B Adjoint Operation in Shankar's "Principles of Quantum Mechanics

[Getterdog]

TL;DR Summary

Confused

In R Shankar text on “principles of quantum mechanics’ discussing the adjoint operation, 1.3.8 shows that a|V> => <V|a*. Then 1.3.9 then states

that <aV| = <V|a*. Is this a typo error?

 

[PeroK]

Summary:: Confused

In R Shankar text on “principles of quantum mechanics’ discussing the adjoint operation, 1.3.8 shows that a|V> => <V|a*. Then 1.3.9 then states

that <aV| = <V|a*. Is this a typo error?

It might help if you were more explicit about what all these symbols mean. For example, what does this mean?
$$a|V \rangle \ \Rightarrow \ \langle V|a^*$$

 

[Getterdog]

It might help if you were more explicit about what all these symbols mean. For example, what does this mean?
$$a|V \rangle \ \Rightarrow \ \langle V|a^*$$

Well.a is a complex number, <v| is a row vector, |V> is a column vector.

 

[PeroK]

Well.a is a complex number, <v| is a row vector, |V> is a column vector.

Ah! When you said "adjoint operation" I thought $V$ was an operator. That makes sense.

I would say that $\langle V|$ is a bra and $|V \rangle$ is a ket.

It's not a typo. What are $\langle aV|$ and $\langle V| a^*$? Before you can see they are equal you first have to understand fully what they are. Can you describe each precisely?

 

[PeroK]

Well.a is a complex number, <v| is a row vector, |V> is a column vector.

And, I guess $\Rightarrow$ means "is associated with" via the mapping of kets to bra?

I would tend to write $\leftrightarrow$, as it's a two-way association. But, if that's Shankar's notation, then stick with it.

 

[PeterDonis]

a is a complex number

A complex number or an operator? I think it's the latter (since otherwise $a^*$ would not need to multiply the bra $\langle V \vert$ from the right).

 

[PeroK]

A complex number or an operator? I think it's the latter (since otherwise $a^*$ would not need to multiply the bra $\langle V \vert$ from the right).

I think I've seen that convention where complex scalars are shown on the right of bras.

 

[Getterdog]

He states that a is a scalar. And I presume complex.

 

[Getterdog]

Ah! When you said "adjoint operation" I thought $V$ was an operator. That makes sense.

I would say that $\langle V|$ is a bra and $|V \rangle$ is a ket.

It's not a typo. What are $\langle aV|$ and $\langle V| a^*$? Before you can see they are equal you first have to understand fully what they are. Can you describe each precisely?

Inside the bra ,<aV |would be a complex scalar times V which is presumed to be the complex conjugate in row form of some other vector v in column form. Since complex multiplication is communitive, I assume <Va| is the same. I’m not clear why taking the a outside of the bra,creates the scalars complex conjugate

 

[PeroK]

Inside the bra ,<aV |would be a complex scalar times V which is presumed to be the complex conjugate in row form of some other vector v in column form. Since complex multiplication is communitive, I assume <Va| is the same. I’m not clear why taking the a outside of the bra,creates the scalars complex conjugate

$\langle aV |$ is the bra associated with the ket $|aV \rangle$

If you don't immediately see that, just let $U = aV$.

And, what can you now say about $|aV \rangle$?

 

[Getterdog]

$\langle aV |$ is the bra associated with the ket $|aV \rangle$

If you don't immediately see that, just let $U = aV$.

And, what can you now say about $|aV \rangle$?

Ok, doesn’t that mean that in <aV|, both a and V are complex conjugates of |aV> ?

 

[PeroK]

Ok, doesn’t that mean that in <aV|, both a and V are complex conjugates of |aV> ?

Complex conjugate applies only to complex numbers. You don't have the complex conjugate of a ket.

Let me explain this then.

First, as I already said: $\langle aV |$ is the bra associated with the ket $|aV \rangle$

And, we know that $|aV \rangle = a |V \rangle$.

Then, we know from your original post that the bra associated with $a |V \rangle$ is $\langle V|a^*$.

Putting this all together we have: $\langle aV | = \langle V|a^*$

Note that bras and kets are intended to form the inner product, which in QM is linear in the second argument. In other words we have:
$$\langle V|aU \rangle = a \langle V|U \rangle \ \text{and} \ \langle aV|U \rangle = a^* \langle V|U \rangle$$
And that's another reason that when you take a complex scalar outside of a bra you get its complex conjugate.

 

[Getterdog]

Ok, doesn’t that mean that in <aV|, both a and V are complex conjugates of |aV> ?

Yep got the obvious,, now. I guess my confusion what why it’s not marked inside the bra as it is outside. Presumably to avoid confusion when doing further inner product calculation. I seems obvious now. I feel stupid! Thanks

 

[Getterdog]

Complex conjugate applies only to complex numbers. You don't have the complex conjugate of a ket.

Let me explain this then.

First, as I already said: $\langle aV |$ is the bra associated with the ket $|aV \rangle$

And, we know that $|aV \rangle = a |V \rangle$.

Then, we know from your original post that the bra associated with $a |V \rangle$ is $\langle V|a^*$.

Putting this all together we have: $\langle aV | = \langle V|a^*$

Note that bras and kets are intended to form the inner product, which in QM is linear in the second argument. In other words we have:
$$\langle V|aU \rangle = a \langle V|U \rangle \ \text{and} \ \langle aV|U \rangle = a^* \langle V|U \rangle$$
And that's another reason that when you take a complex scalar outside of a bra you get its complex conjugate.

Yep,got it now. Thanks.your last equation clears it up.