system. In a system with continuous variables, the basis may be determined by position or momentum states.

Can a single state or vector be used as a basis in quantum mechanics?

No, a single state or vector cannot be used as a basis in quantum mechanics. A basis must consist of a set of states or vectors that are linearly independent, meaning they cannot be expressed as a combination of each other. This allows for a unique representation of any state within the system.

How does the concept of basis relate to the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This is related to the concept of basis in quantum mechanics because the choice of basis can affect the precision with which these properties can be measured. The more accurately one property is known, the less accurately the other can be known, and the choice of basis can impact this trade-off.

Similar threads

Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
26
Views
3K
  • Advanced Physics Homework Help
Replies
9
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Back
Top