Are the Lowering and Raising Operators of QM the same as those of QFT?

We know that the lowering and raising operators in quantum mechanics are defined as

$$\begin{array}{l} a =\frac{1}{\sqrt{2}}(X+i P) \\ a^{\dagger} =\frac{1}{\sqrt{2}}(X-i P), \end{array}$$

respectively.

I was reading in this book page 257 about the different quantization schemes and he mentioned that the Wick-ordered quantization scheme is useful for quantum field theory and they are defined as:

\begin{aligned} a &=X+i \alpha P \\ a^{\dagger} &=X-i \alpha P \end{aligned}

He also mentioned that they differ by a constant from the first two raising and lowering operators and the commutator of $$a$$ and $$a^\dagger$$ is not $$I$$ but rather $$2\alpha \hbar I$$.

Does all of this mean that the raising and lowering operators of quantum mechanics are different than those of quantum field theory?

Yes, normally one uses a convenient normalization for creation and annihilation operators. In QM there is usually only one frequency, ω, in your problem, so you incorporate it into the normalization of $$a$$ and $$a^\dagger\equiv a^*$$, (11.10), and it virtually disappears from the problem, $$D^2={\hbar \over m\omega}, \\ a=\frac{1}{\sqrt 2} \left ({X\over\sqrt{D}} +i {\sqrt{D}\over \hbar}P\right ),\\ a^*=\frac{1}{\sqrt 2}\left ({X\over\sqrt{D}} -i {\sqrt{D}\over \hbar}P\right ),\leadsto \\ [a,a^*]=I,$$ given $$[X,P]=i\hbar I$$.
In the case you are asking about, the many frequencies are not irrelevant, and become part of the normalization of each oscillator, so , as stated, \begin{aligned} a &=X+i \alpha P \\ a^{\dagger} &=X-i \alpha P \end{aligned} and, by a similar calculation, $$[a,a^*]=2\alpha \hbar I$$, now. Presumably, he would proceed later down to fix $$\alpha$$, depending on the QFT oscillators he might be dealing with.
N.B. Some QFT books, but not all!, dealing with the infinity of commuting oscillators comprising QFT, normalize them as $$[a_k,a^\dagger _p]=(2\pi)^3 \delta^3(\vec p -\vec k).$$ But some also incorporate the energy of each in the normalization. Caveat lector!